RBSE Solutions Class 12 Chemistry Chapter 6 Principles and Processes of Isolation of E

Get the most accurate RBSE Solutions for Class 12 Chemistry Chapter 6 Principles and Processes of Isolation of E here. Updated for the 2026-27 academic session, these solutions are based on the latest RBSE textbooks for Class 12 Chemistry. Our expert-created answers for Class 12 Chemistry are available for free download in PDF format.

Detailed Chapter 6 Principles and Processes of Isolation of E RBSE Solutions for Class 12 Chemistry

For Class 12 students, solving RBSE textbook questions is the most effective way to build a strong conceptual foundation. Our Class 12 Chemistry solutions follow a detailed, step-by-step approach to ensure you understand the logic behind every answer. Practicing these Chapter 6 Principles and Processes of Isolation of E solutions will improve your exam performance.

Class 12 Chemistry Chapter 6 Principles and Processes of Isolation of E RBSE Solutions PDF

 

Question 1. The formula of Haematite is
(a) \( \text{Fe}_2\text{O}_4 \)
(b) \( \text{Fe}_2\text{O}_3 \)
(c) \( \text{FeS}_2 \)
(d) FeO
Answer: (b) \( \text{Fe}_2\text{O}_3 \)
In simple words: Haematite is a common ore of iron, and its chemical name is ferric oxide. This compound is a major source for extracting iron metal.

🎯 Exam Tip: Remember the chemical formulas for common iron ores like magnetite (\( \text{Fe}_3\text{O}_4 \)) and haematite (\( \text{Fe}_2\text{O}_3 \)).

 

Question 2. Which metal is extracted from its ore by electrolytic method
(a) Ag metal
(b) Pb metal
(c) Cu metal
(d) Fe metal
Answer: (c) Cu metal
In simple words: While copper is often extracted by other methods, electrolytic refining is a crucial step to get very pure copper from its ore concentrates. This process is highly effective for purification.

🎯 Exam Tip: Electrolytic methods are typically used for extracting highly reactive metals like aluminium, sodium, and magnesium, or for refining less reactive metals like copper and silver to very high purity.

 

Question 4. The most abundant metal in the earth's crust is
(a) Fe
(b) Al
(c) Ca
(d) Na
Answer: (b) Al
In simple words: Aluminium is found in large amounts in the Earth's crust, making it the most common metal. It is lighter than iron and very useful for many things.

🎯 Exam Tip: Know the order of elemental abundance in the Earth's crust: Oxygen, Silicon, Aluminium, Iron, Calcium, Sodium, Potassium, Magnesium.

 

Question 5. Fool's gold is
(a) \( \text{As}_2\text{O}_3 \)
(b) \( \text{Sb}_2\text{S}_3 \)
(c) \( \text{FeS}_2 \)
(d) Alloy of Cu - Zn
Answer: (c) \( \text{FeS}_2 \)
In simple words: Fool's gold is another name for iron pyrite, a mineral that looks similar to gold. It is actually iron sulfide.

🎯 Exam Tip: Iron pyrite \( (\text{FeS}_2) \) has a shiny, metallic yellowish appearance, which often causes it to be mistaken for gold. Its crystal structure is cubic.

 

Question 6. Cryolite is an ore of
(a) Fe
(b) Al
(c) Cu
(d) Ag
Answer: (b) Al
In simple words: Cryolite is a mineral that helps in extracting aluminium. Its chemical formula is \( \text{Na}_3\text{AlF}_6 \).

🎯 Exam Tip: Cryolite is not an ore of aluminium itself, but it acts as a solvent for alumina \( (\text{Al}_2\text{O}_3) \) in the Hall-Heroult process, making aluminium extraction possible at lower temperatures.

 

Question 7. Which of the following is not an ore of Aluminium
(a) Bauxite
(b) Corrundum
(c) Diaspore
(d) Azurite
Answer: (d) Azurite
In simple words: Azurite is a copper mineral, not an aluminium ore. Bauxite, corundum, and diaspore are all important ores from which aluminium is extracted.

🎯 Exam Tip: Familiarize yourself with the common ores of important metals and their chemical compositions.

 

Question 9. Magnetite is an ore of
(a) Mg
(b) Ag
(c) Mn
(d) Fe
Answer: (d) Fe
In simple words: Magnetite is an important ore that contains iron. It is known for its magnetic properties.

🎯 Exam Tip: Magnetite (\( \text{Fe}_3\text{O}_4 \)) is one of the richest iron ores, characterized by its strong magnetic properties.

 

Question 10. Froth floatation process is used to concentrate which type of ores ?
(a) Sulphide ores
(b) Phosphate ores
(c) Silicate ores
(d) Oxide ores
Answer: (a) Sulphide ores
In simple words: This method works best for sulphide ores because it separates the ore from impurities based on how well they get wet.

🎯 Exam Tip: Froth floatation relies on the principle of differential wetting, where sulphide ore particles are selectively wetted by oil and float, while gangue particles are wetted by water and sink.

 

Question 11. Gravity separation method is used to concentrate which type of ores ?
(a) Sulphide ores
(b) Phosphate ores
(c) Silicate ores
(d) Oxide ores
Answer: (d) Oxide ores
In simple words: Gravity separation is good for heavy ores, like many oxide ores, because it uses their weight difference to separate them from lighter impurities.

🎯 Exam Tip: Gravity separation methods like hydraulic washing are effective for concentrating heavy oxide and carbonate ores where there is a significant density difference between the ore and the gangue.

 

Question 12. Magnetic separation method is used for
(a) separating magnetic ores or gangue.
(b) separating coloured ores or gangue.
(c) separating colourless ores or gangue.
(d) none of the options.
Answer: (a) separating magnetic ores or gangue.
In simple words: This method uses magnets to pull apart materials where either the ore or the impurities have magnetic properties.

🎯 Exam Tip: Magnetic separation is specifically used when one of the components (either the ore or the gangue) is magnetic, such as in the concentration of magnetite iron ore.

 

Question 13. Iron can be extracted from its oxide ores by which method ?
(a) Carbon reduction method
(b) Electrolytic method
Answer: (a) Carbon reduction method
In simple words: Iron is typically taken out of its oxide ores by heating it with carbon, like coke, in a large furnace. The carbon helps remove the oxygen from the iron oxide.

🎯 Exam Tip: Carbon reduction is a common method for extracting less reactive metals like iron, zinc, and lead from their oxide ores, typically carried out in a blast furnace.

 

Question 15. Those natural occurring components from which element can be extracted profitably and conveniently known as
(a) minerals
(b) gangue
(c) ores
(d) flux
Answer: (c) ores
In simple words: Ores are specific types of minerals from which we can get metals easily and at a good cost. Not all minerals are ores.

🎯 Exam Tip: The key distinction between a mineral and an ore is economic viability: an ore must allow for profitable and convenient extraction of a metal.

 

Question 16. Which type of furnace can create high temperature?
(a) Blast furnace
(b) Reverberatory furnace
(c) Electric furnace
(d) Muffle furnace
Answer: (a) Blast furnace
In simple words: A blast furnace is designed to reach very high temperatures, especially in its lower zones, which is needed for extracting iron.

🎯 Exam Tip: Blast furnaces operate at extremely high temperatures (up to 2100K) required for the reduction of iron oxides, while electric furnaces can achieve even higher, more controlled temperatures for specialized applications.

 

Question 17. In blast furnace, iron oxide is reduced by
(a) \( \text{SiO}_2 \)
(b) CO
(c) C
(d) \( \text{CaCO}_3 \)
Answer: (b) CO
In simple words: In the iron-making blast furnace, carbon monoxide gas does most of the work to take oxygen away from the iron oxide, turning it into iron metal.

🎯 Exam Tip: While coke (carbon) is used in the blast furnace, carbon monoxide (CO) gas is the primary reducing agent, especially in the upper and middle zones where temperatures are lower.

 

Question 18. When limestone is heated then \( \text{CO}_2 \) gas is released. This process is known as in the metallurgy of metals
(a) Smelting
(b) Roasting
(c) Calcination
(d) Concentration
Answer: (c) Calcination
In simple words: Heating limestone to release carbon dioxide is called calcination. This process prepares ores by removing volatile substances.

🎯 Exam Tip: Calcination involves heating a carbonate or hydroxide ore strongly in the absence or limited supply of air to remove volatile impurities like \( \text{CO}_2 \) or water, converting the ore into a porous oxide.

 

Question 20. The metal which always exists in free or native state is
(a) Au
(b) Ag
(c) Cu
(d) Na
Answer: (a) Au
In simple words: Gold is a very unreactive metal, so it is often found in nature as pure gold, not combined with other elements.

🎯 Exam Tip: Noble metals like gold, platinum, and sometimes silver are found in their native state because they are highly unreactive and resist oxidation and corrosion.

 

Question 21. A mineral is called an ore if
(a) The metal present in the mineral is costly.
(b) A metal can be extracted profitably from it.
(c) A metal can be extracted from it.
(d) A metal cannot be extracted from it.
Answer: (b) A metal can be extracted profitably from it.
In simple words: For a mineral to be called an ore, it must be possible to get the metal from it both easily and affordably.

🎯 Exam Tip: While all ores are minerals, only those minerals from which metals can be extracted economically and conveniently are classified as ores.

 

Question 22. During smelting, a compound is added, which combines with gangue and converted it into fusible product. This compound is known as
(a) Mud
(b) Slag
(c) Flux
(d) Matrix
Answer: (c) Flux
In simple words: During smelting, a special substance called a flux is added to react with unwanted impurities (gangue) and turn them into an easy-to-melt product called slag.

🎯 Exam Tip: Fluxes are essential in smelting to remove gangue by forming a fusible slag that can be easily separated from the molten metal.

 

Question 23. The method of extraction of metal from its ore is known as
(a) Purification
(b) Concentration
(c) Calcination
(d) Metallurgy
Answer: (d) Metallurgy
In simple words: Metallurgy is the entire science and process of getting metals from their ores, cleaning them, and preparing them for use.

🎯 Exam Tip: Metallurgy encompasses all steps from ore concentration to metal refining and alloying, forming the complete sequence of metal production.

 

Question 25. Roasting is done in
(a) Oxide ore
(b) Silicate
(c) Sulphide ore
(d) Carbonate ore
Answer: (c) Sulphide ore
In simple words: Roasting is a process where sulphide ores are strongly heated in the presence of air to change them into oxides.

🎯 Exam Tip: Roasting is specifically applied to sulphide ores to convert them into oxides, which are then easier to reduce to the metal.

 

RBSE Class 12 Chemistry Chapter 6 Very Short Answer Type Questions

 

Question 1. Write name and chemical formula of oxide ore of aluminium and iron.
Answer: Aluminium's oxide ore is Bauxite, with the formula \( \text{Al}_2\text{O}_3\cdot2\text{H}_2\text{O} \). Iron's oxide ore is Haematite, with the formula \( \text{Fe}_2\text{O}_3 \). These ores are crucial sources for extracting these metals.
In simple words: For aluminium, the oxide ore is Bauxite \( (\text{Al}_2\text{O}_3\cdot2\text{H}_2\text{O}) \). For iron, it's Haematite \( (\text{Fe}_2\text{O}_3) \).

🎯 Exam Tip: Always include both the name and the correct chemical formula when asked to describe ores, as both are important for full marks.

 

Question 2. What is slag? Explain with example.
Answer: Slag is a melted waste product formed when unwanted impurities, called gangue, combine with a chemical substance called flux during the metal extraction process. For example, if gangue is silica \( (\text{SiO}_2) \) and flux is limestone \( (\text{CaCO}_3) \), they react to form calcium silicate \( (\text{CaSiO}_3) \), which is the slag. Slag is lighter than molten metal and can be easily removed.
In simple words: Slag is the melted waste formed when flux reacts with impurities (gangue) in an ore. An example is calcium silicate.

🎯 Exam Tip: Clearly define slag and provide a specific chemical example to illustrate its formation from flux and gangue.

 

Question 3. Write name and chemical formula of sulphide and oxide ore of copper.
Answer: Copper has a sulphide ore called Copper Pyrites, with the chemical formula \( \text{CuFeS}_2 \). Its oxide ore is Cuprite, which has the formula \( \text{Cu}_2\text{O} \). These ores are important for getting copper metal.
In simple words: Copper Pyrites \( (\text{CuFeS}_2) \) is a sulphide ore of copper, and Cuprite \( (\text{Cu}_2\text{O}) \) is an oxide ore.

🎯 Exam Tip: Ensure you remember both the common name and the correct chemical formula for key ores, especially for transition metals like copper.

 

Question 4. Write name of any two metals which are found in native state.
Answer: Two metals commonly found in their native, or free, state in nature are Silver (Ag) and Gold (Au). These metals are very unreactive, which is why they do not easily combine with other elements.
In simple words: Silver (Ag) and Gold (Au) are metals often found in nature in their pure form.

🎯 Exam Tip: Metals found in the native state are typically noble metals, which are very unreactive and resistant to oxidation.

 

Question 5. Write the name of most abundant metal in earth's crust.
Answer: The most abundant metal found in the Earth's crust is Aluminium (Al). It is widely used in many industries due to its light weight and strength.
In simple words: Aluminium (Al) is the most common metal found in the Earth's crust.

🎯 Exam Tip: Remember that while oxygen and silicon are more abundant elements, aluminium is the most abundant *metal* in the Earth's crust.

 

Question 7. What is difference between mineral and ore. Explain it.
Answer: A mineral is any natural chemical substance found in the Earth's crust that contains metals. An ore, however, is a special type of mineral from which a metal can be taken out easily and affordably. All ores are minerals, but not all minerals are ores because it might be too expensive or difficult to get the metal from them.
In simple words: A mineral is a natural substance with metal. An ore is a mineral from which we can get metal easily and profitably.

🎯 Exam Tip: The key difference is that an ore allows for *profitable and convenient* extraction, while a mineral simply *contains* the metal.

 

Question 8. How much carbon % is found in cast iron and pig iron ?
Answer: Cast iron contains about 4% carbon. Pig iron, which is an intermediate product, has approximately 3% carbon. These carbon percentages greatly influence the properties and uses of these iron forms.
In simple words: Cast iron has about 4% carbon, and pig iron has roughly 3% carbon.

🎯 Exam Tip: Knowing the carbon content helps differentiate various forms of iron like pig iron, cast iron, and steel, each having different properties and applications.

 

Question 9. Write composition of german silver.
Answer: German silver is an alloy made of copper (Cu), zinc (Zn), and nickel (Ni). It typically contains 25-50% copper, 25-35% zinc, and 10-35% nickel. Despite its name, German silver does not contain any silver.
In simple words: German silver is an alloy made of copper (25-50%), zinc (25-35%), and nickel (10-35%).

🎯 Exam Tip: Highlight the fact that German silver contains no actual silver, which is a common misconception.

 

Question 10. What is anode mud ?
Answer: Anode mud is a collection of less reactive impurities, like zinc, nickel, and iron, that settle down at the bottom near the anode during the electrolytic refining process of metals. These impurities do not dissolve or oxidize into the electrolyte.
In simple words: Anode mud is the sludge of unreactive impurities that collects below the anode during electrolytic refining.

🎯 Exam Tip: Anode mud often contains valuable noble metals like silver, gold, and platinum, making its collection economically important.

 

Question 11. Write name and role of collectors and froth stabilizers in froth floatation method.
Answer:
(i) **Collectors** (also called floatation agents) are chemicals that make the sulphide ore particles water-repelling. This allows the ore particles to stick to air bubbles and float to the surface. Sodium ethyl xanthate is a common example of a collector.
(ii) **Froth stabilizers:** These substances help make the foam strong and stable, so the ore particles stay floating on top for long enough to be collected. Without stabilizers, the froth would break down too quickly. For instance, p-cresol and aniline are often used as froth stabilizers. These compounds ensure the froth doesn't collapse too quickly, allowing for effective separation.
In simple words: Collectors make ore particles water-repelling so they float, like sodium ethyl xanthate. Froth stabilizers (e.g., cresol, aniline) keep the foam steady.

🎯 Exam Tip: Clearly state the function of each component and provide specific examples to show understanding of their roles in froth floatation.

 

Question 12. Write the name of two impurites present in bauxite ore.
Answer: Two common impurities found in bauxite ore are iron oxide \( (\text{Fe}_2\text{O}_3) \) and silica \( (\text{SiO}_2) \). These impurities must be removed before pure aluminium can be extracted.
In simple words: Bauxite ore often contains iron oxide \( (\text{Fe}_2\text{O}_3) \) and silica \( (\text{SiO}_2) \) as impurities.

🎯 Exam Tip: In the Bayer's process for alumina extraction, both \( \text{Fe}_2\text{O}_3 \) and \( \text{SiO}_2 \) need to be removed from bauxite as they interfere with the process.

 

Question 13. Write the chemical reaction of Mond's process for refining of nickel.
Answer: Mond's process purifies nickel through two main reactions. First, impure nickel reacts with carbon monoxide at 330-350K to form a volatile nickel tetracarbonyl complex. Then, this complex is heated to a higher temperature (450-470K) where it breaks down, releasing pure nickel metal and carbon monoxide gas. This method cleverly uses the different stabilities of the nickel carbonyl compound at various temperatures.
\( \text{Ni (impure)} + 4\text{CO} \xrightarrow{ 330-350\text{K} } \text{Ni(CO)}_4 \)
\( \text{Ni(CO)}_4 \xrightarrow{ 450-470\text{K} } \text{Ni (pure)} + 4\text{CO} \)
In simple words: Impure nickel reacts with carbon monoxide to form a gas, nickel tetracarbonyl. This gas is then heated to get pure nickel.

🎯 Exam Tip: Emphasize the two-step temperature-dependent nature of Mond's process, where the formation and decomposition of the volatile carbonyl are key.

 

Question 14. Which complex ions are used for electroplating of silver and gold ?
Answer: For electroplating silver, the complex ion used is dicyanoargentate(I), \( [\text{Ag(CN)}_2]^- \). For gold electroplating, the corresponding complex ion is dicyanoaurate(I), \( [\text{Au(CN)}_2]^- \). Using these complex ions helps in getting a smooth and even layer of the metal during plating.
In simple words: \( [\text{Ag(CN)}_2]^- \) is used for silver plating, and \( [\text{Au(CN)}_2]^- \) is used for gold plating.

🎯 Exam Tip: Using complex ions in electroplating ensures a steady supply of metal ions and helps in achieving a uniform, bright, and adherent coating.

 

Question 15. What is the role of depressants in froth floatation method?
Answer: Depressants are special chemicals used in the froth flotation method to prevent certain minerals from forming froth. They work by selectively interacting with specific ore particles, stopping them from floating and allowing them to sink. For instance, sodium cyanide (NaCN) and sodium carbonate \( (\text{Na}_2\text{CO}_3) \) can be used to separate lead sulfide from zinc sulfide.
In simple words: Depressants stop certain unwanted ore particles from floating, helping to separate different minerals.

🎯 Exam Tip: Depressants are crucial for selective separation in froth floatation, particularly when dealing with mixtures of sulphide ores.

 

Question 16. Gemstones are an impure form of which metal?
Answer: Many gemstones are impure forms of Aluminium, such as ruby and sapphire, which are primarily aluminium oxide with small amounts of other metal ions causing their color. For example, chromium impurities give ruby its red color.
In simple words: Many gemstones, like ruby and sapphire, are impure forms of aluminium oxide.

🎯 Exam Tip: Gemstones are often formed from common metal oxides with trace impurities that give them their characteristic colors.

 

RBSE Class 12 Chemistry Chapter 6 Short Answer Types Questions

 

Question 17. Anode and cathode are made up of which metal in electrolytic refining of metal ?
Answer: In the electrolytic refining of copper, the anode is made of impure copper, and the cathode is made of a thin sheet of pure copper. This setup helps to purify the crude copper.
In simple words: For copper refining, the anode is impure copper, and the cathode is pure copper.

🎯 Exam Tip: In electrolytic refining, the impure metal always forms the anode, and a thin strip of the pure metal acts as the cathode.

 

Question 18. Write chemical reaction of reduction of metal chromium oxide in alumino thermite process.
Answer: In the alumino-thermite process, chromium oxide \( (\text{Cr}_2\text{O}_3) \) is reduced by aluminium (Al) to produce chromium metal. Aluminium is highly reactive and displaces chromium from its oxide. The chemical reaction is:
\( \text{Cr}_2\text{O}_3 + 2\text{Al} \xrightarrow{ \text{Heat} } 2\text{Cr} + \text{Al}_2\text{O}_3 \)
This reaction releases a lot of heat.
In simple words: Chromium oxide is reduced by aluminium in the alumino-thermite process, making pure chromium metal and aluminium oxide.

🎯 Exam Tip: The alumino-thermite process is highly exothermic, meaning it produces a lot of heat, which is often utilized in welding applications.

 

Question 19. Write name and formula of example of each acidic and basic flux.
Answer: An example of an acidic flux is silica \( (\text{SiO}_2) \), which reacts with basic impurities. An example of a basic flux is calcium oxide (CaO), which reacts with acidic impurities. Fluxes are used to remove unwanted gangue by forming fusible slag.
In simple words: Silica \( (\text{SiO}_2) \) is an acidic flux, and calcium oxide (CaO) is a basic flux.

🎯 Exam Tip: Choose a flux that is opposite in chemical nature to the gangue; an acidic gangue needs a basic flux, and vice versa.

 

Question 20. What is the significance of leaching in the extraction of Al metal?
Answer: Leaching is a very important step in extracting aluminium from bauxite ore. It helps remove impurities such as silica \( (\text{SiO}_2) \), iron oxide \( (\text{Fe}_2\text{O}_3) \), and titanium dioxide \( (\text{TiO}_2) \). In this process, the bauxite is treated with a chemical that dissolves only the aluminium oxide, leaving the impurities behind. This makes the subsequent extraction steps much more efficient.
In simple words: Leaching helps clean bauxite ore by dissolving aluminium oxide and leaving behind impurities like silica and iron oxide, making aluminium extraction easier.

🎯 Exam Tip: Leaching enhances the purity of the ore concentrate, which significantly improves the efficiency and economics of the subsequent metallurgical processes.

 

Question 21. Explain roasting and calcination with example.
Answer:
**Roasting:** This is a process where an ore is heated strongly in the presence of extra air. It is mainly used for sulfide ores to change them into oxides. During roasting, impurities like sulfur, phosphorus, and arsenic are removed as gases (volatile oxides). For instance, sulfur reacts with oxygen to form sulfur dioxide gas:
\( \text{S} + \text{O}_2 \xrightarrow{ \text{Heat} } \text{SO}_2\text{(g)} \)

**Calcination:** This process involves heating an ore strongly in the absence or with a very limited supply of air. It is usually done for carbonate and hydroxide ores to break them down and remove volatile components like carbon dioxide or water. For example, limestone \( (\text{CaCO}_3) \) is calcined to produce quicklime \( (\text{CaO}) \) and carbon dioxide gas. This process helps to convert ores into a more suitable form for further reduction.
In simple words: Roasting heats ore in air to remove impurities and form oxides (e.g., \( \text{S} + \text{O}_2 \rightarrow \text{SO}_2 \)). Calcination heats ore without air to remove volatile parts (e.g., \( \text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2 \)).

🎯 Exam Tip: The main difference lies in the presence or absence of air during heating, which dictates the type of ore treated and the chemical changes occurring.

 

Question 22. Draw labelled figure of zone refining process. Where this process is widely used?
Answer: Zone refining is a purification method used to get extremely pure metals, especially for semiconductors like silicon (Si) and germanium (Ge). The diagram shows a circular heater moving along an impure metal rod. As the heater moves, it melts a small "molten zone" in the rod. Impurities are more soluble in the liquid (molten) metal than in the solid metal. So, as the heater slowly moves, the impurities get pushed into the molten zone and move along with it towards one end of the rod. The pure metal then solidifies behind the molten zone. After several passes, the impurities are concentrated at one end, which can then be cut off, leaving behind a highly pure metal rod.
Recrystallised pure metal Rod of Impure metal Molten zone in which impurities are present Circular Heater
In simple words: Zone refining purifies metals by moving a molten zone along a rod, pushing impurities to one end because they dissolve better in liquid metal. This process is important for making very pure semiconductors like silicon.

🎯 Exam Tip: The fundamental principle of zone refining is the differential solubility of impurities in the molten and solid phases of the metal.

 

Question 23. Draw labelled diagram of electrolytic cell for extraction of aluminium and write complete reaction of it.
Answer: The Hall-Heroult process is used for extracting pure aluminium from alumina \( (\text{Al}_2\text{O}_3) \).
**Diagram:**
Steel Tank Carbon Lining (Cathode) - Molten Alumina + Cryolite Electrolyte Graphite Anodes + Molten Aluminium Tap for Al
**Reactions in Hall-Heroult Process:**
The Hall-Heroult process involves two main ideas for the reactions:
1. **Direct ionization of alumina:** Aluminium oxide \( (\text{Al}_2\text{O}_3) \) ionizes in the molten cryolite to produce \( \text{Al}^{3+} \) and \( \text{O}^{2-} \) ions.
\( \text{Al}_2\text{O}_3 \xrightarrow{ \text{Electrolysis} } 2\text{Al}^{3+} + 3\text{O}^{2-} \)
    * **At Cathode (Negative electrode):** Aluminium ions gain electrons and turn into molten aluminium metal.
        \( \text{Al}^{3+} + 3\text{e}^- \rightarrow \text{Al(l)} \)
    * **At Anode (Positive electrode):** Oxide ions lose electrons to form oxygen gas. This oxygen then reacts with the carbon anodes to form carbon monoxide and carbon dioxide, causing the anodes to slowly wear away.
        \( \text{O}^{2-} \rightarrow \frac{1}{2}\text{O}_2 + 2\text{e}^- \)
        \( \text{C(s)} + \text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} \)
        \( \text{C(s)} + \text{CO}_2\text{(g)} \rightarrow 2\text{CO(g)} \)
    The anodes need regular replacement because they are consumed.
2. **Ionization of cryolite:** In this view, cryolite \( (\text{Na}_3\text{AlF}_6) \) breaks down first, forming \( \text{AlF}_3 \) which then ionizes to provide \( \text{Al}^{3+} \) ions. Fluoride ions are also produced, which then react at the anode. However, the direct reaction of oxide ions with carbon at the anode is the most widely accepted mechanism for anode consumption.

The overall reaction is:
\( 2\text{Al}_2\text{O}_3 + 3\text{C} \rightarrow 4\text{Al} + 3\text{CO}_2 \)
Molten aluminium collects at the bottom of the cell and is periodically tapped out. This method produces aluminium metal with about 99.5% purity. This process is highly energy-intensive, making aluminium production expensive.
In simple words: The Hall-Heroult process uses an electrolytic cell to get aluminium from alumina. Aluminium ions go to the cathode to become liquid aluminium, while oxide ions react with the carbon anodes to form carbon oxides. This process uses a lot of electricity.

🎯 Exam Tip: Clearly label all parts of the diagram and specify the reactions at both the anode and cathode, as well as the overall reaction, for a complete answer.

 

Question 24. Explain with equation how does copper purify by electrolytic method ?
Answer: Copper is purified using an electrolytic refining method to get very pure copper. In this setup, an impure block of copper is used as the anode (positive electrode), and a thin strip of pure copper acts as the cathode (negative electrode). The electrolyte is a solution of copper sulfate \( (\text{CuSO}_4) \) mixed with dilute sulfuric acid.
When electricity passes through:
* **At the anode (impure copper):** Copper from the impure anode oxidizes and enters the solution as \( \text{Cu}^{2+} \) ions. More reactive metals present as impurities (like zinc or iron) also oxidize and dissolve into the electrolyte. Less reactive metals (like silver, gold, or platinum) do not oxidize and fall off to collect as "anode mud" at the bottom.
    \( \text{Cu(s) (impure)} \rightarrow \text{Cu}^{2+}\text{(aq)} + 2\text{e}^- \)
* **At the cathode (pure copper):** Only the \( \text{Cu}^{2+} \) ions from the electrolyte are attracted to the pure copper cathode. They gain electrons and deposit as pure copper metal.
    \( \text{Cu}^{2+}\text{(aq)} + 2\text{e}^- \rightarrow \text{Cu(s) (pure)} \)
This process continuously deposits pure copper at the cathode while impurities are either dissolved or collected as anode mud.
In simple words: Impure copper becomes the anode, pure copper the cathode. When current passes, impure copper dissolves, and pure copper deposits on the cathode. Less reactive impurities collect as anode mud.

🎯 Exam Tip: Clearly distinguish between the fate of more reactive and less reactive impurities at the anode during electrolytic refining.

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RBSE Solutions Class 12 Chemistry Chapter 6 Principles and Processes of Isolation of E

Students can now access the RBSE Solutions for Chapter 6 Principles and Processes of Isolation of E prepared by teachers on our website. These solutions cover all questions in exercise in your Class 12 Chemistry textbook. Each answer is updated based on the current academic session as per the latest RBSE syllabus.

Detailed Explanations for Chapter 6 Principles and Processes of Isolation of E

Our expert teachers have provided step-by-step explanations for all the difficult questions in the Class 12 Chemistry chapter. Along with the final answers, we have also explained the concept behind it to help you build stronger understanding of each topic. This will be really helpful for Class 12 students who want to understand both theoretical and practical questions. By studying these RBSE Questions and Answers your basic concepts will improve a lot.

Benefits of using Chemistry Class 12 Solved Papers

Using our Chemistry solutions regularly students will be able to improve their logical thinking and problem-solving speed. These Class 12 solutions are a guide for self-study and homework assistance. Along with the chapter-wise solutions, you should also refer to our Revision Notes and Sample Papers for Chapter 6 Principles and Processes of Isolation of E to get a complete preparation experience.

FAQs

Where can I find the latest RBSE Solutions Class 12 Chemistry Chapter 6 Principles and Processes of Isolation of E for the 2026-27 session?

The complete and updated RBSE Solutions Class 12 Chemistry Chapter 6 Principles and Processes of Isolation of E is available for free on StudiesToday.com. These solutions for Class 12 Chemistry are as per latest RBSE curriculum.

Are the Chemistry RBSE solutions for Class 12 updated for the new 50% competency-based exam pattern?

Yes, our experts have revised the RBSE Solutions Class 12 Chemistry Chapter 6 Principles and Processes of Isolation of E as per 2026 exam pattern. All textbook exercises have been solved and have added explanation about how the Chemistry concepts are applied in case-study and assertion-reasoning questions.

How do these Class 12 RBSE solutions help in scoring 90% plus marks?

Toppers recommend using RBSE language because RBSE marking schemes are strictly based on textbook definitions. Our RBSE Solutions Class 12 Chemistry Chapter 6 Principles and Processes of Isolation of E will help students to get full marks in the theory paper.

Do you offer RBSE Solutions Class 12 Chemistry Chapter 6 Principles and Processes of Isolation of E in multiple languages like Hindi and English?

Yes, we provide bilingual support for Class 12 Chemistry. You can access RBSE Solutions Class 12 Chemistry Chapter 6 Principles and Processes of Isolation of E in both English and Hindi medium.

Is it possible to download the Chemistry RBSE solutions for Class 12 as a PDF?

Yes, you can download the entire RBSE Solutions Class 12 Chemistry Chapter 6 Principles and Processes of Isolation of E in printable PDF format for offline study on any device.