Samacheer Kalvi Class 12 Chemistry Solutions Chapter 1 Metallurgy

12th Chemistry Guide Metallurgy Text Book Questions and Answers

I. Choose the correct answer

 

Question 1. Bauxite has the composition
(a) \( \text{Al}_2\text{O}_3 \)
(b) \( \text{Al}_2\text{O}_3.\text{nH}_2\text{O} \)
(c) \( \text{Fe}_2\text{O}_3.2\text{H}_2\text{O} \)
(d) None of the options
Answer: (b) \( \text{Al}_2\text{O}_3.\text{nH}_2\text{O} \)
In simple words: Bauxite, which is the main source for making aluminum, is made up of aluminum oxide along with some water molecules attached to it. This hydrated form is important for its properties.

🎯 Exam Tip: Remember the chemical formula of common ores like bauxite as they are frequently asked in exams.

 

Question 2. Roasting of sulphide ore gives the gas (A). (A) is a colorless gas. An aqueous solution of (A) is acidic. The gas (A) is
(a) \( \text{CO}_2 \)
(b) \( \text{SO}_3 \)
(c) \( \text{SO}_2 \)
(d) \( \text{H}_2\text{S} \)
Answer: (c) \( \text{SO}_2 \)
In simple words: When sulfide ores are heated in air (roasted), they usually release sulfur dioxide gas. This gas is colorless and forms an acidic solution when dissolved in water, which is a key characteristic.

🎯 Exam Tip: Recall the products of roasting reactions. Sulfur dioxide (\( \text{SO}_2 \)) is a common gaseous product from sulfide ores, and it forms sulfurous acid in water.

 

Question 3. Which one of the following reaction represents calcinations?
(a) \( 2\text{Zn} + \text{O}_2 \rightarrow 2\text{ZnO} \)
(b) \( 2\text{ZnS} + 3\text{O}_2 \rightarrow 2\text{ZnO} + 2\text{SO}_2 \)
(c) \( \text{MgCO}_3 \rightarrow \text{MgO} + \text{CO}_2 \)
(d) Both (a) and (c)
Answer: (c) \( \text{MgCO}_3 \rightarrow \text{MgO} + \text{CO}_2 \)
In simple words: Calcination is a process where a carbonate ore is heated strongly without air, causing it to break down into a metal oxide and carbon dioxide gas. Option (c) shows magnesium carbonate breaking into magnesium oxide and carbon dioxide, which is an example of calcination.

🎯 Exam Tip: Distinguish between roasting (heating in air, typically for sulfides) and calcination (heating without air, typically for carbonates or hydroxides to remove volatile matter like \( \text{CO}_2 \) or water).

 

Question 4. The metal oxide which cannot be reduced to metal by carbon is
(a) \( \text{PbO} \)
(b) \( \text{Al}_2\text{O}_3 \)
(c) \( \text{ZnO} \)
(d) \( \text{FeO} \)
Answer: (b) \( \text{Al}_2\text{O}_3 \)
In simple words: Aluminum oxide is a very stable compound and cannot be easily reduced to pure aluminum using carbon. More active metals like aluminum often need different methods for extraction, like electrolysis, because carbon is not strong enough to remove the oxygen.

🎯 Exam Tip: Remember that carbon reduction is effective for less reactive metals. For highly reactive metals like aluminum, electrochemical methods are often required because their oxides are too stable.

 

Question 5. Which of the metal is extracted by Hall - Heroult process?
(a) Al
(b) Ni
(c) Cu
(d) Zn
Answer: (a) Al
In simple words: The Hall-Heroult process is a special way to get aluminum metal from its ore, alumina. It uses electricity to separate the aluminum, because aluminum is very reactive.

🎯 Exam Tip: The Hall-Heroult process is specific to aluminum extraction and is a key example of electrolytic reduction.

 

Question 6. Which of the following statements, about the advantage of roasting of sulphide ore before the reduction is not true?
(a) \( \Delta \text{G}^\circ \text{f} \) of sulphide is greater than those for \( \text{CS}_2 \) and \( \text{H}_2\text{S} \)
(b) \( \Delta \text{G}^\circ \text{r} \) is negative for roasting of sulphide ore to oxide
(c) Roasting of the sulphide to its oxide is thermodynamically feasible.
(d) Carbon and hydrogen are suitable reducing agents for metal sulphides.
Answer: (d) Carbon and hydrogen are suitable reducing agents for metal sulphides.
In simple words: Roasting converts sulfide ores into oxides, which are then easier to reduce. Carbon and hydrogen are generally used to reduce metal oxides, not metal sulfides directly. Metal sulfides often need other methods, or conversion to oxides first.

🎯 Exam Tip: Understanding the thermodynamic feasibility and reaction products for roasting is crucial. Remember that carbon and hydrogen are primarily for oxide reduction, not sulfide reduction directly.

 

Question 7. Match items in Column I - with the items of Column - II and assign the correct code.

Answer: (c) (iv) (ii) (iii) (i)
In simple words: This question tests your knowledge of different metallurgical processes and their applications. Cyanide process is used for gold, froth floatation for zinc sulfide, electrolytic reduction for aluminum, and zone refining for ultrapure germanium.

🎯 Exam Tip: Memorize the key metallurgical processes and the specific metals or ores they are used for, as matching questions are common.

 

Question 8. Wolframite ore is separated from tinstone by the process of
(a) Smelting
(b) Calcination
(c) Roasting
(d) Electromagnetic separation
Answer: (d) Electromagnetic separation
In simple words: Wolframite contains iron, which is magnetic, while tinstone (cassiterite) is not. So, using a magnetic separator, the magnetic wolframite can be pulled away from the non-magnetic tinstone.

🎯 Exam Tip: Understand the principle behind different concentration methods. Electromagnetic separation is used when one component of the ore mixture is magnetic and the other is not.

 

Question 9. Which one of the following is not feasible
(a) \( \text{Zn(s)} + \text{Cu}^{2+}(\text{aq}) \rightarrow \text{Zn}^{2+}(\text{aq}) + \text{Cu(s)} \)
(b) \( \text{Cu(S)} + \text{Zn}^{2+}(\text{aq}) \rightarrow \text{Zn(s)} + \text{Cu}^{2+}(\text{aq}) \)
(c) \( \text{Cu(S)} + 2\text{Ag}^{+}(\text{aq}) \rightarrow \text{Ag(s)} + \text{Cu}^{2+}(\text{aq}) \)
(d) \( \text{Fe(S)} + \text{Cu}^{2+}(\text{aq}) \rightarrow \text{Cu(s)} + \text{Fe}^{2+}(\text{aq}) \)
Answer: (b) \( \text{Cu(S)} + \text{Zn}^{2+}(\text{aq}) \rightarrow \text{Zn(s)} + \text{Cu}^{2+}(\text{aq}) \)
In simple words: A less reactive metal cannot displace a more reactive metal from its salt solution. Zinc is more reactive than copper, so copper cannot push zinc out of its solution. Only more reactive metals can replace less reactive ones.

🎯 Exam Tip: Use the activity series of metals (reactivity series) to determine the feasibility of displacement reactions. A metal higher in the series can displace a metal lower in the series from its salt solution.

 

Question 10. Electrochemical process is used to extract
(a) Iron
(b) Lead
(c) Sodium
(d) Silver
Answer: (c) Sodium
In simple words: Very reactive metals, like sodium, are hard to get from their ores using regular chemical methods. Instead, electricity is used to break down their molten compounds to get the pure metal. This is known as an electrochemical process or electrolysis.

🎯 Exam Tip: Electrochemical reduction (electrolysis) is the preferred method for extracting highly electropositive metals that are difficult to reduce by common reducing agents.

 

Question 11. Flux is a substance which is used to convert
(a) Mineral into silicate
(b) Infusible impurities to soluble impurities
(c) Soluble impurities to infusible impurities
(d) All of the options
Answer: (b) Infusible impurities to soluble impurities
In simple words: Flux helps to remove unwanted impurities (called gangue) from an ore by reacting with them to form a new substance called slag. This slag can melt easily and be separated. So, it changes impurities that don't melt well into ones that can be easily removed.

🎯 Exam Tip: Understand the role of flux in metallurgy: it chemically combines with infusible gangue to form a fusible slag, which can be easily separated from the molten metal.

 

Question 12. Which one of the following ores is best concentrated by froth floatation method?
(a) Magnetite
(b) Hematite
(c) Galena
(d) Cassiterite
Answer: (c) Galena
In simple words: Froth flotation is a method especially good for sulfide ores because sulfide particles can easily attach to oil and air bubbles. Galena is a sulfide ore (lead sulfide), so it's perfectly suited for this process.

🎯 Exam Tip: Froth flotation is almost exclusively used for concentrating sulfide ores due to the selective wetting properties of sulfide minerals with oil and water.

 

Question 13. In the extraction of aluminium from alumina by electrolysis, cryolite is added to
(a) Lower the melting point of alumina
(b) Remove impurities from alumina
(c) Decrease the electrical conductivity
(d) Increase the rate of reduction
Answer: (a) Lower the melting point of alumina
In simple words: Alumina melts at a very high temperature, which makes it hard to use in electrolysis. Adding cryolite helps to lower this melting point, making the process more practical and energy-efficient. It also helps in dissolving alumina.

🎯 Exam Tip: Cryolite (\( \text{Na}_3\text{AlF}_6 \)) plays a crucial role in the Hall-Heroult process by reducing the melting point of alumina and improving its electrical conductivity, not decreasing it.

 

Question 14. Zinc is obtained from ZnO by
(a) Carbon reduction
(b) Reduction using silver
(c) Electrochemical process
(d) Acid leaching
Answer: (a) Carbon reduction
In simple words: Zinc oxide can be turned into pure zinc metal by heating it with carbon. Carbon takes the oxygen away from the zinc oxide, which is a common way to get less reactive metals.

🎯 Exam Tip: Carbon reduction (smelting) is a cost-effective method for reducing oxides of moderately reactive metals like zinc, iron, and lead.

 

Question 15. Extraction of gold and silver involves leaching with cyanide ion. silver is later recovered by (NEET - 2017)
(a) Distillation
(b) Zone refining
(c) Displacement with zinc
(d) Liquation
Answer: (c) Displacement with zinc
In simple words: After gold and silver are dissolved using cyanide (leaching), they can be recovered by adding a more reactive metal like zinc. Zinc will push the gold or silver out of the solution, allowing the precious metals to be collected.

🎯 Exam Tip: The cyanide leaching process (MacArthur-Forrest process) for gold and silver is followed by displacement with a more electropositive metal like zinc to recover the noble metals. This is also known as cementation.

 

Question 16. Considering the Ellingham diagram, which of the following metals can be used to reduce alumina? (NEET - 2018)
(a) Fe
(c) Mg
(d) Zn
Answer: (c) Mg
In simple words: On an Ellingham diagram, a metal can reduce the oxide of another metal if its free energy line is below that of the other metal's oxide. Magnesium's line is below alumina's line at certain temperatures, meaning magnesium can reduce alumina.

🎯 Exam Tip: To reduce a metal oxide, the reducing agent's free energy line on the Ellingham diagram must be below the metal oxide's formation line at the desired temperature. This indicates a more negative \( \Delta \text{G} \) for the overall reduction reaction.

 

Question 17. The following set of reactions are used in refining Zirconium
\( \text{Zr (impure)} + 2\text{I}_2 \xrightarrow{523 \text{ K}} \text{ZrI}_4 \)
\( \text{ZrI}_4 \xrightarrow{1800 \text{ K}} \text{Zr (pure)} + 2\text{I}_2 \)
This method is known as
(a) Liquation
(b) Van Arkel process
(c) Zone refining
(d) Mond's process
Answer: (b) Van Arkel process
In simple words: This process makes very pure metals like zirconium. First, the impure metal reacts with iodine to form a gas, which leaves impurities behind. Then, this gas is heated to a very high temperature, causing it to break down and leave behind the pure metal.

🎯 Exam Tip: The Van Arkel process is a vapor phase refining method primarily used for metals like titanium and zirconium, involving the formation and decomposition of a volatile iodide.

 

Question 18. Which of the following is used for concentrating ore in metallurgy?
(a) Leaching
(b) Roasting
(c) Froth floatation
(d) Both (a) and (c)
Answer: (d) Both (a) and (c)
In simple words: Leaching and froth flotation are both processes used to increase the amount of the desired metal in the ore by removing unwanted materials. Roasting is a chemical change, not a concentration method.

🎯 Exam Tip: Concentration methods aim to increase the percentage of the metal in the ore by physically or chemically separating the gangue. Leaching and froth flotation achieve this, while roasting is a chemical treatment of the concentrated ore.

 

Question 19. The incorrect statement among the following is
(a) Nickel is refined by Mond's process
(b) Titanium is refined by Van Arkel's process
(c) Zinc blende is concentrated by froth floatation
(d) In the metallurgy of gold, the metal is leached with a dilute sodium chloride solution
Answer: (d) In the metallurgy of gold, the metal is leached with a dilute sodium chloride solution
In simple words: Gold is usually leached with a dilute sodium cyanide solution, not sodium chloride solution. Sodium cyanide forms a soluble complex with gold, which helps extract it from its ore.

🎯 Exam Tip: Correctly recall the reagents used in different refining and extraction processes. Cyanide solutions are specific for gold and silver leaching, not chloride solutions.

 

Question 20. In the electrolytic refining of copper, which one of the following is used as anode?
(a) Pure copper
(b) Impure copper
(c) Carbon rod
(d) Platinum electrode
Answer: (b) Impure copper
In simple words: In electrolytic refining, the impure metal itself is used as the anode. When electricity passes through, the impure metal dissolves from the anode and pure metal deposits onto the cathode.

🎯 Exam Tip: For electrolytic refining, always remember that the impure metal serves as the anode, and a thin strip of the pure metal acts as the cathode.

 

Question 21. Which of the following plot gives Ellingham diagram
(a) \( \Delta \text{S} \) Vs \( \text{T} \)
(b) \( \Delta \text{G}^\circ \) Vs \( \text{T} \)
(c) \( \Delta \text{G}^\circ \) Vs \( 1/\text{T} \)
(d) \( \Delta \text{G}^\circ \) Vs \( \text{P} \)
Answer: (b) \( \Delta \text{G}^\circ \) Vs \( \text{T} \)
In simple words: An Ellingham diagram shows how the standard Gibbs free energy change (\( \Delta \text{G}^\circ \)) for the formation of metal oxides changes with temperature (\( \text{T} \)). This helps us understand which metal can reduce another metal's oxide at different temperatures.

🎯 Exam Tip: The Ellingham diagram is a plot of \( \Delta \text{G}^\circ \) (standard Gibbs free energy change) for oxide formation against temperature (\( \text{T} \)). This relationship is vital for understanding metallurgical reductions.

 

Question 22. In the Ellingham diagram, for the formation of carbon monoxide
(a) \( \left( \frac{\Delta \text{S}^\circ}{\Delta \text{T}} \right) \) is negative
(b) \( \left( \frac{\Delta \text{G}^\circ}{\Delta \text{T}} \right) \) is positive
(c) \( \left( \frac{\Delta \text{G}^\circ}{\Delta \text{T}} \right) \) is negative
(d) initially \( \left( \frac{\Delta \text{G}^\circ}{\Delta \text{T}} \right) \) is positive, after \( 700^\circ\text{C} \), \( \left( \frac{\Delta \text{G}^\circ}{\Delta \text{T}} \right) \) is negative
Answer: (c) \( \left( \frac{\Delta \text{G}^\circ}{\Delta \text{T}} \right) \) is negative
In simple words: The slope of the line in an Ellingham diagram tells us about the change in entropy. For carbon monoxide formation, as temperature goes up, the free energy becomes more negative, meaning the slope is negative. This indicates that carbon monoxide becomes a better reducing agent at higher temperatures.

🎯 Exam Tip: Remember that the slope of an Ellingham line is approximately equal to \( -\Delta \text{S}^\circ \). For reactions where \( \Delta \text{S}^\circ \) is positive (like the formation of CO from C and \( \text{O}_2 \)), the slope \( -\Delta \text{S}^\circ \) will be negative.

 

Question 23. Which of the following reduction is not thermodynamically feasible?
(a) \( \text{Cr}_2\text{O}_3 + 2\text{Al} \rightarrow \text{Al}_2\text{O}_3 + 2\text{Cr} \)
(b) \( \text{Al}_2\text{O}_3 + 2\text{Cr} \rightarrow \text{Cr}_2\text{O}_3 + 2\text{Al} \)
(c) \( 3\text{TiO}_2 + 4\text{Al} \rightarrow 2\text{Al}_2\text{O}_3 + 3\text{Ti} \)
(d) None of these
Answer: (b) \( \text{Al}_2\text{O}_3 + 2\text{Cr} \rightarrow \text{Cr}_2\text{O}_3 + 2\text{Al} \)
In simple words: A metal can reduce the oxide of another metal only if it is more reactive. Aluminum is more reactive than chromium, so aluminum can reduce chromium oxide. However, chromium cannot reduce aluminum oxide because aluminum is more reactive.

🎯 Exam Tip: A reduction reaction is thermodynamically feasible if the reducing agent (the metal in its elemental form) is more reactive (its Gibbs free energy line for oxide formation is lower) than the metal in the oxide being reduced.

 

Question 24. Which of the following is not true with respect to the Ellingham diagram?
(a) Free energy changes follow a straight line. The deviation occurs when there is a phase change.
(b) The graph for the formation of \( \text{CO}_2 \) is a straight line almost parallel to the free energy axis.
(c) Negative slope of CO shows that it becomes more stable with an increase in temperature.
(d) Positive slope of metal oxides shows that their stabilities decrease with an increase in temperature.
Answer: (b) The graph for the formation of \( \text{CO}_2 \) is a straight line almost parallel to the free energy axis.
In simple words: The Ellingham diagram for forming carbon dioxide ( \( \text{CO}_2 \) ) is not parallel to the free energy axis. It actually has a slight positive slope. The line that is nearly parallel to the temperature axis is for reactions where there is almost no change in entropy, which is not the case for \( \text{CO}_2 \) formation.

🎯 Exam Tip: Pay attention to the slopes of the lines in the Ellingham diagram. The slope indicates \( -\Delta \text{S}^\circ \). For \( \text{CO}_2 \) formation from carbon, there is a small positive \( \Delta \text{S}^\circ \), leading to a slightly negative slope, not one parallel to the free energy axis.

II. Answer the following questions

 

Question 1. What is the difference between minerals and ores?
Answer: Minerals are naturally occurring substances from which metals or their compounds can be obtained. Ores are specific minerals from which a metal can be extracted profitably and easily on a large scale. All ores are minerals, but not all minerals are ores. For example, clay (\( \text{Al}_2\text{O}_3.\text{SiO}_2.2\text{H}_2\text{O} \)) is a mineral of aluminum, but bauxite (\( \text{Al}_2\text{O}_3.2\text{H}_2\text{O} \)) is the ore because aluminum is extracted from it profitably.
In simple words: Minerals are natural rocks with metal. Ores are special minerals from which we can get a metal easily and make money.

🎯 Exam Tip: Clearly define both terms and highlight the key distinction: economic feasibility. Provide a common example to illustrate the difference.

 

Question 2. What are the various steps involved in the extraction of pure metals from their ores?
Answer: The extraction of pure metals from their ores generally involves three main steps:
1. Concentration of the ore: This step removes unwanted rocky materials and impurities (gangue) from the ore.
2. Extraction of the crude metal: After concentration, the metal is separated from its chemical compound form in the ore, resulting in an impure or crude metal.
3. Refining of the crude metal: In this final step, the crude metal is purified to remove any remaining impurities, achieving the desired purity level. For example, zinc can be extracted from zinc blende by roasting, carbon reduction, and then electrolytic refining.
In simple words: First, we clean the ore to remove dirt. Next, we get the metal out of the cleaned ore. Finally, we make the metal even purer.

🎯 Exam Tip: List the three main stages clearly: concentration, extraction, and refining. Briefly explain the purpose of each stage.

 

Question 3. What is the role of Limestone in the extraction of iron from its oxide Fe2O3?
Answer: In the extraction of iron from its oxide \( \text{Fe}_2\text{O}_3 \) in a blast furnace, limestone (\( \text{CaCO}_3 \)) is added. It acts as a flux. Limestone decomposes into calcium oxide (\( \text{CaO} \)) and carbon dioxide (\( \text{CO}_2 \)) at high temperatures. The calcium oxide is a basic flux that reacts with acidic impurities like silica (\( \text{SiO}_2 \)) present in the ore. This reaction forms a fusible slag, calcium silicate (\( \text{CaSiO}_3 \)), which is lighter than the molten iron and can be easily removed. This ensures the purity of the iron produced.
\( \text{CaO (s)} + \text{SiO}_2 \text{ (s)} \rightarrow \text{CaSiO}_3 \text{ (s)} \)
Flux     Gangue       Slag
In simple words: Limestone helps clean the iron ore. It melts with dirt (silica) in the furnace to form a waste material called slag, which floats on top of the molten iron and can be easily taken out.

🎯 Exam Tip: Explain the dual role of limestone: it decomposes to form \( \text{CaO} \) (a basic flux) and then reacts with acidic gangue (like \( \text{SiO}_2 \)) to form fusible slag.

 

Question 4. Which type of ores can be concentrated by froth floatation method? Give two examples for such ores.
Answer: The froth flotation method is mainly used to concentrate sulfide ores. This process works because sulfide minerals are preferentially wetted by oil and thus get attached to the froth (foam) when air is blown through the mixture, while water wets the gangue particles, causing them to settle. This allows for a good separation of the valuable sulfide minerals from the unwanted gangue. Two examples of such ores are:
1. Copper pyrites (\( \text{CuFeS}_2 \))
2. Zinc blende (\( \text{ZnS} \))
3. Galena (\( \text{PbS} \))
In simple words: This method is best for ores that contain sulfur. The sulfur parts stick to oil and float as foam, while the rocky parts sink. Examples are copper pyrites and zinc blende.

🎯 Exam Tip: Remember that froth flotation is specific to sulfide ores due to their surface properties. Provide at least two common examples of sulfide ores to score full marks.

 

Question 5. Describe a method for refining nickel Mond process for refining nickel: (PTA - 3)
Answer: The Mond process is a special way to purify nickel, especially for making very pure nickel. It uses a gas called carbon monoxide to separate nickel from impurities. First, impure nickel is heated in a stream of carbon monoxide gas at around \( 350\text{K} \). Nickel reacts with carbon monoxide to form a highly volatile gas called nickel tetracarbonyl (\( \text{Ni(CO)}_4 \)), while the solid impurities are left behind.
\( \text{Ni(s)} + 4\text{CO(g)} \rightarrow \text{Ni(CO)}_4\text{(g)} \)
Then, this nickel tetracarbonyl gas is heated to a higher temperature, about \( 460\text{K} \). At this higher temperature, the nickel tetracarbonyl gas breaks down, leaving behind pure nickel metal and releasing carbon monoxide gas, which can be reused. This effectively removes all impurities from the nickel.
\( \text{Ni(CO)}_4\text{(g)} \rightarrow \text{Ni(s)} + 4\text{CO(g)} \)
In simple words: The Mond process purifies nickel. Impure nickel is heated with carbon monoxide gas to make a new gas. This new gas is then heated more to break it apart, leaving behind pure nickel metal.

🎯 Exam Tip: Focus on the two main steps: formation of the volatile nickel carbonyl complex at a lower temperature and its decomposition at a higher temperature. Include the balanced chemical equations.

 

Question 6. Explain the zone refining process with an example. (PTA - 6 MARCH 2020)
Answer: The zone refining process is used to make very pure metals, especially for semiconductors. This method works because impurities are more soluble (dissolve better) in the molten (liquid) part of a metal than in its solid part. The impure metal is shaped into a long rod. A small section of this rod is heated using a mobile induction heater, creating a molten zone. This molten zone is slowly moved from one end of the rod to the other. As the molten zone moves, the pure metal crystallizes out of the melt, leaving the impurities behind in the liquid phase. The impurities move along with the molten zone towards one end of the rod. This process is repeated several times by moving the heater back and forth. This makes the impurities gather at one end of the rod, which can then be cut off. This method is done in a gas that does not react with the metal to prevent it from reacting with air. Elements like germanium (\( \text{Ge} \)), silicon (\( \text{Si} \)), and gallium (\( \text{Ga} \)), used in making electronic devices, are refined using this process.
In simple words: Zone refining makes metals very pure. A melted part moves along a metal rod, carrying dirt with it. The clean metal solidifies behind the melt, and the dirt ends up at one end. This process is used for materials like silicon.

🎯 Exam Tip: Emphasize the principle that impurities are more soluble in the melt than in the solid. Describe the movement of the molten zone and mention examples of elements (semiconductors) for which this process is used.

 

Question 7. Using the Ellingham diagram (fig 1.4) indicates the lowest temperature at which i) Aluminium might be expected to reduce magnesia. ii) Magnesium could reduce alumina. B) It is possible to reduce Fe2O3 by coke at a temperature around 1200K
Answer: The Ellingham diagram helps us to predict which metal can reduce the oxide of another metal at different temperatures.
A) i) For aluminum to reduce magnesia (\( \text{MgO} \)), the line for the formation of aluminum oxide (\( \text{Al}_2\text{O}_3 \)) must be below the line for magnesium oxide (\( \text{MgO} \)). These two lines intersect around \( 1600\text{K} \). Above \( 1600\text{K} \), the aluminum line is below the magnesium line, so aluminum can reduce magnesia at temperatures above \( 1600\text{K} \).
ii) For magnesium to reduce alumina (\( \text{Al}_2\text{O}_3 \)), the magnesium oxide formation line must be below the aluminum oxide line. Below \( 1600\text{K} \), the magnesium line is below the aluminum line, meaning magnesium can reduce alumina at temperatures below \( 1600\text{K} \).
B) To reduce iron oxide (\( \text{Fe}_2\text{O}_3 \)) with coke (carbon), the carbon line must be below the iron line on the Ellingham diagram. Above \( 1000\text{K} \), the carbon line (for \( \text{C} + \text{O}_2 \rightarrow \text{CO} \)) drops below the iron line (\( \text{Fe} + \text{O}_2 \rightarrow \text{FeO} \)). This means that it is possible to reduce \( \text{Fe}_2\text{O}_3 \) by coke at temperatures around \( 1200\text{K} \) or higher. Carbon becomes a more effective reducing agent at higher temperatures.
In simple words: The Ellingham diagram shows when one metal can take oxygen from another metal's oxide. If a metal's line is lower on the chart than another metal's oxide line at a certain temperature, it means the first metal can reduce the second metal's oxide. For example, above \( 1600\text{K} \), aluminum can reduce magnesium oxide, but below \( 1600\text{K} \), magnesium can reduce aluminum oxide. Carbon can reduce iron oxide above \( 1000\text{K} \).

🎯 Exam Tip: When using the Ellingham diagram, always compare the relative positions of the \( \Delta \text{G}^\circ \) lines for the formation of the reducing agent's oxide and the metal oxide to be reduced. The lower line indicates greater stability and the ability to reduce oxides above it.

 

Question 8. Give the uses of zinc. (PTA - 4)
Answer: Zinc is a versatile metal with many important uses:

• Metallic zinc is primarily used in galvanization. This process coats iron and steel structures with a layer of zinc to protect them from rusting and corrosion.
• Zinc is also used to make die-castings for various industries, including automobiles, electrical components, and hardware.
• Zinc oxide, a compound of zinc, is widely used in making paints, rubber, cosmetics, pharmaceuticals, plastics, inks, batteries, textiles, and electrical equipment due to its unique properties.
• Zinc sulfide is important for making luminous paints, fluorescent lights, and X-ray screens, as it glows when exposed to radiation.
• Brass, an alloy of zinc and copper, is highly resistant to corrosion and is therefore used in water valves and communication equipment.

Zinc plays a key role in protecting other metals and in various industrial and daily products.
In simple words: Zinc is used to protect iron from rust (galvanizing). It's also in car parts, paints, and special lights. Brass, a mix with zinc, is used for water valves.

🎯 Exam Tip: List at least four distinct uses of zinc. Focus on its major applications like galvanization, alloys, and compounds in various industries.

 

Question 9. Explain the electrometallurgy of aluminium.
Answer: The electrometallurgy of aluminum is also known as the Hall-Heroult process. This process extracts pure aluminum from alumina (\( \text{Al}_2\text{O}_3 \)) using electrolysis.

• Cathode: An iron tank lined with carbon acts as the cathode (negative electrode).
• Anode: Several carbon blocks immersed in the electrolyte serve as the anode (positive electrode).
• Electrolyte: The electrolyte is a molten mixture containing about \( 20\% \) alumina (\( \text{Al}_2\text{O}_3 \)) obtained from bauxite, molten cryolite (\( \text{Na}_3\text{AlF}_6 \)), and about \( 10\% \) calcium fluoride (\( \text{CaF}_2 \)). Cryolite lowers the melting point of alumina (from over \( 2000^\circ\text{C} \) to about \( 950^\circ\text{C} \)) and increases its electrical conductivity.
• Temperature: The process operates at a high temperature, typically above \( 1270\text{K} \) (about \( 997^\circ\text{C} \)).

The process involves the ionization of alumina in the molten cryolite:
\( \text{Al}_2\text{O}_3 \rightarrow 2\text{Al}^{3+} + 3\text{O}^{2-} \)
At the electrodes, the following reactions occur:
Reaction at cathode: Aluminum ions gain electrons and are reduced to molten aluminum metal.
\( \text{Al}^{3+} (\text{melt}) + 3\text{e}^- \rightarrow \text{Al(l)} \)
Reaction at anode: Oxide ions lose electrons and are oxidized to oxygen gas.
\( 2\text{O}^{2-} (\text{melt}) \rightarrow \text{O}_2 \text{ (melt)} + 4\text{e}^- \)
Since the carbon anodes react with the oxygen produced, they are slowly consumed and must be replaced regularly. The carbon reacts to form carbon monoxide and carbon dioxide.
\( \text{C(s)} + \text{O}_2 \text{ (melt)} \rightarrow \text{CO(g)} + 2\text{e}^- \)
\( \text{C(s)} + 2\text{O}_2 \text{ (melt)} \rightarrow \text{CO}_2\text{(g)} + 4\text{e}^- \)
During electrolysis, the pure aluminum, being denser, settles at the bottom of the cell and is periodically tapped off. The net electrolysis reaction can be written as:
\( 4\text{Al}^{3+} (\text{melt}) + 6\text{O}^{2-} (\text{melt}) + 3\text{C(s)} \rightarrow 4\text{Al(l)} + 3\text{CO}_2\text{(g)} \)
In simple words: Electrometallurgy of aluminum uses electricity to get pure aluminum from alumina. Impure aluminum oxide is dissolved in molten cryolite, then electricity makes the aluminum separate out. Carbon parts are used as electrodes, but they get used up because they react with oxygen.

🎯 Exam Tip: Explain the key components (cathode, anode, electrolyte) and their materials. Mention the role of cryolite, the operating temperature, and the reactions at both electrodes, including the overall reaction and anode consumption.

 

Question 10. Explain the following terms with suitable examples, i) Gangue ii) Slag (PAT - 2)
Answer: Here's an explanation of two important terms in metallurgy:
i) Gangue: This refers to the non-metallic impurities, such as rocky materials and siliceous matter (like sand), that are present along with the valuable metal compounds in an ore. Gangue is unwanted material that needs to be removed before the metal can be extracted. For example, silica (\( \text{SiO}_2 \)) is a common gangue present in iron ore (\( \text{Fe}_2\text{O}_3 \)).
ii) Slag: Slag is a fusible (easily melting) chemical substance that is formed when gangue reacts with a flux during the smelting process. The flux is added to combine with the infusible gangue to create this molten slag, which can then be easily separated from the molten metal. This process helps to purify the metal. For example, when basic calcium oxide (\( \text{CaO} \)) acts as a flux and reacts with acidic silica (\( \text{SiO}_2 \)), calcium silicate (\( \text{CaSiO}_3 \)) slag is formed:
\( \text{CaO (s)} + \text{SiO}_2 \text{ (s)} \rightarrow \text{CaSiO}_3 \text{ (s)} \)
Flux     Gangue       Slag
In simple words: Gangue is the unwanted dirt or rock in an ore. Slag is a new, easily melted substance formed when this dirt reacts with a cleaning chemical (flux), which then helps to remove the dirt from the metal.

🎯 Exam Tip: Clearly define both gangue and slag, explaining their relationship. Provide a specific chemical example (like the formation of calcium silicate slag from silica gangue and limestone flux) to illustrate the concept.

 

Question 11. Give the basic requirement for vapour phase refining.
Answer: For vapour phase refining, two main requirements are needed. First, the metal must react with a suitable chemical agent to create a compound that easily turns into a gas. This compound should be volatile. Second, this volatile compound must be able to break down easily at a different temperature to give back the pure metal. This method is especially useful for purifying very reactive metals.
In simple words: A metal needs to form a gas with a chemical, and then that gas must easily break apart to give back the pure metal.

🎯 Exam Tip: Remember that the intermediate compound must be volatile and easily decomposable for this method to work effectively.

 

Question 12. Describe the role of the following in the process mentioned.
(i) Silica in the extraction of copper.
(ii) Cryolite in the extraction of aluminium.
(iii) Iodine in the refining of Zirconium.
(iv) Sodium cyanide in froth floatation.
Answer:
(i) In the extraction of copper, silica (\( \text{SiO}_2 \)) acts as an acidic flux. It helps to remove iron oxide (\( \text{FeO} \)) impurities by reacting with them to form a molten slag of iron silicate (\( \text{FeSiO}_3 \)). This slag is lighter and can be easily separated from the molten copper, making the copper purification process more efficient.
\( \text{FeO (s)} + \text{SiO}_2 \text{ (s)} \rightarrow \text{FeSiO}_3 \text{ (s)} \)
Flux is added to remove gangue, forming a fusible slag.
(ii) In the extraction of aluminium, cryolite (\( \text{Na}_3\text{AlF}_6 \)) is added to alumina (\( \text{Al}_2\text{O}_3 \)). Its main roles are to lower the melting point of alumina from about 2072 °C to around 950 °C, which saves a lot of energy. It also increases the electrical conductivity of the electrolyte, which is essential for the electrolysis process to happen effectively. Cryolite itself acts as a solvent for alumina.
(iii) In the refining of Zirconium, iodine (\( \text{I}_2 \)) is used in the Van Arkel method. Impure zirconium metal is heated with iodine to form a volatile zirconium tetraiodide (\( \text{ZrI}_4 \)). This compound is then decomposed at a higher temperature on a hot tungsten filament, leaving behind pure zirconium metal. Impurities do not react with iodine and are left behind.
\( \text{Zr(s)} + \text{2I}_2\text{(s)} \rightarrow \text{ZrI}_4\text{(Vapour)} \)
The pure metal is then collected after decomposition:
\( \text{ZrI}_4\text{(Vapour)} \rightarrow \text{Zr(s)} + \text{2I}_2\text{(S)} \)
(iv) In the froth flotation process, sodium cyanide (\( \text{NaCN} \)) acts as a depressing agent. It helps to prevent certain metal sulphides, like zinc sulphide (\( \text{ZnS} \)), from coming into the froth along with other desired sulphides, such as galena (\( \text{PbS} \)). It does this by forming a layer of zinc complex (\( \text{Na}_2[\text{Zn}(\text{CN})_4] \)) on the surface of zinc sulphide, making it less likely to attach to the froth bubbles.
In simple words: Silica helps remove iron from copper as a waste. Cryolite lowers the melt temperature and helps electricity flow when making aluminium. Iodine helps purify zirconium by forming a gas that then breaks down. Sodium cyanide stops unwanted zinc ore from floating up with other ores.

🎯 Exam Tip: When describing roles, always mention both *what* the substance does and *why* it's important for the specific process, including any chemical reactions if applicable.

 

Question 13. Explain the principle of electrolytic refining with an example. (PTA - 5)
Answer: Electrolytic refining is a process used to get very pure metals from impure ones. It works on the principle that during electrolysis, the impure metal acts as the anode, a thin strip of pure metal acts as the cathode, and a salt solution of the metal acts as the electrolyte. When electricity is passed, the impure metal at the anode dissolves into the solution, while pure metal ions from the solution get deposited on the cathode. More reactive impurities stay in the solution, and less reactive impurities settle down as "anode mud." This is a highly effective way to purify metals that need high purity for their applications.
An excellent example is the electrorefining of silver:
In this process:
- The cathode is a thin sheet of pure silver.
- The anode is a block of impure silver.
- The electrolyte is an acidified aqueous solution of silver nitrate.
When current is passed:
At the anode (impure silver): Silver atoms lose electrons and enter the solution as ions.
\( \text{Ag(s)} \rightarrow \text{Ag}^+\text{(aq)} + \text{e}^- \)
At the cathode (pure silver): Silver ions from the solution gain electrons and deposit as pure silver metal.
\( \text{Ag}^+\text{(aq)} + \text{e}^- \rightarrow \text{Ag(s)} \)
As a result, pure silver is collected at the cathode, and impurities either dissolve in the electrolyte or settle as anode mud.

🎯 Exam Tip: Clearly distinguish between the anode, cathode, and electrolyte, and state the reactions occurring at each electrode for the given example.

 

Question 14. The selection of reducing agent depends on the thermodynamic factor: Explain with an example.
Answer: The choice of a reducing agent in metallurgy is based on thermodynamic factors, primarily the Gibbs free energy change (\( \Delta G \)) of the reaction. For a reduction reaction to happen on its own (spontaneously), the overall Gibbs free energy change for the coupled reaction (the reduction of the metal oxide by the reducing agent) must be negative. Essentially, a good reducing agent is one that can form an oxide that is more stable (has a more negative \( \Delta G^\circ \) of formation) than the metal oxide being reduced, at a specific temperature. The Ellingham diagram helps visualize these thermodynamic stabilities.
For example, let's consider the reduction of iron oxide (\( \text{FeO} \)) by carbon (coke). The Ellingham diagram for the formation of \( \text{FeO} \) and \( \text{CO} \) intersects around 1000K. Below this temperature, the carbon line is above the iron line, meaning \( \text{FeO} \) is more stable than \( \text{CO} \), so carbon cannot reduce \( \text{FeO} \). However, above 1000K, the carbon line falls below the iron line. This indicates that above 1000K, \( \text{CO} \) becomes more stable than \( \text{FeO} \), making the reduction of \( \text{FeO} \) by carbon thermodynamically possible. So, coke can be used as a reducing agent above this temperature.
The reactions are:
1. \( \text{2Fe(s)} + \text{O}_2\text{(g)} \rightarrow \text{2FeO(g)} \quad \Delta G^\circ_1 = +350 \text{ KJmol}^{-1} \)
2. \( \text{2C(s)} + \text{O}_2\text{(g)} \rightarrow \text{2CO(g)} \quad \Delta G^\circ_2 = -480 \text{ KJmol}^{-1} \)
To reduce \( \text{FeO} \), we need to reverse reaction 1 and add it to reaction 2:
\( \text{2FeO(g)} \rightarrow \text{2Fe(s)} + \text{O}_2\text{(g)} \quad \Delta G^\circ = -350 \text{ KJmol}^{-1} \) (reversed reaction 1)
Adding this to reaction 2 gives the overall reduction reaction:
\( \text{2FeO(s)} + \text{2C(s)} \rightarrow \text{2Fe(s)} + \text{2CO(g)} \)
The total \( \Delta G^\circ \) for this coupled reaction is \( -350 + (-480) = -830 \text{ KJmol}^{-1} \). This large negative value confirms the thermodynamic feasibility of reducing \( \text{FeO} \) by carbon at temperatures above 1000K.
In simple words: A reducing agent is chosen based on how easily it can take oxygen from a metal. The overall reaction must naturally happen (have a negative energy change). For example, carbon can remove oxygen from iron oxide at high temperatures because, at those temperatures, carbon likes oxygen more than iron does.

🎯 Exam Tip: When explaining thermodynamic factors, refer to the Ellingham diagram, the sign of \( \Delta G \), and how a reducing agent's oxide stability compares to the metal's oxide. Always give a relevant example.

 

Question 15. Give the limitations of Ellingham diagram.
Answer: The Ellingham diagram is a very useful tool, but it has certain limitations. Firstly, it is constructed solely based on thermodynamic considerations, meaning it predicts whether a reaction is *possible* but doesn't tell us anything about how *fast* the reaction will happen (the reaction rate). Secondly, it assumes that reactants and products are always in a state of equilibrium, which is not always true in real-world metallurgical processes. Lastly, the diagram typically doesn't provide information about other possible reactions that might occur during the process, which could affect the final outcome. Despite these, it remains a valuable guide for choosing appropriate reducing agents.
In simple words: The Ellingham diagram tells if a reaction *can* happen, not how *fast* it will. It also assumes everything is perfectly balanced, which isn't always true.

🎯 Exam Tip: Focus on the difference between thermodynamics (what's possible) and kinetics (how fast it happens) as a primary limitation.

 

Question 16. Write a short note on electrochemical principles of metallurgy.
Answer: Electrochemical principles are very important in metallurgy, especially for extracting and refining highly reactive metals like sodium or potassium, whose oxides cannot be easily reduced by carbon. These methods involve using electricity to either reduce metal ions from their molten salts or solutions, or to purify metals through electrolytic refining. The core idea is that a chemical reaction that is not spontaneous can be driven by supplying electrical energy. The Gibbs free energy change for electrolysis is related to the electrode potential, given by the formula \( \Delta G^\circ = -nFE^\circ \), where \( n \) is the number of electrons, \( F \) is Faraday's constant, and \( E^\circ \) is the standard electrode potential. For a process to be feasible, the net cell potential \( E^\circ \) must be positive. This principle allows the extraction of metals like aluminium from alumina and the refining of copper or silver.
In simple words: Some metals are so reactive that they need electricity to separate them from their ores. This uses electrochemical rules where electrical energy helps change metal ions into pure metal.

🎯 Exam Tip: Mentioning the extraction of highly reactive metals and the role of \( \Delta G^\circ \) or \( E^\circ \) is key to explaining electrochemical principles in metallurgy.

 

III. Evaluate Yourself

 

Question 1. Write the equation for the extraction of silver by leaching with sodium cyanide and show that the leaching process is a redox reaction.
Answer: The extraction of silver by cyanide leaching involves reacting silver metal with sodium cyanide solution in the presence of oxygen and water to form a soluble dicyanoargentate(I) complex. This process is a redox reaction because silver is oxidized and oxygen is reduced. Here's the equation:
\( \text{4Ag(s)} + \text{8CN}^-\text{(aq)} + \text{O}_2\text{(g)} + \text{2H}_2\text{O(l)} \rightarrow \text{4[Ag(CN)}_2]^-\text{(aq)} + \text{4OH}^-\text{(aq)} \)
To show it's a redox reaction:
- Silver (Ag) goes from an oxidation state of 0 in \( \text{Ag(s)} \) to +1 in \( \text{[Ag(CN)}_2]^- \). This increase in oxidation state means silver is oxidized.
- Oxygen (O) in \( \text{O}_2\text{(g)} \) (oxidation state 0) is reduced to oxygen in \( \text{OH}^- \) and \( \text{H}_2\text{O} \) (oxidation state -2). This decrease in oxidation state means oxygen is reduced. This is a crucial step in obtaining silver from low-grade ores.
In simple words: Silver metal reacts with cyanide and oxygen in water to form a dissolved silver compound. In this reaction, silver loses electrons (gets oxidized), and oxygen gains electrons (gets reduced), showing it is a redox process.

🎯 Exam Tip: For redox reactions, explicitly state the oxidation states of elements before and after the reaction to clearly demonstrate oxidation and reduction.

 

Question 2. Magnesite (Magnesium carbonate) is calcined to obtain magnesia, which is used to make refractory bricks. Write the decomposition reaction
Answer: Calcination is the process of heating a substance strongly in the absence or limited supply of air. For magnesite, it decomposes into magnesium oxide (magnesia) and carbon dioxide gas. This magnesia has a high melting point, making it suitable for refractory bricks that resist high temperatures.
\( \text{MgCO}_3\text{(s)} \underrightarrow{\triangle} \text{MgO(s)} + \text{CO}_2\text{(g)} \)
In simple words: When magnesite is heated, it breaks down into magnesia (magnesium oxide) and carbon dioxide.

🎯 Exam Tip: Remember that calcination typically involves the decomposition of carbonates or hydroxides, often releasing \( \text{CO}_2 \) or \( \text{H}_2\text{O} \), and occurs without air.

 

Question 3. Using the Ellingham diagram (fig 1.4) indicates the lowest temperature at which ZnO can be reduced to Zinc metal by carbon. Write the overall reduction reaction at this temperature
Answer: From the Ellingham diagram, the lines for the formation of \( \text{ZnO} \) and \( \text{CO} \) intersect around 1200K. Below this temperature, the carbon line is above the zinc line, meaning \( \text{ZnO} \) is more stable than \( \text{CO} \), so carbon cannot reduce \( \text{ZnO} \). However, above 1200K, the carbon line falls below the zinc line. This indicates that above 1200K, \( \text{CO} \) becomes more stable than \( \text{ZnO} \), making the reduction of \( \text{ZnO} \) by carbon thermodynamically feasible. Therefore, the lowest temperature at which \( \text{ZnO} \) can be reduced to zinc metal by carbon is approximately 1200K. This principle allows for the efficient extraction of zinc in industrial processes.
The overall reduction reaction at this temperature is:
\( \text{2ZnO(s)} + \text{2C(s)} \rightarrow \text{2Zn(s)} + \text{2CO(g)} \)
In simple words: Carbon can reduce zinc oxide into zinc metal and carbon monoxide, but only at temperatures higher than about 1200K. This is because above 1200K, carbon is better at attracting oxygen than zinc is.

🎯 Exam Tip: For Ellingham diagram questions, identify the intersection point of the metal oxide and reducing agent lines; the reducing agent is effective at temperatures where its line is below the metal oxide line.

 

Question 4. Metallic Sodium is extracted by the electrolysis of brine (aq.NaCl). After electrolysis, the electrolytic solution becomes basic in nature. Write the possible electrode reactions.
Answer: The extraction of metallic sodium by the electrolysis of aqueous sodium chloride (brine) produces sodium hydroxide, hydrogen gas, and chlorine gas. The solution becomes basic because hydroxide ions are formed at the cathode. This process, also known as the chlor-alkali process, is industrially important for producing these chemicals.
The possible electrode reactions are:
At the anode (positive electrode): Chloride ions lose electrons to form chlorine gas.
\( \text{2Cl}^-\text{(aq)} \rightarrow \text{Cl}_2\text{(g)} + \text{2e}^- \)
At the cathode (negative electrode): Water molecules gain electrons to form hydrogen gas and hydroxide ions.
\( \text{2H}_2\text{O(l)} + \text{2e}^- \rightarrow \text{H}_2\text{(g)} + \text{2OH}^-\text{(aq)} \)
The overall reaction for the electrolysis of brine is:
\( \text{2NaCl(aq)} + \text{2H}_2\text{O(l)} \rightarrow \text{H}_2\text{(g)} + \text{Cl}_2\text{(g)} + \text{2NaOH(aq)} \)
The sodium ions (\( \text{Na}^+ \)) remain in the solution as spectator ions and combine with the newly formed hydroxide ions (\( \text{OH}^- \)) to produce sodium hydroxide, making the solution basic.
In simple words: When salt water is split using electricity, chlorine gas forms at one side, hydrogen gas and hydroxide (which makes the water basic) form at the other. Sodium stays in the water.

🎯 Exam Tip: Clearly show the electron transfer at both the anode (oxidation) and cathode (reduction), and ensure the overall equation balances both atoms and charge.

 

12th Chemistry Guide Metallurgy Additional Questions And Answers

 

Part - II - Additional Questions One Mark

 

I. Match The Following

 

Question 1. Match the following ores with their correct formulas.

Answer:
The correct matching of ores with their formulas is:
1. Magnetite - c) \( \text{Fe}_3\text{O}_4 \)
2. Cuprite - e) \( \text{Cu}_2\text{O} \)
3. Calamine - a) \( \text{ZnCO}_3 \)
4. Cerrusite - b) \( \text{PbCO}_3 \)
5. Cassiterite - d) \( \text{SnO}_2 \)
This table clearly shows the chemical composition for common ores, which helps in understanding their extraction processes.
In simple words: This matches common ores like magnetite and cuprite with their correct chemical names.

🎯 Exam Tip: Memorize the common names and chemical formulas of important ores as they are frequently asked in exams.

 

Question 2. Match the following ores with the metal they contain.

Answer:
The correct matching of ores with the metals they contain is:
1. Ore of copper - c) Malachite
2. Ore of aluminium - a) Diaspore
3. Ore of iron - d) Limonite
4. Ore of lead - e) Anglesite
5. Ore of silver - b) Chlorargyrite
Knowing which ore belongs to which metal is fundamental for understanding the extraction processes involved in metallurgy. These names are important for identifying sources of different metals.
In simple words: This connects different metal ores like malachite and limonite to the metals they contain, such as copper and iron.

🎯 Exam Tip: Familiarize yourself with common ores for each metal, as this helps in quickly identifying the source material for extraction processes.

 

Question 3. Match the following concentration methods with the ores they are suitable for.

Answer:
The correct matching of concentration methods with suitable ores is:
Gravity separation - d) Tinstone
Froth floatation - c) Zinc blende
Cyanide leaching - e) Gold
Alkali leaching - b) Alumina
Magnetic separation - a) Pyrolusite
Each concentration method is chosen based on the physical and chemical properties of the ore and the impurities, ensuring the most efficient separation for each specific case.
In simple words: This matches ways to clean ores, like using water for heavy tinstone or bubbles for zinc blende, with the right ore type.

🎯 Exam Tip: Understand the underlying principle of each concentration method (e.g., density difference, surface properties, magnetic properties) to correctly match it with the appropriate ore.

 

Question 4. Match the following purification methods with the metals they are used for.

Answer:
The correct matching of purification methods with the metals they are used for is:
1. Distillation - b) Zinc
2. Liquation - d) Tin
3. Electrolytic refining - e) Silver
4. Zone refining - a) Silicon
5. Mond process - c) Nickel
These purification methods are tailored to the specific properties of each metal and its impurities, ensuring a high degree of purity for various industrial applications. For instance, distillation is used for low boiling point metals.
In simple words: This connects different ways to purify metals, like heating to separate zinc or melting to purify tin, with the specific metal they work best for.

🎯 Exam Tip: Understand the principle behind each purification method (e.g., boiling point difference for distillation, melting point for liquation, volatility for Mond) to easily recall the metals they refine.

 

II. Fill In The Blanks

 

Question 1. ______ shows high resistance to corrosion and used in the design of Chemical reactors is ______.
Answer: **Aluminium** shows high resistance to corrosion and used in the design of chemical reactors is **Aluminium**.
In simple words: Aluminium metal resists rust well and is used to build chemical reactors.

🎯 Exam Tip: Remember key properties like corrosion resistance for important metals and their common applications in industries.

 

Question 2. ______ are used for increasing the efficiency of the solar cells.
Answer: **Gold nanoparticles** are used for increasing the efficiency of the solar cells.
In simple words: Tiny bits of gold help solar cells work better.

🎯 Exam Tip: Note emerging applications of advanced materials like nanoparticles in fields such as renewable energy.

 

Question 3. The removal of gangue from ores is called as ______.
Answer: The removal of gangue from ores is called as **concentration of ores**.
In simple words: Taking out the unwanted dirt from ores is called concentrating the ores.

🎯 Exam Tip: Understand "gangue" as the undesirable earthy or rocky materials in an ore, and "concentration" as the process to remove them.

 

Question 4. ______ is the process in which concentrated ore is strongly heated in the absence of air.
Answer: **Calcination** is the process in which concentrated ore is strongly heated in the absence of air.
In simple words: Calcination is when you heat up an ore a lot without air.

🎯 Exam Tip: Differentiate calcination (heating without air, usually for carbonates/hydroxides) from roasting (heating with air, usually for sulfides).

 

III. Find The Odd Man Out

 

Question 1. a) Sphalerite b) Galena c) Azurite d) Iron pyrite
Answer: (c) Azurite
Azurite is the odd one out because it is a basic carbonate ore of copper (\( \text{Cu}_3(\text{CO}_3)_2(\text{OH})_2 \)). Sphalerite (\( \text{ZnS} \)), Galena (\( \text{PbS} \)), and Iron pyrite (\( \text{FeS}_2 \)) are all sulphide ores. This distinction is important for choosing the correct concentration method.
In simple words: Azurite is different because it's a carbonate ore, while the others are sulfide ores.

🎯 Exam Tip: Classify ores based on their chemical composition (e.g., sulphide, carbonate, oxide) to identify patterns and odd ones out.

 

Question 2. a) Malachite b) Limonite c) Siderite d) Haematite
Answer: (a) Malachite
Malachite is the odd one out because it is an ore of copper (\( \text{Cu}_2(\text{CO}_3)(\text{OH})_2 \)). Limonite (\( \text{Fe}_2\text{O}_3 \cdot \text{nH}_2\text{O} \)), Siderite (\( \text{FeCO}_3 \)), and Haematite (\( \text{Fe}_2\text{O}_3 \)) are all ores of iron. Recognizing the specific metal each ore belongs to is crucial in metallurgy.
In simple words: Malachite is a copper ore, but the others (limonite, siderite, haematite) are all iron ores.

🎯 Exam Tip: A good understanding of the composition and classification of common ores is essential for these types of questions.

 

IV. Choose The Incorrect Pair

 

Question 1. a) Malachite, Azurite b) Ruby silver, Horn silver c) Zincite, Cuprite d) Anglesite, Cerrusite
Answer: (c) Zincite, Cuprite
This is the incorrect pair in the context of grouping ores of the same metal. Zincite is an ore of zinc (\( \text{ZnO} \)), while Cuprite is an ore of copper (\( \text{Cu}_2\text{O} \)). The other options pair ores of the same metal: Malachite and Azurite are both copper ores; Ruby silver and Horn silver are both silver ores; Anglesite and Cerrusite are both lead ores. So, Zincite and Cuprite is the pair that contains ores of different metals. This highlights the importance of precise ore classification.
In simple words: The pair "Zincite, Cuprite" is incorrect because Zincite is for zinc and Cuprite is for copper, while all other pairs listed are for the same metal.

🎯 Exam Tip: When asked to identify an "incorrect pair" for ores, check if both ores in the pair belong to the same metal, or if they are correctly matched based on another common characteristic.

 

Question 2. a) Kaolinite, Aluminium b) Stefinite, Silver c) Galena, Lead d) Prousitite, Tin
Answer: (d) Prousitite, Tin
The pair "Prousitite, Tin" is incorrect. Prousitite is actually an ore of silver (\( \text{Ag}_3\text{AsS}_3 \)), not tin. The correct pair for Prousitite would be Prousitite, Silver. The other options correctly pair an ore with its corresponding metal: Kaolinite for Aluminium, Stefinite for Silver, and Galena for Lead. This emphasizes knowing the metal associated with each specific ore name.
In simple words: "Prousitite, Tin" is the wrong pair because Prousitite is an ore of silver, not tin.

🎯 Exam Tip: Accurately linking ore names to their principal metal is critical for avoiding errors in matching questions.

 

V. Choose The Correct Pair

 

Question 1. a) Cerrusite, Cassiterite b) Siderite, Limonite c) Anglesite, Zincite d) Azurite, Kaolinite
Answer: (b) Siderite, Limonite
The pair "Siderite, Limonite" is the correct one. Both Siderite (\( \text{FeCO}_3 \)) and Limonite (\( \text{Fe}_2\text{O}_3 \cdot \text{nH}_2\text{O} \)) are important ores of iron. The other options contain ores of different metals or incorrectly paired elements. For example, Cerrusite is a lead ore, while Cassiterite is a tin ore. Anglesite is a lead ore, while Zincite is a zinc ore. Azurite is a copper ore, and Kaolinite is an aluminium mineral. Correct identification of common ores for major metals is essential.
In simple words: "Siderite, Limonite" is the right pair because both are ores from which we get iron.

🎯 Exam Tip: Focus on the main metal extracted from each ore to quickly identify correct pairings, especially when multiple ores for a single metal are common.

 

Question 2. a) Diaspore, Copper b) Galena, Tin c) Stefinite, Silver d) Malachite, Aluminium
Answer: (c) Stefinite, Silver
The correct pair is "Stefinite, Silver." Stefinite is indeed an ore from which silver is extracted. Diaspore is an aluminium ore, not copper. Galena is a lead ore, not tin. Malachite is a copper ore, not aluminium. Knowing these specific associations is key to metallurgy. This kind of knowledge helps trace the origin of metals.
In simple words: "Stefinite, Silver" is the correct match because Stefinite is an ore of silver.

🎯 Exam Tip: Practice associating each common ore name with the specific metal it yields to improve accuracy in matching questions.

 

VI. Assertion And Reason

 

Question 1. Assertion (A): Tinstone ore is concentrated by magnetic separation. Reason (R): Wolframite impurities are magnetic
(i) A and R are correct, R explains A.
(ii) A is correct, R is wrong
(iii) A is wrong, R is correct
(iv) A and R are correct but R does not explain A.
Answer: (i) A and R are correct, R explains A.
Both the assertion and the reason are correct, and the reason accurately explains the assertion. Tinstone (Cassiterite, \( \text{SnO}_2 \)) is a non-magnetic ore, while Wolframite (\( (\text{Fe,Mn})\text{WO}_4 \)), a common impurity in tinstone, is magnetic. Therefore, magnetic separation is used to effectively remove the magnetic wolframite impurities from the non-magnetic tinstone ore, thereby concentrating the tinstone. This is a practical application of differences in physical properties for ore beneficiation.
In simple words: Tinstone ore is cleaned using magnets because the unwanted bits (wolframite) stick to the magnets, which is why the method works.

🎯 Exam Tip: In assertion-reason questions, first determine if each statement is true, then check if the reason logically explains the assertion.

 

Question 2. Assertion (A): Aluminium can be commercially extracted from china clay which is a profitable one Reason (R): China clay is a mineral of aluminium.
(i) A and R are correct, R explains A.
(ii) A is correct, R is wrong
(iii) A is wrong, R is correct
(iv) A and R are correct but R does not explain A.
Answer: (iii) A is wrong, R is correct
The Assertion (A) is incorrect. While China clay (Kaolinite, \( \text{Al}_2\text{Si}_2\text{O}_5(\text{OH})_4 \)) is indeed a mineral containing aluminium, aluminium is *not* commercially extracted from it. The primary and most profitable ore for commercial extraction of aluminium is bauxite (\( \text{Al}_2\text{O}_3 \cdot \text{nH}_2\text{O} \)). The Reason (R) is correct; China clay is a mineral that contains aluminium. Therefore, Assertion is wrong but Reason is correct. This is a common misunderstanding in introductory metallurgy.
In simple words: The idea that aluminium is taken from china clay for money is false, even though china clay does have aluminium. Bauxite is the correct ore used.

🎯 Exam Tip: Distinguish between a mineral that *contains* a metal and an ore that is *economically viable* for its extraction. Not all minerals containing a metal are considered ores for that metal.

 

Question 3. Assertion (A): Zinc blende can be concentrated by the froth floatation method. Reason (R): Metallic ore particles are preferentially wetted by water and settle at the bottom.
(i) A and R are correct, R explains A.
(ii) A is correct, R is wrong
(iii) A is wrong, R is correct
(iv) A and R are correct but R does not explain A.
Answer: (ii) A is correct, R is wrong
The Assertion (A) is correct. Zinc blende (\( \text{ZnS} \)), being a sulphide ore, is indeed concentrated by the froth flotation method, which separates it from gangue. However, the Reason (R) is incorrect. In froth flotation, metallic ore particles are *preferentially wetted by oil* (and made water repellent), causing them to rise to the surface with the froth. The gangue particles are preferentially wetted by water and settle at the bottom. Therefore, Assertion is correct, but Reason is wrong. This is the fundamental principle of the froth flotation process.
In simple words: Zinc blende is cleaned using the froth flotation method. But the reason given is wrong, because in this method, the ore sticks to oil and floats, it doesn't get wet by water and sink.

🎯 Exam Tip: Clearly understand the principles of froth flotation: ore particles are made water-repellent (hydrophobic) and attach to air bubbles, while gangue particles remain water-wettable (hydrophilic) and sink.

 

Question 4. Assertion (A): \( \text{Cr}_2\text{O}_3 \) is reduced into chromium by aluminothermic process. Reason (R): Aluminium acts as the reducing agent.
(i) A and R are correct, R explains A.
(ii) A is correct, R is wrong
(iii) A is wrong, R is correct
(iv) A and R are correct but R does not explain A.
Answer: (i) A and R are correct, R explains A.
Both the assertion and the reason are correct, and the reason provides a direct explanation for the assertion. In the aluminothermic process, highly reactive aluminium metal is used as a strong reducing agent to reduce metal oxides of less reactive metals, such as chromium oxide (\( \text{Cr}_2\text{O}_3 \)). Aluminium itself gets oxidized to aluminium oxide, and a large amount of heat is released, which drives the reaction to produce molten chromium. This demonstrates aluminium's effectiveness as a reducing agent due to its high affinity for oxygen.
In simple words: Chromium oxide is turned into chromium metal using aluminium powder. This happens because aluminium is a good reducing agent, meaning it easily takes oxygen from other substances.

🎯 Exam Tip: Recall that the aluminothermic process utilizes aluminium's strong reducing power and the highly exothermic nature of \( \text{Al}_2\text{O}_3 \) formation to extract metals like chromium and manganese.

 

VII. Choose The Correct Statement

 

Question 1.
(a) Metals having more chemical reactivity occur as native elements.
(b) Removal of gangue from ores is called refining.
(c) Tin stone ore is concentrated by gravity separation.
(d) Silver glance is a carbonate ore.
Answer: (c) Tin stone ore is concentrated by gravity separation.
This statement is correct. Tin stone (cassiterite, \( \text{SnO}_2 \)) is much heavier than its associated gangue, making gravity separation an effective method for its concentration. Highly reactive metals (like sodium, potassium, calcium) do not occur as native elements but in combined forms (ores) due to their reactivity. Refining is the purification of an extracted metal, not the removal of gangue (which is concentration). Silver glance (\( \text{Ag}_2\text{S} \)) is a sulphide ore, not a carbonate ore. This shows a good understanding of ore concentration methods.
In simple words: The correct statement is that tin stone ore is cleaned by gravity separation, meaning the heavier tin ore is separated from lighter impurities using water.

🎯 Exam Tip: Pay close attention to definitions like native elements, refining, and concentration, and the specific properties that make certain concentration methods suitable for particular ores.

 

Question 2.
(a) In froth floatation sodium ethyl xanthate acts as a collector.
(b) In leaching the ore is converted into insoluble salt or complex and the gangue remains in the solution.
(c) Ammonia leaching is suitable for gold and silver.
(d) Bauxite ore is subjected to acid leaching
Answer: (a) In froth floatation sodium ethyl xanthate acts as a collector.
This statement is correct. Sodium ethyl xanthate is a commonly used collector in froth flotation. Collectors attach to the ore particles, making them water-repellent and enabling them to float with the froth. In leaching, the ore is converted into a *soluble* complex, and the gangue remains *insoluble*. Ammonia leaching is suitable for copper, nickel, and cobalt, not typically gold and silver (cyanide leaching is used for gold/silver). Bauxite ore is subjected to *alkali* leaching (Bayer's process), not acid leaching. This highlights the precise roles of reagents in different metallurgical processes.
In simple words: The correct fact is that a chemical called sodium ethyl xanthate helps collect the ore in froth flotation, making it float with the bubbles.

🎯 Exam Tip: Understand the specific functions of different reagents (collectors, frothers, depressants) in froth flotation and the correct leaching agents for various ores.

 

Question 3.
(a) Calcination is the process in which concentrated ore is strongly heated in the presence of air.
(b) Flux is a chemical substance that forms an easily fusible slag with gangue.
(c) In aluminothermic process the ignition mixture used is magnesium peroxide and barium.
(d) Any metal can reduce the oxides of other metals that are located below it in Ellingham diagram.
Answer: (b) Flux is a chemical substance that forms an easily fusible slag with gangue.
This statement is correct and accurately defines a flux's role in metallurgy. Calcination is heating in the *absence* of air, not presence. In the aluminothermic process, the ignition mixture typically contains magnesium powder and barium peroxide to start the reaction, but this option lists two substances, so this is also incorrect as per a complete definition of the ignition mixture. A metal can reduce oxides of other metals that are located *above* it in the Ellingham diagram, not below it. This ensures a clear understanding of metallurgical terminology and principles.
In simple words: The correct statement is that flux is a chemical that mixes with unwanted parts of the ore (gangue) to form a melted waste material (slag).

🎯 Exam Tip: Differentiate between key metallurgical terms like calcination, roasting, flux, and slag. Remember that a metal can reduce any metal oxide whose Ellingham line is *above* its own.

 

Question 4.
(a) In electrorefining pure metal is taken as anode and impure metal is taken as cathode.
(b) Distillation is employed for high boiling nonvolatile metals.
(c) Zone refining is based on the principle of fractional crystallisation.
(d) Mond's process is used for refining titanium.
Answer: (c) Zone refining is based on the principle of fractional crystallisation.
This statement is correct. Zone refining relies on the principle that impurities are more soluble in the molten state of a metal than in its solid state, which is a form of fractional crystallization. In electrorefining, the impure metal is the *anode*, and the pure metal is the *cathode*. Distillation is used for *low* boiling volatile metals, not high boiling non-volatile ones. Mond's process is used for refining *nickel*, not titanium (titanium is refined by Van Arkel method). This demonstrates a solid grasp of metal purification techniques.
In simple words: The correct statement is that zone refining works by freezing and melting metal repeatedly to make it pure, like how salt separates from water when it freezes.

🎯 Exam Tip: Understand the specific principles behind each refining method and the types of metals or impurities they are best suited for. This prevents common mix-ups.

VIII. Choose the incorrect statement

 

Question 1.
i) In cyanide leaching gold is converted into an insoluble cyanide complex.
ii) In ammonia leaching nickel forms a soluble complex.
iii) In alkali leaching aluminum forms an insoluble complex,
(a) i & ii
(b) i & iii
(c) ii & iii
(d) i, ii, iii
Answer: (d) i, ii & iii
In simple words: All three statements provided are incorrect. Gold forms a soluble complex during cyanide leaching, not an insoluble one. Nickel forms a soluble complex in ammonia leaching, which is true. Aluminum forms a soluble complex, not an insoluble one, in alkali leaching. So, the source implies all three statements are to be considered incorrect.

🎯 Exam Tip: Always remember the chemical nature of complexes formed during leaching processes – whether they are soluble or insoluble, as this dictates the separation.

 

Question 2.
i) In the Ellingham diagram for most of the metal oxide forming the slope is negative.
ii) Oxygen gas is consumed during the formation of metal oxides resulting in the increase of randomness.
iii) As temperature increases value for the formation of the metal oxide become more negative
(a) i & ii
(b) i & iii
(c) ii & iii
Answer: (c) ii & iii
In simple words: Based on the provided answer, statements (ii) and (iii) are considered incorrect. Statement (ii) is incorrect because consuming gas to form a solid generally decreases randomness (entropy). Statement (iii) is incorrect because for most metal oxides, increasing temperature makes the Gibbs free energy for formation less negative (or more positive), not more negative.

🎯 Exam Tip: Understand the thermodynamic principles behind Ellingham diagrams, especially how temperature affects Gibbs free energy and entropy changes for metal oxide formation.

 

Question 3.
i) The reduction of oxides of active metals such as sodium, potassium, etc. by carbon is thermodynamically feasible
ii) When a more reactive metal is added to the solution containing less reactive metal, the less reactive metal will go into the solution.
iii) Copper displaces zinc from zinc salt solution.
(a) i & ii
(b) i & iii
(c) ii & iii
(d) i, ii & iii
Answer: (d) i, ii & iii
In simple words: All three statements are incorrect. Active metals are hard to reduce with carbon. A more reactive metal displaces a less reactive one, meaning the less reactive metal comes out of solution. Copper is less reactive than zinc, so it cannot displace zinc from its salt solution. These are fundamental reactivity principles.

🎯 Exam Tip: Remember the activity series of metals to predict displacement reactions and the feasibility of reductions by carbon for different metals.

 

Question 4.
i) When an impure metal is melted and allowed to solidify, the impurities will prefer to be in the solid region.
ii) Zone refining is carried out in an inert gas atmosphere to prevent the reduction of metals.
iii) Elements such as germanium, silicon, and gallium are refined by zone refining.
(a) i & ii
(b) i & iii
(c) ii & iii
(d) i, ii & iii
Answer: (a) i & ii
In simple words: Statements (i) and (ii) are incorrect. In zone refining, impurities prefer to stay in the molten part, not the solid part. Also, zone refining uses an inert atmosphere to prevent *oxidation*, not reduction, of metals. Statement (iii) is correct, as zone refining is ideal for semiconductors.

🎯 Exam Tip: Understand the specific principles of each refining method. For zone refining, impurities prefer the liquid phase, and an inert atmosphere prevents unwanted reactions with air.

IX. Choose the best answer.

 

Question 1. Which of the following is not an oxide ore?
(a) Cuprite
(b) Siderite
(c) Cassiterite
(d) Zincite
Answer: (b) Siderite
In simple words: Siderite is an iron carbonate ore (\( FeCO_3 \)), not an oxide. All the other options are examples of oxide ores.

🎯 Exam Tip: Classify ores based on their chemical composition (oxide, carbonate, sulphide, etc.) to correctly identify them.

 

Question 2. Which of the following is an oxide ore?
(a) Sphalerite
(b) Calamine
(c) Cassiterite
(d) Stefinite
Answer: (c) Cassiterite
In simple words: Cassiterite is a tin oxide ore (\( SnO_2 \)). The other options are sulphide, carbonate, and silver antimony sulphide ores, respectively.

🎯 Exam Tip: Memorize common ores and their chemical formulas to easily identify their type.

 

Question 3. The process of converting hydrated alumina into anhydrous alumina is called
(a) Roasting
(b) Smelting
(c) Auto-reduction
(d) Calcination
Answer: (d) Calcination
In simple words: Calcination is a heating process where hydrated compounds lose water and carbonate ores lose carbon dioxide without melting. Here, hydrated alumina loses water.

🎯 Exam Tip: Distinguish between calcination (heating below melting point, removing volatile impurities like water or \( CO_2 \)) and roasting (heating sulphide ores in air to convert them to oxides).

 

Question 4. Which of the following is a sulphide ore?
(a) Pyrargyrite
(b) Malachite
(c) Limonite
(d) Kaolinite
Answer: (a) Pyrargyrite
In simple words: Pyrargyrite is a silver antimony sulphide ore (\( Ag_3SbS_3 \)). Malachite and limonite are carbonate and hydrated oxide ores, respectively, and kaolinite is a silicate.

🎯 Exam Tip: Recognize the 'sulfide' suffix for sulphide ores (e.g., galena, sphalerite, pyrites) and learn the specific names like pyrargyrite.

 

Question 5. Which of the following is not a carbonate ore?
(a) Siderite
(b) Calamine
(c) Cerrusite
(d) Cassiterite
Answer: (d) Cassiterite
In simple words: Cassiterite is an oxide ore of tin (\( SnO_2 \)). Siderite, calamine, and cerrusite are all carbonate ores.

🎯 Exam Tip: Pay close attention to negative phrasing like "not a carbonate ore" to avoid common mistakes. This requires knowing what each ore *is*.

 

Question 6. Which of the following is a carbonate ore?
(a) Limonite
(b) Siderite
(c) Magnetite
(d) Haematite
Answer: (b) Siderite
In simple words: Siderite is an iron carbonate ore (\( FeCO_3 \)). Limonite and haematite are hydrated oxide and oxide ores of iron respectively, and magnetite is also an iron oxide.

🎯 Exam Tip: A quick way to identify carbonate ores is to look for names ending with '-ite' that aren't sulphides, but memorization is best.

 

Question 7. Which of the following is the ore of iron?
(a) Limonite
(b) Azurite
(c) Stefinite
(d) Cerrusite
Answer: (a) Limonite
In simple words: Limonite is an important hydrated oxide ore of iron (\( Fe_2O_3.nH_2O \)). The other options are copper, silver, and lead ores.

🎯 Exam Tip: Group ores by the metal they contain (e.g., iron ores, copper ores) to make recall easier.

 

Question 8. Which of the following is not an ore of iron?
(a) Haematite
(b) Magnetite
(d) Anglesite
Answer: (d) Anglesite
In simple words: Anglesite is an ore of lead (\( PbSO_4 \)), meaning it is not an iron ore. Haematite and magnetite are both important ores of iron.

🎯 Exam Tip: Practice identifying the primary metal for common ores to easily distinguish them.

 

Question 9. Which of the following is an ore of silver?
(a) Azurite
(b) Prousitite
(c) Cerrusite
(d) Limonite
Answer: (b) Prousitite
In simple words: Prousitite is a sulphide ore of silver (\( Ag_3AsS_3 \)). The other options are ores of copper, lead, and iron respectively.

🎯 Exam Tip: Learning the specific chemical formulas helps in identifying the metal content of each ore type.

 

Question 10. Which of the following is a sulphate ore?
(a) Galena
(b) Zinc blende
(c) Cerrusite
(d) Anglesite
Answer: (d) Anglesite
In simple words: Anglesite is a lead sulphate ore (\( PbSO_4 \)). Galena and zinc blende are sulphide ores, while cerrusite is a carbonate ore.

🎯 Exam Tip: Pay attention to the "-ate" suffixes (e.g., sulphate, carbonate) which often indicate the non-metal part of the ore's chemical formula.

 

Question 11. Non-metallic impurities, rocky materials, and siliceous matter which are associated with ores are called as.
(a) Slag
(b) Flux
(c) Gangue
(d) residue
Answer: (c) Gangue
In simple words: The unwanted rocky and earthy materials mixed with an ore are known as gangue. These are removed during ore concentration.

🎯 Exam Tip: Understand the key terms in metallurgy: ore (metal source), gangue (impurities), flux (substance added to remove gangue), and slag (fusible product of flux + gangue).

 

Question 12. Gravity separation is suitable for
(a) Oxide ore
(b) Sulphide ore
(c) Carbonate ore
(d) Sulphate ore
Answer: (a) Oxide ore
In simple words: Gravity separation is best for oxide ores, especially those that are heavy, because it relies on the density difference between the ore and the lighter gangue particles.

🎯 Exam Tip: Gravity separation (or hydraulic washing) works well when there is a significant density difference between the ore and the gangue.

 

Question 13. Froth floatation is suitable for
(a) Oxide ore
(b) Sulphide ore
(c) Carbonate ore
(d) Sulphate ore
Answer: (b) Sulphide ore
In simple words: Froth flotation is a method specifically designed to concentrate sulphide ores because their particles are easily wetted by oil and float, while gangue particles sink.

🎯 Exam Tip: Froth flotation is the primary method for concentrating most sulphide ores, leveraging their hydrophobic (water-repelling) nature when treated with collectors.

 

Question 14. In froth floatation, pine oil is used as a
(a) Collector
(b) depressing agent
(c) Frothing agent
(d) Flux
Answer: (c) Frothing agent
In simple words: Pine oil helps to create a stable foam or froth that can carry the ore particles to the surface during the froth flotation process.

🎯 Exam Tip: Differentiate between collectors (e.g., xanthates, which make ore particles water-repellent) and frothers (e.g., pine oil, cresols, which stabilize the froth).

 

Question 15. In froth floatation sodium ethyl Xanthate is used as a
(a) Collector
(b) depressing agent
(c) frothing agent
(d) Flux
Answer: (a) Collector
In simple words: Sodium ethyl xanthate is a substance that makes the sulphide ore particles hydrophobic, so they can attach to air bubbles and float.

🎯 Exam Tip: Collectors are crucial in froth flotation as they selectively bind to the ore particles, making them water-repellent and enabling them to be carried by the froth.

 

Question 16. In froth floatation sodium cyanide is used as a
(a) Collector
(b) depressing agent
(c) frothing agent
(d) Flux
Answer: (b) depressing agent
In simple words: Sodium cyanide is used to prevent certain sulphide ores, like zinc sulphide, from forming froth, allowing other sulphide ores, like lead sulphide, to be separated.

🎯 Exam Tip: Depressing agents enhance the separation of two sulphide ores by preventing one from coming into the froth.

 

Question 17. The floatation property of the impurity ZnS present in galena is depressed by adding
(a) Pure oil
(b) Eucalyptus oil
(c) Sodium cyanide
(d) Sodium ethyl Xanthate
Answer: (c) Sodium cyanide
In simple words: When separating lead sulphide (galena) and zinc sulphide (ZnS), sodium cyanide is added. It stops the zinc sulphide from floating, allowing the galena to be collected.

🎯 Exam Tip: Specific depressants are used for specific ore separations; NaCN is a classic example for ZnS depression during galena flotation.

 

Question 18. Which method of purification represented by the equation?
\( Zr (impure) + 2I_2 \xrightarrow{550K} ZrI_4 \xrightarrow{1800K} Zr (pure) + 2I_2 \)
(a) Cupellation
(b) Zone refining
(c) Van-Arkel method
(d) Mond's process
Answer: (c) Van-Arkel method
In simple words: This process describes the Van-Arkel method, where impure zirconium reacts with iodine to form a volatile compound, which then decomposes at a higher temperature to give pure zirconium. It's a key refining process for very pure metals.

🎯 Exam Tip: The Van-Arkel method and Mond's process both involve forming volatile compounds of metals for purification, but use different reagents (iodine for Van-Arkel, carbon monoxide for Mond's).

 

Question 19. The refining of gold ore is done by
(a) Cyanide leaching
(b) Ammonia leaching
(c) Alkali leaching
(d) Acid leaching
Answer: (a) Cyanide leaching
In simple words: Gold is extracted from its ore using cyanide leaching, where gold forms a soluble complex with cyanide ions. This method is effective for precious metals like gold and silver.

🎯 Exam Tip: Cyanide leaching is specifically used for gold and silver because these metals can form stable soluble complexes with cyanide ions in the presence of oxygen.

 

Question 20. Ammonia leaching is done for the concentration of the ore of
(a) Silver
(b) Copper
(c) Aluminium
(d) Zinc
Answer: (b) Copper
In simple words: Ammonia leaching is used to concentrate copper ores, as copper can form soluble ammine complexes with ammonia. This is a selective process.

🎯 Exam Tip: Remember that different leaching agents are chosen based on the metal's ability to form soluble complexes with them (e.g., cyanide for Au/Ag, ammonia for Cu).

 

Question 21. During roasting sulphide ores are converted into their
(a) Metals
(b) Oxides
(c) Carbonates
(d) nitrates
Answer: (b) Oxides
In simple words: Roasting is a process where sulphide ores are heated strongly in the presence of air, turning them into metal oxides and releasing sulphur dioxide gas. This conversion helps in further reduction steps.

🎯 Exam Tip: Roasting is a crucial step for sulphide ores as it converts them to a more easily reducible oxide form, also removing sulphur impurities as \( SO_2 \).

 

Question 22. During the calcination of carbonate ore the expelled gas is
(a) Carbon monoxide
(b) Carbon dioxide
(c) Sulphur dioxide
(d) Nitrogen dioxide
Answer: (b) Carbon dioxide
In simple words: Calcination of carbonate ores involves heating them without air, which causes the carbonate to break down and release carbon dioxide gas, leaving behind a metal oxide.

🎯 Exam Tip: Calcination is a thermal decomposition process; carbonate ores release \( CO_2 \), and hydrated ores release water vapour.

 

Question 23. Sulphide ores of metals are usually concentrated by froth floatation process. Which one of the following sulphide ore offers an exception and is concentrated by chemical leaching.
(a) Argentite
(b) galena
(c) Copper pyrites
(d) Sphalerite
Answer: (a) Argentite
In simple words: While most sulphide ores are concentrated by froth flotation, argentite (silver sulphide) is an exception. It is usually concentrated through chemical leaching using a cyanide solution.

🎯 Exam Tip: Be aware of exceptions to general rules in metallurgy; argentite's concentration by leaching is an important detail.

 

Question 24. Cinnabar is converted into mercury by
(a) Reduction by metal
(b) Reduction by hydrogen
(c) Reduction by carbon
(d) Auto reduction
Answer: (d) Auto reduction
In simple words: Cinnabar (\( HgS \)) can be converted to mercury simply by heating it in air. The mercury sulphide first becomes mercury oxide, which then decomposes into mercury and sulphur dioxide, an example of auto-reduction.

🎯 Exam Tip: Auto-reduction (or self-reduction) is common for less reactive metals like mercury, lead, and copper, where the sulphide ore is partially oxidized to an oxide, which then reacts with the remaining sulphide to produce the metal.

 

Question 25. Thermodynamically the reduction of metal oxide with a given reducing agent can occur if the free energy change for the coupled reaction is
(a) Positive
(b) Negative
(c) One
(d) Zero
Answer: (b) Negative
In simple words: For any chemical reaction, including the reduction of metal oxides, to happen on its own (spontaneously), the total change in Gibbs free energy for the reaction must be negative.

🎯 Exam Tip: A negative Gibbs free energy change (\( \Delta G \)) indicates a spontaneous and thermodynamically feasible reaction, which is a core concept in chemical thermodynamics.

 

Question 26. For the reduction of metal oxide into metal a reducing agent is selected in such a way that for the coupled reaction it provides a
(a) Large negative G value
(b) Small positive G value
Answer: (c) Large negative G value
In simple words: A good reducing agent for a metal oxide should cause the combined reaction (reduction of oxide + oxidation of reducing agent) to have a strongly negative Gibbs free energy change, making the reduction very favorable.

🎯 Exam Tip: The more negative the \( \Delta G \) for the coupled reaction, the greater the driving force for the reduction process, ensuring high yield and efficiency.

 

Question 27. For the formation of various metal oxides Ellingham diagram is a graphical representation between
(a) G° & S
(b) G° & H
(c) G° & T
(d) H & S
Answer: (c) G° & T
In simple words: The Ellingham diagram shows how the standard Gibbs free energy change (\( \Delta G^\circ \)) for the formation of metal oxides changes with temperature (T). This helps in choosing suitable reducing agents.

🎯 Exam Tip: Understand that Ellingham diagrams plot \( \Delta G^\circ \) versus T to visualize the thermodynamic stability of oxides and predict reduction feasibility.

 

Question 28. In the Ellingham diagram, for most of the metal oxide formation the slope is
(a) Positive
(b) Negative
(c) Zero
(d) One
Answer: (a) Positive
In simple words: For most metal oxides, the slope of the line in an Ellingham diagram is positive. This is because oxygen gas is consumed to form a solid oxide, leading to a decrease in randomness and thus a negative entropy change (\( \Delta S \)). Since \( \Delta G = \Delta H - T\Delta S \), a negative \( \Delta S \) makes \( -T\Delta S \) positive, contributing to a positive slope.

🎯 Exam Tip: A positive slope generally means that as temperature increases, the oxide becomes less stable and easier to reduce, as \( \Delta G \) becomes less negative.

 

Question 29. Elements like Silicon and Germanium to be used as a semiconductor is purified by (PTA – 1)
(a) heating under Vaccum
(b) Van-Arkel method
(c) Zone refining
(d) Electrolysis
Answer: (c) Zone refining
In simple words: Zone refining is the best method to purify semiconductor materials like silicon and germanium, as it can achieve extremely high purity levels by moving impurities into a molten zone.

🎯 Exam Tip: Zone refining is known for producing ultra-pure metals, especially those used in semiconductors, by concentrating impurities in the molten region.

 

Question 30. If the e.m.f of the net redox reaction is positive, its G is
(a) Positive
(b) Negative
(c) Zero
(d) One
Answer: (b) Negative
In simple words: For a redox reaction to happen on its own, a positive electromotive force (e.m.f.) means that the Gibbs free energy change (\( \Delta G \)) for the reaction will be negative. This relationship is important in electrochemistry.

🎯 Exam Tip: Recall the relationship \( \Delta G = -nFE^\circ \); a positive \( E^\circ \) directly corresponds to a negative \( \Delta G \), indicating spontaneity.

 

Question 31. Which of the following metal is refined by distillation?
(a) Tin
(b) Lead
(c) Zinc
(d) Bismuth
Answer: (c) Zinc
In simple words: Zinc is a low-boiling point metal, so it can be purified by distillation. In this process, the metal is heated, vaporized, and then condensed to separate it from non-volatile impurities.

🎯 Exam Tip: Distillation is suitable for metals with low boiling points (like Zn, Cd, Hg) to separate them from impurities that have much higher boiling points.

 

Question 32. Which of the following is not refined by zone refining?
(a) Germanium
(b) Zirconium
(d) Gallium
Answer: (b) Zirconium
In simple words: Zirconium is not refined by zone refining; it is refined using the Van-Arkel method. Germanium and gallium are typically refined by zone refining for semiconductor applications.

🎯 Exam Tip: Remember that zone refining is ideal for semiconductors, while the Van-Arkel method is used for metals like zirconium and titanium to achieve very high purity by forming and decomposing volatile compounds.

 

Question 33. Which of the following is refined by the Mond process?
(a) Silicon
(b) Copper
(d) Zinc
Answer: (c) Nickel
In simple words: The Mond process is used to refine nickel. It involves reacting impure nickel with carbon monoxide to form a volatile nickel carbonyl, which then decomposes to yield pure nickel.

🎯 Exam Tip: The Mond process is specific to nickel purification, leveraging its ability to form a volatile carbonyl compound at moderate temperatures.

 

Question 34. Which of the following is defined by Van Arkel method?
(a) Gallium
(b) Titanium
(c) Germanium
(d) Silicon
Answer: (b) Titanium
In simple words: The Van-Arkel method is used to purify metals like titanium and zirconium. It involves creating a volatile halide that later breaks down to release the pure metal.

🎯 Exam Tip: Remember that the Van-Arkel method is used for refining high melting point, reactive metals like titanium and zirconium.

 

Question 35. Which of the following metal is used in galvanization?
(a) Copper
(b) Aluminium
(c) Zinc
(d) Gold
Answer: (c) Zinc
In simple words: Zinc is used in galvanization, which is a process of coating iron or steel with a layer of zinc to protect it from rusting.

🎯 Exam Tip: Galvanization is a form of corrosion protection where zinc acts as a sacrificial anode because it is more reactive than iron.

 

Question 36. Which is used in making luminous paints, fluorescent lights, and x-ray screens?
(a) Brass
(b) Zinc sulphide
(c) Cast iron
(d) Gold nanoparticles
Answer: (b) Zinc sulphide
In simple words: Zinc sulphide is used to make luminous paints, fluorescent lights, and x-ray screens because it glows when exposed to certain types of energy, a property called luminescence.

🎯 Exam Tip: Zinc sulphide is a common phosphor, meaning it can absorb energy and re-emit it as light, making it useful in various illuminating applications.

 

Question 37. Which is used for increasing the efficiency of solar cells?
(b) Zinc sulphide
(c) Cast iron
(d) Gold nanoparticles
Answer: (d) Gold nanoparticles
In simple words: Gold nanoparticles can be used to make solar cells work better. They help the cells absorb more sunlight and turn it into electricity.

🎯 Exam Tip: Nanomaterials often have unique optical and electronic properties that can significantly enhance the performance of advanced technologies like solar cells.

 

Question 38. Which is not refined by liquation?
(a) Tin
(b) Zinc
(c) Lead
(d) Bismuth
Answer: (b) Zinc. Zinc is refined by distillation.
In simple words: Liquation is a method for purifying metals with low melting points like tin, lead, and bismuth. Zinc has a low melting point but is refined by distillation, not liquation, because it is more volatile.

🎯 Exam Tip: Liquation separates a low-melting metal from high-melting impurities, while distillation separates metals based on significant differences in boiling points.

 

Question 39. Which is not refined by zone refining?
(a) Silicon
(b) Gallium
(c) Zirconium
(d) Germanium
Answer: (c) Zirconium. Zirconium is refined by Van Arkel method
In simple words: Zirconium is typically purified using the Van-Arkel method, which involves forming volatile compounds. Zone refining, on the other hand, is generally used for semiconductors like silicon, germanium, and gallium.

🎯 Exam Tip: Different refining methods are chosen based on the specific properties of the metal and the type of impurities present.

X. Two Mark Questions

 

Question 1. What is a mineral?
Answer: A mineral is a naturally occurring substance found in the Earth's crust, obtained by mining. It contains a metal either in its pure, free state or combined in compounds like oxides or sulphides. Minerals are essential sources for metals, though not all minerals are suitable for economical extraction. An example is bauxite for aluminum.
In simple words: A mineral is a natural substance found in the ground, dug up by mining. It contains metals, either alone or as compounds.

🎯 Exam Tip: Differentiate between minerals (naturally occurring compounds) and ores (minerals from which metals can be profitably extracted).

 

Question 2. What is an ore?
Answer: An ore is a mineral that contains a high percentage of metal, making it commercially viable to extract the metal conveniently and economically. Not all minerals are ores, as some may contain too little metal or be too difficult to process. For example, bauxite is the main ore of aluminum.
In simple words: An ore is a special rock that has a lot of metal in it, making it cheap and easy to take the metal out.

🎯 Exam Tip: Emphasize the 'economical' and 'convenient' aspects when defining an ore; these are key differentiators from a general mineral.

 

Question 3. What is a concentration of ores?
Answer: Concentration of ores is the process of removing unwanted non-metallic impurities, rocky materials, and siliceous matter, collectively known as gangue, from the ore. This step increases the proportion of the metal-containing compound in the ore, preparing it for further extraction processes. This initial cleaning makes later stages more efficient.
In simple words: Concentration of ores means cleaning the ore to remove unwanted dirt and rocks, so there's more metal part left.

🎯 Exam Tip: Concentration (or beneficiation) is the first step in metallurgy, vital for reducing the bulk of material to be processed and improving efficiency.

 

Question 4. What is leaching?
Answer: Leaching is a chemical process where the metal present in an ore is dissolved into a suitable solvent to form a soluble metal salt or complex. This process leaves the unwanted impurities (gangue) undissolved. It is often used for low-grade ores or those where physical separation methods are not effective, like extracting gold with cyanide. The metal is later recovered from the solution.
In simple words: Leaching is a chemical way to get metal from its ore by dissolving the metal part into a liquid, leaving the dirt behind.

🎯 Exam Tip: Leaching is a hydrometallurgical technique, relying on selective chemical dissolution rather than physical separation methods, and is particularly useful for precious metals or complex ores.

 

Question 5. What is the reaction of Ammonia with Iron and copper salts? (PTA – 4)
Answer: Ammonia reacts differently with iron and copper salts:
For iron salts: Ammonia reacts with metallic salts like \( Fe^{3+} \) to produce insoluble metal hydroxides.
\( Fe^{3+} + 3NH_4^+ \rightarrow Fe(OH)_3 + 3NH_4^+ \)
For copper salts: Ammonia reacts with metallic salts like \( Cu^{2+} \) to form soluble ammine complexes.
\( Cu^{2+} + 4NH_3 \rightarrow [Cu(NH_3)_4]^{2+} \)
This blue colored complex is known as tetraamminecopper(II) ion.
In simple words: Ammonia makes iron salts form a solid called iron hydroxide. But with copper salts, ammonia forms a special blue liquid compound called a complex.

🎯 Exam Tip: Remember that ammonia acts as a base to precipitate hydroxides, but it can also act as a ligand to form soluble ammine complexes, depending on the metal ion.

 

Question 7. What is the role of the depressing agent in the froth flotation process?
Answer: In froth flotation, a depressing agent is used to selectively prevent certain sulfide ore particles from forming froth with the desired metal. For example, if zinc sulfide (ZnS) is an impurity mixed with galena (PbS), sodium cyanide (NaCN) is added. The NaCN forms a special zinc complex, \( \text{Na}_2[\text{Zn}(\text{CN})_4] \), on the surface of the zinc sulfide, which stops it from floating. This selective action ensures that only the desired ore floats to the top. This is important for separating different sulfide minerals from each other.
In simple words: A depressing agent helps to keep unwanted metal sulfide impurities from floating up with the valuable ore during the froth flotation process, making the separation cleaner.

🎯 Exam Tip: Remember that depressing agents are crucial for selective separation of different sulfide minerals, as without them, multiple valuable metals might float together, reducing purity.

 

Question 8. In the extraction of metal, why is the ore first converted into metal oxide before reduction into metal?
Answer: Metal ores are often converted into their oxides before reduction for a few reasons. Firstly, in a concentrated ore, the metal is usually in a positive oxidation state, meaning it has lost electrons. It needs to gain electrons (be reduced) to become an elemental metal. Secondly, when we look at the principles of thermodynamics, it is generally much easier to reduce a metal oxide than other metal compounds, such as sulfides or carbonates. Oxides have simpler chemical bonds to break. Therefore, to make the reduction process more efficient and require less energy, the ore is first changed into its oxide form.
In simple words: It is easier to remove oxygen from a metal compound to get the pure metal than to remove other elements like sulfur. So, ores are first changed into oxides because it makes the next step of getting pure metal much simpler.

🎯 Exam Tip: Always remember that converting to oxide simplifies the reduction step, often making it thermodynamically more favorable due to the relative stabilities of oxides compared to other compounds.

 

Question 9. Write about roasting.
Answer: Roasting is a process where a concentrated ore, especially a sulfide ore, is strongly heated in the presence of excess oxygen. The temperature used is kept below the melting point of the metal in a suitable furnace. The main goal of roasting is to convert sulfide ores into their corresponding metal oxides. For example, lead sulfide changes to lead oxide:
\( \text{2PbS} + \text{3O}_2 \rightarrow \text{2PbO} + \text{2SO}_2 \uparrow \)
Roasting also helps to remove common impurities like arsenic, sulfur, and phosphorus, as they get converted into their volatile oxides and escape as gases. For instance, arsenic converts to arsenic oxide:
\( \text{4As} + \text{3O}_2 \rightarrow \text{2As}_2\text{O}_3 \)
Roasting is a very common first step for many sulfide ores, preparing them for further reduction.
In simple words: Roasting is heating an ore with a lot of air to change sulfide parts into oxide parts and to get rid of impurities that turn into gas.

🎯 Exam Tip: Key aspects of roasting are the presence of oxygen, heating below the metal's melting point, and its use specifically for sulfide ores to form oxides and remove volatile impurities.

 

Question 10. Write about the extraction of metal by the process of reduction by carbon.
Answer: In this method, a metal oxide ore is mixed with carbon (usually in the form of coke) and then strongly heated in a blast furnace. Carbon acts as a strong reducing agent because it easily combines with oxygen to form stable carbon monoxide or carbon dioxide, removing oxygen from the metal oxide. This process is suitable for metals that do not form carbides when they react with carbon at the reduction temperature. For example, zinc oxide can be reduced by carbon to produce zinc metal and carbon monoxide gas:
\( \text{ZnO(s)} + \text{C(s)} \rightarrow \text{Zn(s)} + \text{CO(g)} \uparrow \)
The formation of stable carbon oxides drives this reaction forward, making carbon a very effective reducing agent.
In simple words: Metals can be taken out of their oxide ores by mixing them with carbon (like coke) and heating them a lot. The carbon takes the oxygen away, leaving pure metal.

🎯 Exam Tip: When carbon is used as a reducing agent, remember that it's highly effective at high temperatures and is chosen for metals that don't form carbides with carbon.

 

Question 11. Write about the extraction of metal by the process of reduction by hydrogen.
Answer: The extraction of metals by reduction with hydrogen is a method used for oxides of metals that are less electropositive than hydrogen, such as iron, lead, and copper. In this process, the metal oxide reacts with hydrogen gas, which removes the oxygen to form water, leaving the pure metal behind. For example, silver oxide can be reduced by hydrogen:
\( \text{Ag}_2\text{O(s)} + \text{H}_2\text{(g)} \rightarrow \text{2Ag(s)} + \text{H}_2\text{O(l)} \uparrow \)
Additionally, nickel oxide can be reduced to nickel using a mixture of hydrogen and carbon monoxide, often called water gas:
\( \text{2NiO(s)} + \text{CO(g)} + \text{H}_2\text{(g)} \rightarrow \text{2Ni(s)} + \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)} \uparrow \)
Hydrogen is particularly useful for reducing oxides of metals that are not easily reduced by carbon or for producing very pure metals.
In simple words: Some metal oxides can be turned into pure metal by heating them with hydrogen gas. The hydrogen takes the oxygen, making water and leaving the metal.

🎯 Exam Tip: Hydrogen reduction is generally used for oxides of less reactive metals and often produces high-purity metals due to hydrogen's clean reducing action.

 

Question 13. How is \( \text{Cr}_2\text{O}_3 \) reduced to \( \text{Cr} \) by Al powder?
Answer: Chromium oxide (\( \text{Cr}_2\text{O}_3 \)) can be reduced to chromium metal using aluminium powder through a process known as the aluminothermic process. In this method, \( \text{Cr}_2\text{O}_3 \) is mixed with aluminium powder in a fire clay crucible. To start the reaction, an ignition mixture, usually made of magnesium and barium peroxide (\( \text{BaO}_2 \)), is used. The reaction is highly exothermic, meaning it releases a large amount of heat, reaching temperatures up to 2400°C. This intense heat (852 kJmol^-1 liberated) helps to reduce the \( \text{Cr}_2\text{O}_3 \) into pure chromium. The main chemical reaction is:
\( \text{Cr}_2\text{O}_3 + \text{2Al} \underrightarrow { \triangle } \text{2Cr} + \text{Al}_2\text{O}_3 \)
Aluminium is a very reactive metal and strongly attracts oxygen, making it an excellent reducing agent for other metal oxides. The heat generated ensures the reaction goes to completion.
In simple words: Chromium oxide is mixed with aluminum powder and heated very strongly. The aluminum takes the oxygen from the chromium oxide, turning it into pure chromium metal.

🎯 Exam Tip: For aluminothermic reactions, remember that aluminium acts as a powerful reducing agent, and the process is highly exothermic, producing enough heat for the reduction to occur effectively.

 

Question 14. What is auto reduction of metallic ores?
Answer: Auto-reduction, also known as self-reduction, is a process where certain metallic ores can be reduced to crude metal simply by roasting them, without the need for an additional reducing agent. This happens because the ore itself helps in its reduction. A common example is the roasting of cinnabar (HgS), which is an ore of mercury. When cinnabar is heated in air, it converts directly into liquid mercury and sulfur dioxide gas. The ore partially oxidizes to form a small amount of oxide, which then reacts with the remaining sulfide to produce the metal.
\( \text{HgS(s)} + \text{O}_2\text{(g)} \rightarrow \text{Hg(l)} + \text{SO}_2\text{(g)} \uparrow \)
This method is self-sufficient because the ore effectively reduces itself.
In simple words: Auto-reduction means some metal ores can turn into metal just by heating them, without needing another chemical to take away the oxygen.

🎯 Exam Tip: Auto-reduction is a unique process for specific ores (like cinnabar) where the ore itself contains elements that facilitate its own reduction, eliminating the need for external reducing agents.

 

Question 15. What is the role of graphite rods in the electrometallurgy of Aluminium?
Answer: In the electrometallurgy of aluminium, specifically the Hall-Heroult process, graphite rods play a crucial role as anodes. The electrolysis is carried out in a large iron tank that is lined with carbon, which functions as the cathode. The carbon blocks or graphite rods are immersed into the electrolyte (molten mixture of alumina, cryolite, and calcium fluoride) and act as the anode. At the cathode, aluminium ions gain electrons to form pure aluminium metal. At the anode, oxygen ions from the molten alumina lose electrons and react with the carbon anode to form carbon monoxide and carbon dioxide gases. Because the carbon anode reacts with oxygen, it is slowly consumed during the process and needs to be replaced regularly. This shows its active participation in the chemical reaction.
In simple words: Graphite rods act as the positive part (anode) in the process of making aluminum using electricity. They react with oxygen and get used up while the pure aluminum is formed.

🎯 Exam Tip: Remember that in the Hall-Heroult process, the carbon lining of the tank is the cathode, while the graphite rods act as the consumable anode due to their reaction with oxygen from the alumina.

 

Question 16. Write about the distillation process of refining a metal?
Answer: Distillation is a refining process used to purify metals with low boiling points, especially when impurities have much higher boiling points. In this method, the impure metal is heated until it evaporates and turns into vapor. The metal vapor is then collected and condensed back into its pure liquid form, leaving behind the non-volatile impurities. This technique is particularly effective for refining volatile metals such as zinc and mercury, which have relatively low boiling points, making them easy to separate from impurities that stay solid or liquid at higher temperatures. This difference in boiling points is key to the separation.
In simple words: Distillation purifies metals by heating the impure metal until it turns into a gas, then cooling the gas to get pure liquid metal, leaving the impurities behind. It works best for metals that boil easily.

🎯 Exam Tip: The effectiveness of distillation relies entirely on a significant difference in boiling points between the pure metal and its impurities, allowing for selective vaporization.

 

Question 17. Write about the liquation process of refining a metal?
Answer: Liquation is a refining method used for metals that have a relatively low melting point, especially when they contain impurities with much higher melting points. In this process, the crude metal is placed on a sloping hearth of a reverberatory furnace and heated gently, just above the melting point of the main metal, in the absence of air. As the impure metal heats up, the pure metal melts first and forms a fusible liquid. This molten pure metal then flows down the sloping surface, leaving behind the solid impurities, which have higher melting points and remain on the hearth. This method is commonly used for refining metals such as tin, lead, mercury, and bismuth, allowing for their separation from less fusible impurities.
In simple words: Liquation is a way to purify metals by gently heating them on a slope. The pure metal melts and flows away, leaving behind solid impurities that need much higher heat to melt.

🎯 Exam Tip: The principle of liquation hinges on the difference in melting points, where the desired metal melts and flows away, separating from high-melting-point impurities.

 

Question 18. Write the applications or uses of copper.
Answer: Copper is a very important metal that has been used by humans for a very long time, leading to the 'Bronze Age' when its alloy, bronze, was widely adopted. Today, copper has many uses due to its excellent properties. It is used to make coins and ornaments, often mixed with gold and other metals. Copper and its alloys, such as brass and bronze, are widely used for making electrical wires because copper is an excellent conductor of electricity. They are also used for water pipes and various other electrical and plumbing components. Its high electrical conductivity makes it an indispensable material for wiring and electronics.
In simple words: Copper is used for making coins, jewelry, electrical wires, and water pipes because it conducts electricity well and has been important for a long time.

🎯 Exam Tip: When listing applications of copper, always highlight its excellent electrical conductivity and ductility, which are key to its widespread use in electrical components and wiring.

 

Question 19. Write the applications or uses of gold.
Answer: Gold is a highly valuable and precious metal known for its beautiful luster and resistance to corrosion. Historically, it has been used extensively for making coins and as a standard for monetary systems in many countries. Today, gold is widely used in jewelry, often in alloy forms with copper to increase its hardness. It is also used in electroplating to cover other metals with a thin layer of gold for decorative purposes or to protect against corrosion, such as in watches, artificial limb joints, cheap jewelry, dental fillings, and electrical connectors. In modern technology, gold nanoparticles are used to improve the efficiency of solar cells and act as catalysts in various chemical reactions. Gold's resistance to corrosion and its aesthetic appeal make it highly valued.
In simple words: Gold is used for jewelry, as money, to coat other metals, and in new technologies like solar cells, because it is expensive, beautiful, and does not rust.

🎯 Exam Tip: Emphasize gold's high value, corrosion resistance, and excellent electrical conductivity when discussing its applications, as these properties drive its diverse uses from jewelry to electronics.

 

Question 20. Describe the underlying principle of froth flotation process.
Answer: The froth flotation process is a method commonly used to concentrate sulfide ores, such as galena (PbS) and zinc blende (ZnS). The underlying principle is based on the difference in the wetting properties of the ore particles and the gangue (impurities). In this method, metallic ore particles are preferentially wetted by oil, while gangue particles are preferentially wetted by water. When the crushed ore is mixed with water, a frothing agent (like pine oil), and a collector (like sodium ethyl xanthate), and air is blown through the mixture, a froth is formed. The collector molecules attach to the ore particles, making them water-repellent and causing them to cling to the air bubbles. These ore-laden bubbles rise to the surface, forming a froth that is then skimmed off. Meanwhile, the gangue particles, being water-wetted, settle at the bottom. This selective wetting allows the valuable metal particles to float, leaving impurities behind.
In simple words: Froth flotation works by using oil to make the metal ore particles sticky so they float with air bubbles, while the unwanted rock pieces stay in the water and sink, separating them.

🎯 Exam Tip: The core principle of froth flotation is the selective wetting of ore particles by oil and impurities by water, allowing for their separation via froth formation.

 

XI. Three Mark Questions

 

Question 1. Write about gravity separation or hydraulic wash?
Answer: Gravity separation, also known as hydraulic washing, is a method used to concentrate ores based on the difference in specific gravities (densities) between the ore particles and the lighter gangue (impurities). In this process, the finely powdered ore is placed in a trough and washed with a rapidly flowing current of water. Because the metal ore particles are much heavier (have a high specific gravity), they settle down quickly. The lighter gangue particles, on the other hand, are easily carried away and washed off by the flowing water. This simple physical separation technique is commonly used for concentrating native ores like gold, and for oxide ores such as haematite and tinstone. This simple method is effective for ores where the metal and impurities have very different densities.
In simple words: Gravity separation or hydraulic washing uses flowing water to separate heavy metal particles from lighter dirt particles in an ore. The heavy particles sink, and the light ones are washed away.

🎯 Exam Tip: For gravity separation, remember that it's a physical method effective for ores where the density difference between the ore and gangue is significant, using water to separate them.

 

Question 2. What is cyanide leaching?
Answer: Cyanide leaching is a chemical process primarily used for the extraction of precious metals like gold and silver from their low-grade ores. In this method, the crushed ore is treated with a dilute, aerated solution of sodium cyanide (NaCN). The gold metal reacts with the cyanide and oxygen to form a soluble dicyanoaurate(I) complex.
\( \text{4Au(s)} + \text{8CN}^-\text{(aq)} + \text{O}_2\text{(g)} + \text{2H}_2\text{O(l)} \rightarrow \text{4}[\text{Au(CN)}_2]^-\text{(aq)} + \text{4OH}^-\text{(aq)} \)
The gangue, typically alumino silicate impurities, remains insoluble and is filtered off. After leaching, the gold is recovered from the deoxygenated complex solution by reacting it with zinc powder, a process called cementation. The more reactive zinc displaces the gold, reducing it back to its elemental state (zero oxidation state).
\( \text{Zn(s)} + \text{2}[\text{Au(CN)}_2]^-\text{(aq)} \rightarrow [\text{Zn(CN)}_4]^{2-}\text{(aq)} + \text{2Au(s)} \)
Cyanide forms a stable complex with gold, allowing it to dissolve, which is crucial for extraction from low-grade ores.
In simple words: Cyanide leaching is a process where gold is dissolved from its ore using a special liquid containing cyanide. Then, zinc is used to get the pure gold back out of the liquid.

🎯 Exam Tip: In cyanide leaching, remember the two key steps: forming a soluble gold-cyanide complex and then recovering gold by displacement with a more reactive metal like zinc (cementation).

 

Question 3. Write about alkali leaching?
Answer: Alkali leaching is a chemical method of ore concentration where the ore is heated with an aqueous (water-based) alkali solution to form a soluble complex. This process is commonly used for extracting aluminium from bauxite ore. The bauxite is heated with a concentrated solution of sodium hydroxide (or sodium carbonate) at a high temperature (470K - 520K) and pressure (35 atm). This treatment converts the aluminium oxide in bauxite into soluble sodium meta aluminate:
\( \text{Al}_2\text{O}_3\text{(s)} + \text{2NaOH(aq)} + \text{3H}_2\text{O(l)} \rightarrow \text{2Na}[\text{Al(OH)}_4]\text{(aq)} \)
During this process, impurities like iron oxide and titanium oxide do not dissolve and are left behind. The hot, clear solution containing the sodium meta aluminate is then decanted, cooled, and diluted. Finally, carbon dioxide gas is passed through this solution to neutralize it, causing hydrated aluminium oxide (\( \text{Al}_2\text{O}_3 \cdot \text{xH}_2\text{O} \)) to precipitate out:
\( \text{2Na}[\text{Al(OH)}_4]\text{(aq)} + \text{4CO}_2\text{(g)} \rightarrow \text{Al}_2\text{O}_3 \cdot \text{xH}_2\text{O(s)} + \text{2NaHCO}_3\text{(aq)} \)
This precipitate is filtered off and then heated to about 1670K to obtain pure, anhydrous alumina (\( \text{Al}_2\text{O}_3 \)). This method specifically targets aluminum oxide, leaving behind common impurities like iron and titanium oxides.
In simple words: Alkali leaching uses a strong alkaline liquid to dissolve aluminum from its ore, bauxite. The unwanted parts stay behind, and then the dissolved aluminum is processed to get pure aluminum oxide.

🎯 Exam Tip: Remember that alkali leaching is essential for purifying bauxite (aluminium ore) by selectively dissolving aluminium oxide and then reprecipitating it, leaving insoluble impurities behind.

 

Question 4. Write about magnetic separation.
Answer: Magnetic separation is a method used to concentrate ores based on the differences in magnetic properties between the ore and its impurities. This technique is applicable to ferromagnetic ores, which are attracted to a magnet, or for separating magnetic impurities from a non-magnetic ore. For example, non-magnetic tin stone (cassiterite) can be separated from magnetic impurities like wolframite. Similarly, magnetic ores such as chromite and pyrolusite can be separated from non-magnetic siliceous impurities. The process involves pouring crushed ore onto a conveyor belt that moves over two rollers. One of these rollers is magnetic. As the powdered ore reaches the end of the belt, the non-magnetic particles fall straight off. However, the magnetic particles are attracted by the magnetic roller for a longer time, so they follow a curved path and fall closer to the magnetic wheel, into a separate heap. This physical separation is very effective for mixtures containing magnetic and non-magnetic components.
In simple words: Magnetic separation uses a magnet to pull magnetic ore particles away from non-magnetic impurities. Crushed ore moves on a belt, and the magnetic roller separates them into different piles.

🎯 Exam Tip: Remember that magnetic separation is a physical method suitable for ores where the mineral and gangue have different magnetic properties, often used for separating iron-containing impurities.

 

Question 5. Write about calcination.
Answer: Calcination is a thermal treatment process where a concentrated ore is strongly heated in the absence or limited supply of air. The primary purpose of calcination is to remove volatile matter from the ore. During this process, water of crystallization present in hydrated oxide ores (like bauxite) escapes as moisture. Any organic matter present in the ore also gets expelled, making the ore porous. For carbonate ores, calcination leads to the decomposition of carbonates, liberating carbon dioxide gas.
For example, calcium carbonate decomposes into calcium oxide and carbon dioxide:
\( \text{CaCO}_3 \underrightarrow { \triangle } \text{CaO} + \text{CO}_2 \uparrow \)
And hydrated alumina decomposes into anhydrous alumina and water vapor:
\( \text{Al}_2\text{O}_3 \cdot \text{2H}_2\text{O} \underrightarrow { \triangle } \text{Al}_2\text{O}_3\text{(s)} + \text{2H}_2\text{O(g)} \uparrow \)
This process makes the ore porous, which helps in subsequent metallurgical steps. It's important to control the air supply to prevent unwanted oxidation.
In simple words: Calcination is when an ore is heated very strongly without much air. This helps to remove water, gases like carbon dioxide, and other unwanted stuff, making the ore lighter and more porous.

🎯 Exam Tip: The key differentiator for calcination is heating in the absence or limited supply of air, primarily to remove volatile impurities like water and carbon dioxide, and making the ore porous.

 

Question 6. Write about smelting.
Answer: Smelting is a high-temperature process used to extract a metal from its concentrated ore by melting it in a furnace. In this process, the concentrated ore is mixed with two important substances: a flux and a reducing agent. The reducing agent (often carbon, carbon monoxide, or aluminium) reacts with the metal oxide to remove oxygen, yielding the pure metal. For example, iron oxide can be reduced by carbon monoxide:
\( \text{Fe}_2\text{O}_3\text{(s)} + \text{3CO(g)} \rightarrow \text{2Fe(s)} + \text{3CO}_2\text{(g)} \uparrow \)
The flux is a chemical substance added to remove unwanted rocky impurities, called gangue. It reacts with the gangue to form an easily fusible (easily melted) substance called slag. For instance, if acidic silica (\( \text{SiO}_2 \)) is the gangue, a basic flux like lime (\( \text{CaO} \)) is added to form calcium silicate slag:
\( \text{CaO (s)} + \text{SiO}_2\text{ (s)} \rightarrow \text{CaSiO}_3\text{ (s)} \)
The molten slag, being lighter, floats on top of the molten metal, allowing for easy separation. Smelting is crucial for separating the metal from both oxygen and other solid impurities.
In simple words: Smelting is a process where metal ore is melted at high heat with special chemicals to separate the pure metal from its oxygen and rocky impurities, which form a waste product called slag.

🎯 Exam Tip: Smelting is characterized by its use of high temperatures, a reducing agent to remove oxygen, and a flux to form slag with the gangue, enabling efficient metal extraction.

 

Question 7. Write about the Ellingham diagram.
Answer: The Ellingham diagram is a graphical representation that shows how the standard Gibbs free energy change (\( \Delta G^\circ \)) for the formation of various metal oxides varies with temperature (T). It is a crucial tool in metallurgy for understanding the thermodynamic feasibility of reduction processes. The change in Gibbs free energy is related to enthalpy change (\( \Delta H \)) and entropy change (\( \Delta S \)) by the equation: \( \Delta G = \Delta H - T \Delta S \).
For a reaction at equilibrium, \( \Delta G^\circ \) can also be calculated using the equilibrium constant \( K_p \) as \( \Delta G^\circ = -RT \ln K_p \).
Harold Ellingham used this relationship to plot \( \Delta G^\circ \) values against temperature. In the diagram, temperature is plotted on the x-axis and \( \Delta G^\circ \) for oxide formation is plotted on the y-axis. The lines on the diagram generally have \( \Delta S \) as their slope and \( \Delta H \) as their y-intercept. This diagram is a powerful tool for selecting the right reducing agent and temperature for different metal oxides.
In simple words: The Ellingham diagram is a graph that shows how easy or hard it is to make metal oxides at different temperatures. It helps scientists decide the best way to separate metals from their ores.

🎯 Exam Tip: Always recall that the Ellingham diagram plots \( \Delta G^\circ \) versus temperature, making it a thermodynamic predictor for the feasibility of reducing metal oxides, with its slope related to entropy change.

 

Question 8. Write the observations from the Ellingham diagram.
Answer: Several key observations can be made from the Ellingham diagram:
1. For most metal oxides, the line representing their formation has a positive slope. This is because oxygen gas is consumed during the formation of metal oxides, leading to a decrease in randomness (\( \Delta S \) is negative). Since \( \Delta G = \Delta H - T \Delta S \), a negative \( \Delta S \) makes \( -T \Delta S \) positive, causing \( \Delta G \) to become less negative (or more positive) as temperature increases.
2. However, the line for the formation of carbon monoxide (\( \text{2C} + \text{O}_2 \rightarrow \text{2CO} \)) has a negative slope. This occurs because two moles of CO gas are formed by consuming one mole of oxygen gas, leading to an increase in randomness (\( \Delta S \) is positive). This makes \( -T \Delta S \) negative, and thus \( \Delta G \) for CO formation becomes more negative at higher temperatures, indicating carbon monoxide is more stable at high temperatures.
3. As temperature increases, the \( \Delta G \) for the formation of most metal oxides generally becomes less negative and eventually becomes zero at a particular temperature. Below this temperature, \( \Delta G \) is negative, and the oxide is stable. Above this temperature, \( \Delta G \) is positive, and the oxide becomes less stable, making its decomposition easier.
4. Sudden changes in the slope of a line occur at specific temperatures due to phase transitions (like melting or evaporation) of the metal or its oxide. Examples include changes observed for MgO and HgO.
These observations help predict the conditions under which a metal oxide can be easily reduced, a core principle in metallurgy.
In simple words: The Ellingham diagram shows that most metal oxides become less stable at higher temperatures, except for carbon monoxide, which becomes more stable. This helps us know when a metal can be easily extracted.

🎯 Exam Tip: When interpreting Ellingham diagrams, pay attention to the slope (positive means oxide less stable at high T, negative means oxide more stable at high T), crossover points (which indicate favored reducing agents), and sudden changes in slope (phase transitions).

 

Question 9. Which method is used for refining zirconium/titanium?
Answer: The method used for refining zirconium and titanium is the Van Arkel process. This purification technique is based on the principle of thermal decomposition of metal compounds to yield pure metals. Here's how it works for titanium (the process is similar for zirconium):
1. Impure titanium is first heated in an evacuated vessel with iodine at a moderate temperature of about 550K. This reaction forms a volatile (easily evaporated) titanium tetraiodide compound:
\( \text{Ti(s)} + \text{2I}_2\text{(s)} \rightarrow \text{TiI}_4\text{(vapour)} \)
Crucially, any impurities present in the titanium do not react with iodine under these conditions and are left behind as solids.
2. The volatile titanium tetraiodide vapor is then passed over a hot tungsten filament heated to a much higher temperature, around 1800K. At this high temperature, the titanium tetraiodide decomposes, depositing pure titanium metal onto the tungsten filament:
\( \text{TiI}_4\text{(vapour)} \rightarrow \text{Ti(s)} + \text{2I}_2\text{(s)} \)
The iodine that is released is then reused in the process. This two-step process leverages the different volatilities of the metal halide and the pure metal to achieve very high purity, which is especially important for metals used in high-tech applications.
In simple words: The Van Arkel method refines titanium and zirconium. It turns the impure metal into a gas using iodine, then heats the gas to get very pure metal, leaving impurities behind.

🎯 Exam Tip: The Van Arkel process is a specialized refining technique known for producing ultra-pure metals like titanium and zirconium by forming a volatile intermediate compound that is later decomposed.

 

Question 10. Write the applications or uses of aluminium.
Answer: Aluminium is a versatile metal with numerous applications due to its unique properties, such as being lightweight, strong, corrosion-resistant, and a good conductor. Here are some of its key uses:
1. It is used in manufacturing heat exchangers and sinks, which are vital in various cooling systems.
2. Aluminium is widely used for making everyday vessels and utensils because it is light and does not rust easily.
3. It is commonly used to make aluminium foils for packaging food items, keeping them fresh and protected.
4. Aluminium alloys, formed by mixing aluminium with metals like copper, manganese, magnesium, and silicon, are lightweight and strong. These alloys are extensively used in the design of aeroplanes, cars, and other forms of transport to reduce weight and improve fuel efficiency.
5. Due to its high resistance to corrosion, aluminium is used in the design of chemical reactors, medical equipment, refrigeration units, and gas pipelines, where chemical stability is important.
6. It is a good electrical conductor and relatively inexpensive, making it suitable for electrical overhead cables, often with a steel core for added strength.
Aluminium's excellent strength-to-weight ratio makes it indispensable for industries like aerospace.
In simple words: Aluminum is used in airplanes, cooking pots, food packaging, electrical wires, and chemical machines because it is light, strong, does not rust, and conducts heat and electricity well.

🎯 Exam Tip: When discussing aluminium's uses, always highlight its low density, high strength (in alloys), excellent corrosion resistance, and good electrical/thermal conductivity, as these properties are behind its widespread applications.

 

XII. Five Mark Questions

 

Question 12. Out of coke and CO, which is a better reducing agent for the reduction of ZnO? why?
Answer: At higher temperatures, coke (carbon) is generally a better reducing agent than carbon monoxide (CO) for the reduction of zinc oxide (ZnO). This can be understood by looking at the Ellingham diagram. The line representing the formation of carbon monoxide from carbon (\( \text{2C} + \text{O}_2 \rightarrow \text{2CO} \)) has a negative slope, meaning that \( \Delta G^\circ \) for this reaction becomes more negative at higher temperatures. In contrast, the line for the formation of zinc oxide (\( \text{2Zn} + \text{O}_2 \rightarrow \text{2ZnO} \)) has a positive slope.
On the Ellingham diagram, at temperatures above approximately 1200K, the carbon line (for CO formation) lies below the zinc line (for ZnO formation). When the line for the formation of a reducing agent's oxide lies below the metal oxide line, it indicates that the reducing agent can effectively reduce that metal oxide. Therefore, at these high temperatures, carbon is more stable as an oxide (CO) than zinc is as an oxide (ZnO).
The reduction of zinc oxide by carbon at high temperatures proceeds as:
\( \text{ZnO(s)} + \text{C(s)} \rightarrow \text{Zn(s)} + \text{CO(g)} \uparrow \)
During this process, oxygen from ZnO combines with carbon, forming stable carbon monoxide gas and liberating zinc metal. This reaction is thermodynamically favorable because carbon has a stronger affinity for oxygen than zinc at these elevated temperatures. The ability of carbon to form highly stable carbon monoxide or dioxide makes it a very effective reducing agent for many metal oxides.
In simple words: At high temperatures, coke (carbon) is better than carbon monoxide for taking oxygen away from zinc oxide. This is because carbon is very good at grabbing oxygen to form CO at high heat, which is shown by its line being lower than zinc's line on the Ellingham diagram.

🎯 Exam Tip: To compare reducing agents using the Ellingham diagram, find the temperature at which the line for the reducing agent's oxide formation lies below the metal oxide line; that reducing agent will be more effective.

 

Question 1. Explain the froth flotation method.
Answer: The froth flotation method is a widely used process for concentrating sulfide ores, such as galena (\( \text{PbS} \)) and zinc blende (\( \text{ZnS} \)), due to their unique surface properties. The principle behind this method relies on the preferential wetting of ore particles by oil and gangue (unwanted impurities) particles by water.
Here’s a breakdown of the process:
1. **Preparation:** The crude ore is finely crushed and then mixed with water to form a slurry.
2. **Adding Agents:** To this slurry, specific chemicals are added:
- **Frothing agents:** Substances like pine oil or eucalyptus oil are added to create a stable froth (foam).
- **Collectors:** A small amount of chemicals like sodium ethyl xanthate is added. These molecules attach to the metallic ore particles, making their surfaces water-repellent (hydrophobic) and oil-attracting (oleophilic).
3. **Froth Formation:** Air is blown vigorously through the mixture, usually with a mechanical stirrer. The hydrophobic ore particles, now coated with oil, selectively cling to the rising air bubbles.
4. **Separation:** These ore-laden air bubbles rise to the surface, forming a stable froth layer. This froth, rich in concentrated ore, is then skimmed off.
5. **Tailings:** The gangue particles, being preferentially wetted by water (hydrophilic), sink to the bottom of the tank and are discharged as tailings.
6. **Depressing Agents:** If the sulfide ore contains other metal sulfides as impurities (e.g., \( \text{ZnS} \) in galena \( \text{PbS} \)), depressing agents like sodium cyanide (\( \text{NaCN} \)) or sodium carbonate are used. These agents selectively prevent certain sulfide impurities from forming froth. For instance, \( \text{NaCN} \) depresses the flotation property of \( \text{ZnS} \) by forming a soluble zinc complex \( \text{Na}_2[\text{Zn}(\text{CN})_4] \) on its surface, preventing it from attaching to air bubbles, while \( \text{PbS} \) continues to float.
This method is highly effective because it leverages the surface properties of minerals to achieve a clear separation between the valuable ore and gangue.
In simple words: Froth flotation separates metal ores from dirt by mixing them with water, oil, and air. The metal pieces stick to the oil and float as foam, while the dirt sinks. Special chemicals can also stop certain unwanted metals from floating.

🎯 Exam Tip: For froth flotation, ensure you mention the key components: frothing agent (pine oil), collector (xanthates), air, and the principle of preferential wetting of ore by oil.

 

Question 2. How is copper extracted from its ore.
Answer: Copper is primarily extracted from its principal ore, copper pyrites (\( \text{CuFeS}_2 \)), through a multi-stage process involving concentration, roasting, smelting, converting, and refining.
**1. Concentration:** The copper pyrites ore is first concentrated using the froth flotation method, which separates the valuable sulfide ore from the gangue.
**2. Roasting:** The concentrated ore is then heated strongly in a reverberatory furnace with an acidic flux, typically silica (\( \text{SiO}_2 \)), in the presence of air. During roasting, copper pyrites partially oxidizes:
\( \text{2CuFeS}_2\text{ (s)} + \text{O}_2\text{ (g)} \rightarrow \text{2FeS (l)} + \text{Cu}_2\text{S (l)} + \text{SO}_2\text{ (g)} \)
And some ferrous sulfide further reacts with oxygen:
\( \text{2FeS (l)} + \text{3O}_2\text{ (g)} \rightarrow \text{2FeO (l)} + \text{2SO}_2\text{ (g)} \)
**3. Smelting to Matte:** The ferrous oxide (\( \text{FeO} \)) formed is basic and reacts with the acidic silica flux to create an easily removable slag called ferrous silicate:
\( \text{FeO (s)} + \text{SiO}_2\text{ (s)} \rightarrow \text{FeSiO}_3\text{ (s)} \)
The remaining mutually soluble metal sulfides, \( \text{Cu}_2\text{S} \) and \( \text{FeS} \), form a molten mixture called copper matte. This matte is then separated from the slag.
**4. Converting to Blister Copper:** The copper matte, rich in \( \text{Cu}_2\text{S} \) and \( \text{FeS} \), is transferred to a converting furnace. Here, more air is blown through, which oxidizes any remaining \( \text{FeS} \) to \( \text{FeO} \), which again forms slag with added silica. The remaining copper sulfide is oxidized to cuprous oxide (\( \text{Cu}_2\text{O} \)):
\( \text{2Cu}_2\text{S(l,s)} + \text{3O}_2\text{(g)} \rightarrow \text{2Cu}_2\text{O(l,s)} + \text{2SO}_2\text{(g)} \)
Then, the cuprous oxide reacts with the remaining copper sulfide to produce metallic copper:
\( \text{2Cu}_2\text{O(l)} + \text{Cu}_2\text{S(l)} \rightarrow \text{6Cu(l)} + \text{SO}_2\text{(g)} \)
The \( \text{SO}_2 \) gas liberated creates blisters on the surface of the solidifying copper, hence it's called blister copper.
**5. Electrorefining:** Blister copper is about 98% pure and needs further refining. This is done by electrorefining, where:
- **Cathode:** A thin strip of pure copper.
- **Anode:** Impure blister copper.
- **Electrolyte:** An aqueous solution of copper sulfate (\( \text{CuSO}_4 \)) with dilute sulfuric acid (\( \text{H}_2\text{SO}_4 \)).
When electric current passes, copper from the impure anode dissolves into the electrolyte, and an equivalent amount of pure copper is deposited onto the cathode. More reactive metals in the anode dissolve but do not deposit on the cathode, while less reactive impurities settle as anode mud. Copper extraction is a multi-step process because copper is often found in complex sulfide ores, requiring several treatments to achieve high purity.
In simple words: Copper is taken from its ore in many steps: first, the ore is concentrated, then heated to remove sulfur and iron, which makes an impure copper called blister copper. Finally, electricity is used to make it very pure.

🎯 Exam Tip: When explaining copper extraction, detail each stage: concentration (froth flotation), roasting (sulfide to oxide), smelting (matte formation, slag removal), converting (blister copper), and electrorefining (pure copper), noting the purpose of each step and key reactions.

 

Question 3. Explain the thermodynamic principle of metallurgy.
Answer: Metals can be taken out from their ores using various reducing chemicals. This process involves the reduction of metal oxides. For example, if we consider a metal oxide like \( \text{M}_x\text{O}_y \), it can be reduced to pure metal \( \text{M} \) and oxygen gas, represented as \( \text{M}_x\text{O}_y\text{(s)} \rightarrow \frac{2}{Y}\text{M(s)} + \text{O}_2\text{(g)} \). This reduction process can often be carried out using carbon. The goal is to separate the metal from oxygen, often needing energy to break the bonds.
In simple words: Metallurgy uses chemicals to remove metals from their ores. These chemicals, called reducing agents, take oxygen away from metal oxides to give us pure metals.

🎯 Exam Tip: When explaining thermodynamic principles, clearly state the role of reducing agents and how they facilitate the conversion of metal oxides into pure metals, often by forming more stable oxides themselves.

 

Question 4. Write the applications of Ellingham diagram.
Answer:

• The Ellingham diagram helps us pick the right chemical to remove oxygen from a metal oxide and find the best temperature for it.
• Taking oxygen away from a metal oxide is like a competition between the metal and the reducing agent, both wanting to combine with oxygen.
• If a metal oxide is very stable, the oxygen will tend to stay with the metal.
• However, if the reducing agent forms a more stable oxide, it will take the oxygen from the metal oxide.
• This diagram also shows how stable different metal oxides are at various temperatures. Understanding these stability differences helps chemists design efficient ways to purify metals from their natural compounds.
• For compounds like silver oxide (Ag\( _2 \)O) and mercury oxide (HgO), their lines are at the top of the Ellingham diagram. This means they are not very stable and easily break down into metal and oxygen when heated, even without a reducing agent.

In simple words: The Ellingham diagram is a map for chemists. It shows which chemicals can pull oxygen from metal oxides and at what heat. It helps find the best way to get pure metals by understanding how strong the oxygen bond is for different metal oxides.

🎯 Exam Tip: Focus on how the Ellingham diagram guides the selection of reducing agents and optimal temperatures for metallurgical processes, highlighting the stability of oxides and their decomposition points.