CBSE Class 12 Chemistry General Principles And Processes Of Isolation Of Elements Worksheet

One mark Questions

1.Differentiate between a mineral and an ore.

Ans: the naturally occurring chemical substances present in the earth’s crust which can be obtained by mining are called minerals while minerals from which metals can be extracted economically are called ores.

2. Why is it that only sulphide ores are concentrated by froth floatation process.

Ans: This is because sulphide ores particles are wetted by oil and gangue particles are wetted by water.

3. Name one acidic flux and one basic flux.

Ans: silica and lime

4. Name the chief ore of silver.

Ans: argentite or silver glance

5. Name a reagent used during leaching of bauxite ore.

Ans : NaOH (sodium hydroxide)

6. Why is silica added to sulphide ore of copper in the reverberatory furnace?

Ans : in order to remove the iron impurity as slag

7. What is the role of flux in metallurgical processes?

Ans : flux is used for making the molten mass more conducting.

8. What is the thermodynamic relation between Gibbs free energy and emf of the cell.

Ans: ΔG⁰= —nFE⁰

9. What is the relation between gibbs free energy and equilibrium constant?

Ans: ΔG⁰= —RTlnK

10. Give the expression for Gibbs Helmholtz equation. 

Ans : ΔG= ΔH—TΔS

11. Name one chemical which can be used to concentrate galena selectively by froth floatation process.

Ans: sodium cyanide (NaCN)

12. What type of ores are roasted?

Ans : sulphide ores

13. Out of C and CO which is a better reducing agent for ZnO?

Ans : the free energy of formation of CO from C becomes lower at temp. above 1120K whereas that of CO₂ from C becomes lower above 1323K than free energy of formation of ZnO. The free energy of formation of from CO is always higher than that of ZnO. Therefore, C can reduce ZnO to Zn better than CO. CO₂ from CO is always higher than that of ZnO. Therefore, C can reduce ZnO to Zn better than CO.

14. What is the chemical principle on which chromatography separation based on?

Ans : Adsorption

15. What are the products obtained during the electrolysis of brine solution? Also write the name of this process.

Ans : chlorine, hydrogen and sodium hydroxide. The process is popularly known as chlor-alkali process.

Important Questions for NCERT Class 12 Chemistry General Principles Process of Isolation of Elements

Question. Nitriding is the process of surface hardening of steel by treating it in an atmosphere of
(a) NH₃
(b) O₃
(c) N₂
(d) H₂S

Question. Aluminium is extracted from alumina (Al₂O₃) by electrolysis of a molten mixture of
(a) Al₂O₃ + HF + NaAlF₄
(b) Al₂O₃ + CaF₂ + NaAlF₄
(c) Al₂O₃ + Na₃AlF₆ + CaF₂
(d) Al₂O₃ + KF + Na₃AlF₆ 

Question. Purification of aluminium, by electrolytic refining, is known as
(a) Hoope’s process
(b) Baeyer’s process
(c) Hall’s process
(d) Serpeck’s process. 

Question. Calcium is obtained by
(a) reduction of calcium chloride with carbon
(b) electrolysis of molten anhydrous calcium chloride
(c) roasting of limestone
(d) electrolysis of solution of calcium chloride in H₂O.

Question. Extraction of gold and silver involves leaching with
CN– ion. Silver is later recovered by
(a) distillation
(b) zone refining
(c) displacement with Zn
(d) liquation

Question. Identify the correct statement from the following :
(a) Wrought iron is impure iron with 4% carbon.
(b) Blister copper has blistered appearance due to evolution of CO₂.
(c) Vapour phase refining is carried out for Nickel by van Arkel method.
(d) Pig iron can be moulded into a variety of shapes.

Question. Which of the following pairs of metals is purified by van Arkel method?
(a) Ga and In
(b) Zr and Ti
(c) Ag and Au
(d) Ni and Fe 

Question. The method of zone refining of metals is based on the principle of
(a) greater mobility of the pure metal than that of the impurity
(b) higher melting point of the impurity than that of the pure metal
(c) greater noble character of the solid metal than that of the impurity
(d) greater solubility of the impurity in the molten state than in the solid.

Question. 2CuFeS₂ +O₂ →Cu₂S + 2FeS + SO₂ Which process of metallurgy of copper is represented by above equation? 
(a) Concentration
(b) Roasting
(c) Reduction
(d) Purification

ONE MARK QUESTIONS

1.Out of C and CO, which is a better reducing agent at 673 K?

TWO MARK QUESTIONS

1.Describe the role of

a) Iodine in the refining of titanium.

b) Collector in the froth floatation process.

2.Describe how the following changes are brought out:

a) Pig iron into steel

b) Zinc oxide into zinc metal.

THREE MARK QUESTIONS

1.Differentiate between

a) Calcination and roasting

b) Electrolytic reduction and electrolytic refining

c) Flux and slag

2.Write the chemical reactions which take place in the following operations:

a) Electrolytic reduction of Al₂O₃.

b) Isolation of Zn from zinc blende.

c) Mond’s process for refining of Ni.

3.Give reasons:

a) Copper matte is put in silica lined convertor.

b) Cryolite is added to alumina during electrolytic reduction.

c) Pine oil is used in the froth floatation process

4.a) Name the method used for the refining of titanium. b) What is the role of Zn in the extraction of silver? c) Reduction of metal oxide to metal becomes easier if the metalobtained is in liquid state. Why?

VALUE BASED QUESTION

1.Sam owns sites from where copper with other metals are mined. At sites he found the low grade copper ores are available with zinc and iron scraps.

a) Which of the two scraps should Sam use for the reduction of leached copper and why?

b) Mention the value in Sam’s choice.

c) What do you mean by blister copper?

d) Why is extraction of copper from copper pyrite more difficult than that from its oxide ore through reduction?

Important Questions for NCERT Class 12 Chemistry General Principles Process of Isolation of Elements

Question. The metal oxide which cannot be reduced to metal by carbon is
(a) Al2O₃
(b) PbO
(c) ZnO
(d) Fe2O₃

Question. Carbon and CO gas are used to reduce which of the following pairs of metal oxides for extraction of metals? 
(a) FeO, SnO
(b) SnO, ZnO
(c) BaO, Na2O₂
(d) FeO, ZnO

Question. Which of the following elements is present as the impurity to the maximum extent in the pig iron?
(a) Manganese
(b) Carbon
(c) Silicon
(d) Phosphorus

Question. In metallurgical process of aluminium, cryolite is mixed with alumina in its molten state, because it 
(a) decreases the amount of alumina
(b) oxidises the alumina
(c) increases the melting point of alumina
(d) decreases the melting point of alumina

Question. The following reactions take place in the blast furnace in the preparation of impure iron. Identify the reaction pertaining to the formation of the slag.
(a) Fe₂O₃(s) + 3CO(g) → 2Fe(l) + 3CO₂(g)
(b) CaCO₃(s) → CaO(s) + CO₂(g)
(c) CaO(s) + SiO₂(s) → CaSiO₃(s)
(d) 2C(s) + O₂(g) → 2CO(g)

Question. Sulfide ores are common for the metals 
(a) Ag, Cu and Pb
(c) Ag, Cu and Sn
(b) Ag, Mg and Pb
(d) Al, Cu and Pb

Question. Which of the following statements, about the advantage of roasting of sulphide ore before reduction is not true?
(a) The DGf° of the sulphide is greater than those for CS2 and H2S.
(b) The DGf° is negative for roasting of sulphide ore to oxide.
(c) Roasting of the sulphide to the oxide is thermodynamically feasible.
(d) Carbon and hydrogen are suitable reducing agents for metal sulphides.

Question. The main reactions occurring in blast furnace during extraction of iron from haematite are________.
(i) Fe₂O₃ + 3CO → 2Fe + 3CO₂
(ii) FeO + SiO₂ → FeSiO₃
(iii) Fe2O3 + 3C → 2Fe + 3CO
(iv) CaO + SiO₂ → CaSiO₃
(a) (i) and (iii)
(b) (ii) and (iv)
(c) (i) and (iv)
(d) (i), (ii) and (iii)

One Mark Questions

Question. Differentiate between a mineral and an ore.
Answer: Naturally occurring chemical substances present in the Earth's crust that can be obtained by mining are called minerals. On the other hand, minerals from which metals can be extracted both economically and conveniently are called ores.
In simple words: Minerals are natural chemical compounds found in the earth. Ores are specific minerals that contain enough metal to be extracted profitably.

Exam Tip: Remember the fundamental distinction: "All ores are minerals, but all minerals are not ores."

Question. Why is it that only sulphide ores are concentrated by froth floatation process.
Answer: This is because sulphide ore particles are preferentially wetted by pine oil, whereas the gangue particles are wetted by water, allowing the ore to attach to air bubbles and float as froth.
In simple words: Sulphide ore sticks to oil and floats to the top with the bubbles, while the dirt waste sticks to water and sinks to the bottom.

Exam Tip: Cite the difference in wetting properties between the ore and gangue as the main physical principle of this process.

Question. Name one acidic flux and one basic flux.
Answer: Silica (\( \text{SiO}_2 \)) is commonly used as an acidic flux, while lime (\( \text{CaO} \)) is used as a basic flux.
In simple words: Silica is an acidic flux, and lime is a basic flux.

Exam Tip: Acidic flux is added to remove basic impurities (like \( \text{FeO} \)), whereas basic flux is added to remove acidic impurities (like \( \text{SiO}_2 \)).

Question. Name the chief ore of silver.
Answer: The chief ore of silver is argentite, which is also commonly referred to as silver glance (\( \text{Ag}_2\text{S} \)).
In simple words: The main ore used to extract silver is called argentite, or silver glance.

Exam Tip: Be sure to write down both the name and chemical formula (\( \text{Ag}_2\text{S} \)) of the ore.

Question. Name a reagent used during leaching of bauxite ore.
Answer: Concentrated sodium hydroxide (\( \text{NaOH} \)) solution is used as the leaching reagent during the purification of bauxite ore.
In simple words: Sodium hydroxide is the chemical used to selectively dissolve alumina from bauxite ore.

Exam Tip: Mention that the bauxite leaching process under hot concentrated NaOH is also known as Baeyer's process.

Question. Why is silica added to sulphide ore of copper in the reverberatory furnace?
Answer: Silica is added to act as an acidic flux, which reacts with basic iron oxide (\( \text{FeO} \)) impurities to form fusible iron silicate slag (\( \text{FeSiO}_3 \)) that can be easily skimmed off.
In simple words: Silica combines with iron impurities to form a liquid waste called slag, allowing the copper to be separated cleanly.

Exam Tip: Always support your explanation with the slag-formation chemical equation: \( \text{FeO} + \text{SiO}_2 \rightarrow \text{FeSiO}_3 \).

Question. What is the role of flux in metallurgical processes?
Answer: Flux is added to combine with infusible gangue impurities present in the ore to form a fusible compound called slag, which can then be easily separated. It also makes the molten mass more conducting and lowers its melting point.
In simple words: Flux reacts with solid impurities to melt them into a liquid waste called slag, which can be easily separated from the molten metal.

Exam Tip: Remember: Impurity + Flux = Slag.

Question. What is the thermodynamic relation between Gibbs free energy and emf of the cell.
Answer: The thermodynamic relation is given by the expression: \[ \Delta G^\circ = -nFE^\circ \] where \( \Delta G^\circ \) represents the standard Gibbs free energy change, \( n \) is the number of moles of electrons transferred, \( F \) is Faraday's constant, and \( E^\circ \) is the standard electromotive force (emf) of the cell.
In simple words: This equation links standard free energy to cell voltage. A positive cell voltage gives a negative free energy change, meaning the reaction happens spontaneously.

Exam Tip: Define each variable in the equation clearly to secure full marks.

Question. What is the relation between gibbs free energy and equilibrium constant?
Answer: The thermodynamic relationship is expressed as: \[ \Delta G^\circ = -RT \ln K \] where \( \Delta G^\circ \) is the standard Gibbs free energy change, \( R \) is the universal gas constant, \( T \) is the absolute temperature in Kelvin, and \( K \) is the equilibrium constant of the reaction.
In simple words: This equation shows how the thermodynamic stability of a reaction relates to how far the reaction will go.

Exam Tip: Remember that \( \ln K \) can be written as \( 2.303 \log_{10} K \), which gives the expression \( \Delta G^\circ = -2.303 RT \log K \).

Question. Give the expression for Gibbs Helmholtz equation.
Answer: The Gibbs-Helmholtz equation is expressed as: \[ \Delta G = \Delta H - T\Delta S \] where \( \Delta G \) represents the change in Gibbs free energy, \( \Delta H \) is the change in enthalpy, \( T \) is the temperature in Kelvin, and \( \Delta S \) is the change in entropy.
In simple words: This equation shows how heat changes and molecular disorder combine at a certain temperature to determine if a reaction can happen on its own.

Exam Tip: Highlight that a negative \( \Delta G \) value indicates a spontaneous process.

Question. Name one chemical which can be used to concentrate galena selectively by froth floatation process.
Answer: Sodium cyanide (\( \text{NaCN} \)) is used as a depressant to selectively concentrate galena (\( \text{PbS} \)) from zinc blende (\( \text{ZnS} \)) impurities.
In simple words: Sodium cyanide acts as a blocker (depressant) so that galena floats to the top while zinc blende is held back in the water.

Exam Tip: Mention that NaCN selectively forms a soluble complex on the surface of ZnS to prevent it from forming a froth.

Question. What type of ores are roasted?
Answer: Concentrated sulphide ores are subjected to roasting to convert them into their respective metal oxides.
In simple words: Sulphide ores are heated strongly in plenty of air to replace the sulphur with oxygen.

Exam Tip: Always specify that roasting requires heating "in excess of air below the melting point of the metal."

Question. Out of C and CO which is a better reducing agent for ZnO?
Answer: Carbon (\( \text{C} \)) is a better reducing agent for \( \text{ZnO} \) than carbon monoxide (\( \text{CO} \)). According to the Ellingham diagram, the free energy of formation of \( \text{CO} \) from \( \text{C} \) becomes lower than that of \( \text{ZnO} \) at temperatures above \( 1120\text{ K} \). On the other hand, the free energy of formation of \( \text{CO}_2 \) from \( \text{CO} \) is always higher than that of \( \text{ZnO} \), meaning \( \text{CO} \) cannot reduce \( \text{ZnO} \).
In simple words: At high temperatures (above 1120 K), carbon has a much stronger affinity for oxygen than zinc does, allowing it to easily pull oxygen away from zinc oxide.

Exam Tip: Mention the critical temperature of 1120 K and cite the Ellingham diagram to support your answer.

Question. What is the chemical principle on which chromatography separation based on?
Answer: Chromatography separation is based on the principle of differential adsorption, where different components of a mixture are adsorbed to varying degrees on a stationary phase under the flow of a mobile phase.
In simple words: Chromatography works because different components of a mixture travel at different speeds based on how tightly they stick to the stationary background.

Exam Tip: Identify that components are separated based on their relative affinities for the stationary and mobile phases.

Question. What are the products obtained during the electrolysis of brine solution? Also write the name of this process.
Answer: The electrolysis of a brine solution produces chlorine gas (\( \text{Cl}_2 \)), hydrogen gas (\( \text{H}_2 \)), and sodium hydroxide (\( \text{NaOH} \)). This industrial electrolytic process is widely known as the chlor-alkali process.
In simple words: Running electricity through salty water splits it into chlorine gas, hydrogen gas, and sodium hydroxide. This is called the chlor-alkali process.

Exam Tip: Be sure to specify that chlorine gas is evolved at the anode, while hydrogen gas is evolved at the cathode.

Question. What is roasting?
Answer: Roasting is a metallurgical process where concentrated sulphide ore is heated strongly in a continuous supply of excess air at a temperature below its melting point to convert it into metal oxide.
In simple words: Roasting means heating sulphide ores in plenty of air below their melting point to turn them into oxides so they can be easily reduced.

Exam Tip: Write down a representative reaction like \( 2\text{ZnS} + 3\text{O}_2 \rightarrow 2\text{ZnO} + 2\text{SO}_2 \) to show roasting in action.

Question. What is calcination?
Answer: Calcination is a thermal treatment process where concentrated ore (typically carbonates or hydrated oxides) is heated below its melting point in the absence or a highly limited supply of air to expel water and volatile impurities.
In simple words: Calcination is heating ores without air to drive off moisture and carbon dioxide gas, leaving behind a dry metal oxide.

Exam Tip: Provide a typical equation such as \( \text{CaCO}_3 \xrightarrow{\Delta} \text{CaO} + \text{CO}_2 \) as an example of calcination.

Question. What is smelting?
Answer: Smelting is the high-temperature chemical reduction of a metal oxide ore using carbon or carbon monoxide in the presence of a flux to obtain the metal in a molten state while removing gangue as slag.
In simple words: Smelting is heating metal oxides at very high temperatures with carbon to melt them and extract the pure liquid metal.

Exam Tip: Smelting involves both chemical reduction and phase separation of molten metal from slag.

Question. What is blister copper/ copper matte?
Answer: Copper matte is a molten mixture consisting of cuprous sulfide (\( \text{Cu}_2\text{S} \)) and iron sulfide (\( \text{FeS} \)) produced during smelting. Blister copper is the solidified metal obtained after refining copper matte, which has a bubbly or blistered appearance on its surface caused by the escape of dissolved sulfur dioxide (\( \text{SO}_2 \)) gas during cooling.
In simple words: Copper matte is the raw mixture of copper and iron sulfides. Blister copper is the final metal that has a bumpy, blister-like surface because gas bubbled out of it while it cooled.

Exam Tip: Differentiate these terms clearly: copper matte contains sulfides, whereas blister copper is the metal product containing bubbles of \( \text{SO}_2 \).

Question. What is meant by beneficiation process?
Answer: Beneficiation, also known as ore dressing or concentration, is the preliminary metallurgical process of removing unwanted earthy, sandy, and siliceous impurities (collectively termed gangue) from mined ore.
In simple words: Beneficiation is simply washing and cleaning the raw ore to remove dirt and rocks, leaving behind a highly concentrated metal ore.

Exam Tip: Use "concentration of ore" or "ore dressing" as synonymous terms to explain this step.

2marks questions

Question. Write down the reactions taking place in blast furnace related to the metallurgy of iron.
Answer: Inside the blast furnace, the step-by-step reduction of iron oxides occurs at different temperature zones:
At lower temperature range (\( 500\text{ - }800\text{ K} \)): \[ 3\text{Fe}_2\text{O}_3 + \text{CO} \rightarrow 2\text{Fe}_3\text{O}_4 + \text{CO}_2 \] \[ \text{Fe}_3\text{O}_4 + 4\text{CO} \rightarrow 3\text{Fe} + 4\text{CO}_2 \] \[ \text{Fe}_2\text{O}_3 + \text{CO} \rightarrow 2\text{FeO} + \text{CO}_2 \] At higher temperature range (\( 900\text{ - }1500\text{ K} \)): \[ \text{FeO} + \text{CO} \rightarrow \text{Fe} + \text{CO}_2 \] Additionally, limestone decomposes and reacts with silica to form slag: \[ \text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2 \] \[ \text{CaO} + \text{SiO}_2 \rightarrow \text{CaSiO}_3 \text{ (Slag)} \]
In simple words: Inside the blast furnace, carbon monoxide gas removes oxygen from iron ore in steps to leave behind liquid iron, while limestone binds with sandy waste to form liquid slag.

Exam Tip: Grouping your equations by temperature zones is highly appreciated by examiners.

Question. Describe with chemical equation the extraction of silver from its ore.
Answer: Silver is extracted from its sulfide ore (argentite, \( \text{Ag}_2\text{S} \)) using the MacArthur-Forrest cyanide leaching process:
1. **Leaching:** Finely powdered ore is treated with a dilute solution of sodium cyanide (\( \text{NaCN} \)) in the presence of oxygen to form a soluble dicyanoargentate(I) complex: \[ 2\text{Ag}_2\text{S} + 8\text{CN}^- + \text{O}_2 + 2\text{H}_2\text{O} \rightarrow 4[\text{Ag}(\text{CN})_2]^- + 2\text{S} + 4\text{OH}^- \] 2. **Displacement:** Zinc metal, being more electropositive than silver, acts as a reducing agent and displaces silver from the soluble complex: \[ 2[\text{Ag}(\text{CN})_2]^- + \text{Zn} \rightarrow [\text{Zn}(\text{CN})_4]^{2-} + 2\text{Ag} \]
The crude silver is subsequently refined by fusion with borax or by electrolysis.
In simple words: Silver is dissolved out of the ore by a cyanide solution using air. Then, zinc is added to push the silver out of the solution so it can be collected as pure metal.

Exam Tip: Note that oxygen is essential in the leaching step to drive the equilibrium forward by oxidizing sulfide ions.

Question. Describe the role of the following. (a) NaCN in the extraction of silver from a silver ore (b) Cryolite in the extraction of aluminium from pure alumina.
Answer:
(a) **Role of \( \text{NaCN} \) in Silver Extraction:** It serves as a complexing agent. It reacts with argentite (\( \text{Ag}_2\text{S} \)) to form a soluble sodium dicyanoargentate(I) complex, bringing silver into solution while separating it from insoluble gangue impurities: \[ \text{Ag}_2\text{S} + 4\text{NaCN} \rightarrow 2\text{Na}[\text{Ag}(\text{CN})_2] + \text{Na}_2\text{S} \]
(b) **Role of Cryolite (\( \text{Na}_3\text{AlF}_6 \)) in Aluminium Extraction:** It is added to alumina (\( \text{Al}_2\text{O}_3 \)) to: i. Lower the melting point of the mixture from \( 2323\text{ K} \) to around \( 1173\text{ K} \), saving fuel and energy. ii. Increase the electrical conductivity of the molten electrolyte mixture.
In simple words: Cyanide is used to selectively dissolve silver into a liquid compound. Cryolite is added to alumina to help it melt at a much lower temperature and conduct electricity easily for electrolysis.

Exam Tip: List both of cryolite's roles clearly: (1) lowering the melting point and (2) increasing electrical conductivity.

Question. Explain the role of carbon monoxide in the purification of nickel and iodine in zirconium.
Answer: Both processes utilize vapor phase refining:
- **Role of Carbon Monoxide in Nickel Purification (Mond's Process):** Carbon monoxide reacts with impure nickel at \( 330\text{ - }350\text{ K} \) to form a volatile complex, nickel tetracarbonyl \( [\text{Ni}(\text{CO})_4] \), leaving impurities behind. This complex is then heated to \( 450\text{ - }470\text{ K} \) where it decomposes to yield pure nickel: \[ \text{Ni} + 4\text{CO} \xrightarrow{330\text{-}350\text{ K}} [\text{Ni}(\text{CO})_4] \xrightarrow{450\text{-}470\text{ K}} \text{Ni} + 4\text{CO} \]
- **Role of Iodine in Zirconium Purification (Van Arkel Method):** Impure zirconium is heated with iodine at around \( 870\text{ K} \) to form volatile zirconium tetraiodide (\( \text{ZrI}_4 \)), leaving non-volatile impurities behind. This vapor decomposes on a hot tungsten filament heated to \( 2075\text{ K} \) to deposit pure zirconium: \[ \text{Zr} + 2\text{I}_2 \xrightarrow{870\text{ K}} \text{ZrI}_4 \xrightarrow{2075\text{ K}} \text{Zr} + 2\text{I}_2 \]
In simple words: Carbon monoxide binds with nickel to make a gas that is easily separated from solid impurities, then split back into pure nickel. Iodine does the same thing for zirconium, turning it into a gas that breaks down on a hot wire to deposit pure metal.

Exam Tip: Specify the exact temperatures for both steps in these vapor phase refining processes to score full marks.

Question. (a) Name the method used for refining of (i) nickel (ii) zirconium (b)The extraction of gold by leaching with NaCN involves both oxidation and reduction. Justify giving equations.
Answer:
(a) the refining methods are:
(i) **Nickel:** Mond's process.
(ii) **Zirconium:** Van Arkel method.

(b) The extraction of gold using \( \text{NaCN} \) involves two key redox steps:
1. **Oxidation:** Metallic gold (\( \text{Au}^0 \)) is oxidized to aurous ion (\( \text{Au}^+ \)) by oxygen in the presence of cyanide ions, forming a soluble complex: \[ 4\text{Au} + 8\text{CN}^- + \text{O}_2 + 2\text{H}_2\text{O} \rightarrow 4[\text{Au}(\text{CN})_2]^- + 4\text{OH}^- \] Here, the oxidation state of gold increases from \( 0 \) to \( +1 \).
2. **Reduction:** Zinc metal acts as a reducing agent to displace gold from the complex, reducing the \( \text{Au}^+ \) ions back to elemental gold (\( \text{Au}^0 \)): \[ 2[\text{Au}(\text{CN})_2]^- + \text{Zn} \rightarrow [\text{Zn}(\text{CN})_4]^{2-} + 2\text{Au} \] Here, gold is reduced from \( +1 \) to \( 0 \), while zinc is oxidized to \( +2 \).
In simple words: First, oxygen oxidizes gold so it can dissolve in a cyanide solution. Second, zinc reduces the gold ions back into solid gold metal, taking their place in the solution.

Exam Tip: Be ready to explain the changes in oxidation states (\( 0 \rightarrow +1 \) and \( +1 \rightarrow 0 \)) for gold to support your justification.

Question. What criterion is followed for selection of the stationary phase in chromatography?
Answer: The stationary phase is chosen such that the impurities are either more strongly adsorbed or exhibit greater solubility in it compared to the element being purified. This ensures that the impurities are retained firmly on the column while the pure component moves down faster and is easily washed out (eluted) first.
In simple words: We choose a background material (stationary phase) that holds onto the impurities very tightly but lets the pure metal pass through easily, allowing us to collect the pure metal first.

Exam Tip: Clearly state that the difference in adsorption strength/solubility between the impurities and the metal is the deciding factor.

Question. Explain electrolytic refining of copper with thermodynamic principle involve in the process.
Answer: In this purification method, a thick block of impure copper is set as the anode, and a thin sheet of pure copper acts as the cathode. Both are submerged in an acidic copper sulfate (\( \text{CuSO}_4 \)) bath. Upon passing electric current, copper dissolves from the anode and deposits as pure metal on the cathode:
At Cathode (Reduction): \[ \text{Cu}^{2+}(\text{aq}) + 2\text{e}^- \rightarrow \text{Cu}(\text{s}) \] At Anode (Oxidation): \[ \text{Cu}(\text{s}) \rightarrow \text{Cu}^{2+}(\text{aq}) + 2\text{e}^- \] **Thermodynamic Principle:** The overall process has an electrical energy supply from an external source. The Gibbs free energy change of the non-spontaneous deposition is driven by the applied voltage, related by: \[ \Delta G^\circ = -nFE^\circ \] By applying a carefully regulated potential, only copper (and metals more electropositive than impurities) are deposited at the cathode, while less active impurities (like Ag, Au) fall below the anode as anode mud.
In simple words: Electricity is used to dissolve copper from a dirty block (anode) and deposit it onto a clean plate (cathode). Impurities either dissolve in the bath or drop to the bottom as waste.

Exam Tip: Be sure to write the electrode half-reactions and mention "anode mud" containing valuable metals like silver and gold.

Question. What are the limitations of Ellingham diagram?
Answer: The primary limitations of the Ellingham diagram are:
1. **Kinetics:** The diagram is purely thermodynamic and only determines the feasibility of a reduction reaction (\( \Delta G^\circ < 0 \)). It provides no information regarding the rate or kinetics of the chemical reaction.
2. **Equilibrium Assumption:** The derivation of \( \Delta G^\circ \) relies on the equilibrium constant \( K \), assuming that reactants and products exist in equilibrium, which is rarely the case in practical industrial processes.
In simple words: The diagram tells us if a reaction is chemically possible, but it cannot tell us how fast the reaction will happen or if the reactants will actually reach equilibrium.

Exam Tip: Memorize these two clear points (kinetics/rate of reaction and equilibrium assumption) for 2-mark questions on thermodynamics.

Question. What is the role of depressant in froth floatation process?
Answer: A depressant is used to selectively prevent a specific type of sulphide mineral from adhering to the oil and rising with the froth, enabling the separation of two sulphide ores. For example, sodium cyanide (\( \text{NaCN} \)) is added to separate galena (\( \text{PbS} \)) from zinc blende (\( \text{ZnS} \)). It reacts selectively with \( \text{ZnS} \) to form a soluble complex on its surface, preventing it from floating: \[ \text{ZnS} + 4\text{NaCN} \rightarrow \text{Na}_2[\text{Zn}(\text{CN})_4] + \text{Na}_2\text{S} \]
In simple words: A depressant acts as a selective blocker that stops one metal sulfide from floating so that the other one can be collected by itself.

Exam Tip: Be sure to write down the chemical formula of the zinc complex, \( \text{Na}_2[\text{Zn}(\text{CN})_4] \), to make your answer stand out.

Question. How are metals used as semiconductors refined? What is the principle of the method used?
Answer: Semiconductor metals (such as silicon, germanium, and gallium) are refined to high purity using **Zone Refining**.
**Principle:** This technique is based on the principle that impurities are more soluble in the molten state (melt) of the metal than in its solid state. When a circular heater moves slowly along an impure metal rod, the molten zone moves forward, carrying the dissolved impurities with it, leaving highly pure crystallized metal behind.
In simple words: Semiconductors are purified by a melting heater that rolls along a metal rod. The dirt dissolves in the liquid zone and gets pushed all the way to one end, leaving the rest of the rod extremely pure.

Exam Tip: Mention silicon (Si) or germanium (Ge) as examples when writing about zone refining.

3marks questions:

Question. Describe how the following changes are brought about : (i) Pig iron into steel (ii) Zinc oxide into metallic zinc (iii) Impure titanium into pure titanium
Answer:
(i) **Pig Iron into Steel:** Pig iron is converted into steel by oxidizing its impurities in a converter. A blast of oxygen gas diluted with carbon dioxide is blown through the molten metal. This oxidizes carbon to gaseous carbon monoxide (\( \text{CO} \)), which burns at the converter's mouth, and converts impurities like silicon and manganese into slag. After removing the slag, necessary metals (like Mn, Cr, Ni) are added to obtain steel.
(ii) **Zinc Oxide into Metallic Zinc:** Zinc oxide is reduced by mixing it with powdered coke (carbon) and clay, which are formed into briquettes. These briquettes are heated strongly: \[ \text{ZnO} + \text{C} \xrightarrow{\Delta} \text{Zn} + \text{CO} \] The metal vaporizes, is distilled off, and is collected by rapid condensation.
(iii) **Impure Titanium into Pure Titanium (Van Arkel Method):** Impure titanium is heated with iodine in an evacuated vessel to form volatile titanium tetraiodide (\( \text{TiI}_4 \)). This vapor is then decomposed on a hot tungsten filament heated to high temperatures to deposit pure titanium: \[ \text{Ti} + 2\text{I}_2 \rightarrow \text{TiI}_4 \xrightarrow{\Delta} \text{Ti} + 2\text{I}_2 \]
In simple words: Steel is made by blowing oxygen through melted iron to burn away extra carbon and impurities. Zinc oxide is heated with carbon to pull the oxygen out. Titanium is turned into a gas with iodine, then broken down on a hot wire to leave behind pure titanium.

Exam Tip: Be sure to write the chemical equations for the reduction of zinc and the purification of titanium to ensure maximum marks.

Question. Describe the role of (a) NaCN in the extraction of gold from gold ore. (b) SiO2 in the extraction of copper from copper matte. (c) Iodine in the refining of zirconium
Answer:
(a) **Role of \( \text{NaCN} \) in Gold Extraction:** It acts as a leaching agent. In the presence of atmospheric oxygen, it selectively dissolves elemental gold by converting it into a soluble cyano complex, leaving insoluble gangue behind: \[ 4\text{Au} + 8\text{CN}^- + \text{O}_2 + 2\text{H}_2\text{O} \rightarrow 4[\text{Au}(\text{CN})_2]^- + 4\text{OH}^- \]
(b) **Role of \( \text{SiO}_2 \) in Copper Extraction:** Silica acts as an acidic flux in smelting. It reacts with basic iron oxide (\( \text{FeO} \)) impurities in the copper matte to form a fusible slag (\( \text{FeSiO}_3 \)) that floats and can be separated easily: \[ 2\text{FeS} + 3\text{O}_2 \rightarrow 2\text{FeO} + 2\text{SO}_2 \] \[ \text{FeO} + \text{SiO}_2 \rightarrow \text{FeSiO}_3 \text{ (Slag)} \]
(c) **Role of Iodine in Zirconium Refining:** Iodine acts as a gaseous refining agent. It reacts with impure zirconium at around \( 870\text{ K} \) to form volatile zirconium tetraiodide (\( \text{ZrI}_4 \)), separating it from non-volatile impurities. The vapor decomposes on a hot tungsten filament at \( 2075\text{ K} \) to deposit pure zirconium: \[ \text{Zr} + 2\text{I}_2 \xrightarrow{870\text{ K}} \text{ZrI}_4 \xrightarrow{2075\text{ K}} \text{Zr} + 2\text{I}_2 \]
In simple words: Cyanide dissolves gold into a liquid complex to separate it from dirt. Silica turns iron impurities in copper into a liquid waste called slag. Iodine converts zirconium into a gas so it can be purified on a hot filament.

Exam Tip: For each part, write the complete, balanced chemical equations as they are highly valued by examiners.

Question. Describe how the following changes are brought about : (i) Pig iron into steel (ii) Bauxite into pure alumina (iii) Impure copper into pure copper
Answer:
(i) **Pig Iron into Steel:** Molten pig iron is treated in a converter by blowing a mixture of oxygen and carbon dioxide. This oxidizes excess carbon to carbon monoxide and converts other impurities (like Si, Mn) into slag. After discarding the slag, small quantities of alloying elements are added to produce steel.
(ii) **Bauxite into Pure Alumina (Baeyer's Process):** Powdered bauxite is digested with concentrated aqueous \( \text{NaOH} \) at high temperatures and pressures. Alumina dissolves as sodium aluminate, leaving behind iron oxides and silica: \[ \text{Al}_2\text{O}_3 + 2\text{NaOH} + 3\text{H}_2\text{O} \rightarrow 2\text{Na}[\text{Al}(\text{OH})_4] \] The filtrate is neutralized with \( \text{CO}_2 \) to precipitate hydrated alumina, which is then filtered and heated to get pure alumina: \[ 2\text{Na}[\text{Al}(\text{OH})_4] + \text{CO}_2 \rightarrow \text{Al}_2\text{O}_3 \cdot x\text{H}_2\text{O} + 2\text{NaHCO}_3 \] \[ \text{Al}_2\text{O}_3 \cdot x\text{H}_2\text{O} \xrightarrow{\Delta} \text{Al}_2\text{O}_3 + x\text{H}_2\text{O} \]
(iii) **Impure Copper into Pure Copper:** Electrolytic refining is used. Impure copper serves as the anode, and a pure copper sheet serves as the cathode in an acidified copper sulfate (\( \text{CuSO}_4 \)) bath. Current passage deposits pure copper on the cathode: \[ \text{Cathode: } \text{Cu}^{2+}(\text{aq}) + 2\text{e}^- \rightarrow \text{Cu}(\text{s}) \] \[ \text{Anode: } \text{Cu}(\text{s}) \rightarrow \text{Cu}^{2+}(\text{aq}) + 2\text{e}^- \]
In simple words: Steel is made by burning away carbon from pig iron. Alumina is dissolved using hot sodium hydroxide, filtered, and baked to get pure alumina. Copper is purified by dissolving an impure block electrically and collecting the pure metal on a clean copper plate.

Exam Tip: Be sure to write the full set of reactions for the purification of bauxite, as it is a frequent 3-mark question.

Question. Describe the principle behind each of the following process. (i) Vapour phase refining of a metal (ii) Electrolytic refining of a metal (iii) Recovery of silver after silver ore was leached with NaCN
Answer:
(i) **Vapour Phase Refining:** The metal is reacted with a suitable reagent at lower temperatures to form a volatile compound, while leaving impurities behind. This gaseous compound is collected and then decomposed at a higher temperature to yield the ultra-pure metal.
(ii) **Electrolytic Refining:** Impure metal is used as the anode, and a thin strip of the same pure metal acts as the cathode in an electrolytic bath of its soluble salt. On passing current, metal ions from the anode dissolve into the electrolyte and deposit as pure metal on the cathode, while less active impurities collect as anode mud.
(iii) **Recovery of Silver:** Silver is dissolved as a soluble cyano complex \( [\text{Ag}(\text{CN})_2]^- \) during leaching. It is recovered by a displacement (hydrometallurgical) reaction using a more electropositive metal like zinc, which acts as a reducing agent: \[ 2[\text{Ag}(\text{CN})_2]^- + \text{Zn} \rightarrow [\text{Zn}(\text{CN})_4]^{2-} + 2\text{Ag} \]
In simple words: Vapor phase refining turns metal into a gas to separate it from solid dirt, then decomposes the gas back into pure solid. Electrolytic refining uses electricity to migrate pure metal from a dirty plate to a clean plate. Silver is recovered from solution by using zinc to displace it as a solid.

Exam Tip: State the key requirements for vapor phase refining: (1) the metal must form a volatile compound easily, and (2) this volatile compound must be easily decomposable.

Question. Write the reaction involved in the following process (i) Leaching of bauxite ore to prepare pure alumina (ii) Refining of zirconium by Van Arkel method (iii) Recovery of gold after gold ore has been leached with NaCN
Answer:
(i) **Leaching of Bauxite:** \[ \text{Al}_2\text{O}_3 + 2\text{NaOH} + 3\text{H}_2\text{O} \rightarrow 2\text{Na}[\text{Al}(\text{OH})_4] \] \[ 2\text{Na}[\text{Al}(\text{OH})_4] + \text{CO}_2 \rightarrow \text{Al}_2\text{O}_3 \cdot x\text{H}_2\text{O} + 2\text{NaHCO}_3 \] \[ \text{Al}_2\text{O}_3 \cdot x\text{H}_2\text{O} \xrightarrow{\Delta} \text{Al}_2\text{O}_3 + x\text{H}_2\text{O} \]
(ii) **Van Arkel Method for Zirconium:** \[ \text{Zr} + 2\text{I}_2 \xrightarrow{870\text{ K}} \text{ZrI}_4 \] \[ \text{ZrI}_4 \xrightarrow{2075\text{ K}} \text{Zr} + 2\text{I}_2 \]
(iii) **Recovery of Gold:** \[ 4\text{Au} + 8\text{CN}^- + \text{O}_2 + 2\text{H}_2\text{O} \rightarrow 4[\text{Au}(\text{CN})_2]^- + 4\text{OH}^- \] \[ 2[\text{Au}(\text{CN})_2]^- + \text{Zn} \rightarrow [\text{Zn}(\text{CN})_4]^{2-} + 2\text{Au} \]
In simple words: These are the key equations showing how bauxite is purified to alumina, how zirconium is turned into volatile iodide gas and decomposed on hot wire, and how gold is dissolved by cyanide and then precipitated out using zinc.

Exam Tip: Ensure that all chemical formulas and complexes (like \( \text{Na}[\text{Al}(\text{OH})_4] \) and \( [\text{Au}(\text{CN})_2]^- \)) are written with their correct coordination numbers and charges.

Question. Write the reactions involved in the following process: (i) Mond’s process (ii) Mac Arthur forest cyanide process (iii) Hall heroult’s process
Answer:
(i) **Mond's Process:** \[ \text{Ni} + 4\text{CO} \xrightarrow{330\text{-}350\text{ K}} [\text{Ni}(\text{CO})_4] \] \[ [\text{Ni}(\text{CO})_4] \xrightarrow{450\text{-}470\text{ K}} \text{Ni} + 4\text{CO} \]
(ii) **MacArthur-Forrest Cyanide Process (\( \text{M} = \text{Ag} \text{ or } \text{Au} \)):** \[ 4\text{M} + 8\text{CN}^- + \text{O}_2 + 2\text{H}_2\text{O} \rightarrow 4[\text{M}(\text{CN})_2]^- + 4\text{OH}^- \] \[ 2[\text{M}(\text{CN})_2]^- + \text{Zn} \rightarrow [\text{Zn}(\text{CN})_4]^{2-} + 2\text{M} \]
(iii) **Hall-Heroult's Process:** Electrolyte ionization: \[ \text{Al}_2\text{O}_3 \rightarrow 2\text{Al}^{3+} + 3\text{O}^{2-} \] At Cathode (Reduction): \[ \text{Al}^{3+} + 3\text{e}^- \rightarrow \text{Al}(\text{l}) \] At Anode (Oxidation): \[ \text{C} + \text{O}^{2-} \rightarrow \text{CO} + 2\text{e}^- \] \[ \text{C} + 2\text{O}^{2-} \rightarrow \text{CO}_2 + 4\text{e}^- \] Overall reaction: \[ 2\text{Al}_2\text{O}_3 + 3\text{C} \rightarrow 4\text{Al} + 3\text{CO}_2 \]
In simple words: Mond's process purifies nickel via carbon monoxide. MacArthur-Forrest process extracts silver/gold via cyanide complexation and zinc displacement. Hall-Heroult's process reduces molten alumina to liquid aluminium using carbon anodes.

Exam Tip: Be sure to write the overall cell reaction for the Hall-Heroult process, as it clearly shows why the carbon anodes are consumed over time and must be replaced.

Question. Account for the following facts : (a) Reduction of a metal oxide is easier if the metal formed is in the liquid state at the temperature of reduction (b) The reduction of Cr2O3 with aluminium is thermodynamically feasible, yet it does not occur at room temperature (c) Pine oil is used in froth floatation method
Answer:
(a) When the metal is produced in the liquid state rather than the solid state, its entropy (\( S \)) is significantly higher. This increase in entropy (\( \Delta S > 0 \)) makes the value of \( T\Delta S \) more positive, which consequently makes \( \Delta G \) (\( \Delta H - T\Delta S \)) more negative, facilitating the reduction reaction.
(b) Although the reduction of \( \text{Cr}_2\text{O}_3 \) with Al has a negative \( \Delta G^\circ \) at room temperature, it does not proceed due to a high activation energy barrier. Heating the system provides the required activation energy, allowing the reactant molecules to cross this energy barrier and proceed with the reaction.
(c) Pine oil serves as a collector in froth flotation. It selectively wets the sulphide ore particles, rendering them hydrophobic (water-repellent) so that they attach to air bubbles and float to the surface as froth.
In simple words: Liquid metal has higher entropy, which makes the reaction more spontaneous. Aluminum reduction of chromium oxide needs heat to overcome a start-up energy barrier. Pine oil is used because it selectively coats the metal ore, making it waterproof so it can ride air bubbles to the top.

Exam Tip: Use the Gibbs free energy equation \( \Delta G = \Delta H - T\Delta S \) to mathematically justify why the liquid state (with higher entropy) makes reduction easier.

Question. State briefly the principles which serve as basis for the following operation in metallurgy . (a) Froth floatation process (b) Zone refining (c) Refining by liquation
Answer:
(a) **Froth Flotation Process:** It is based on the difference in wetting properties of the ore and gangue particles. Sulphide ore particles are preferentially wetted by pine oil (becoming hydrophobic), while gangue particles are wetted by water.
(b) **Zone Refining:** This method relies on the principle of fractional crystallization, specifically that impurities are more soluble in the molten state (melt) than in the solid state of the metal.
(c) **Refining by Liquation:** This method is based on the difference in melting points of the metal and its impurities. It is used when the metal has a low melting point (such as tin or lead) compared to its infusible impurities. The impure metal is heated on a sloping hearth; the low-melting metal melts and flows down, leaving high-melting impurities behind.
In simple words: Froth flotation separates ore by making it float on oil while waste sinks in water. Zone refining works because impurities prefer to stay in the liquid melt rather than freeze with the solid metal. Liquation separates low-melting metals by melting them off a sloping ramp, leaving solid impurities behind.

Exam Tip: Differentiate these principles clearly in your answers as they are often tested in conceptual matching or direct definition questions.

Question. Explain the basic principles of the following metallurgical operations (a) Zone refining (b) Vapour phase refining (c) Electrolytic refining
Answer:
(a) **Zone Refining:** It is based on the principle that impurities are far more soluble in the molten state (melt) of the metal than in its crystallized solid state. As a mobile heater passes along an impure metal rod, impurities concentrate in the molten zone and get pushed to the end.
(b) **Vapour Phase Refining:** This method relies on converting the metal into a volatile compound by reacting it with a suitable reagent at a low temperature, separating it from non-volatile impurities, and then thermally decomposing the volatile complex at a higher temperature to recover the pure metal.
(c) **Electrolytic Refining:** The principle is to pass an electric current through an electrolyte bath of a metal salt, using the impure metal as the anode and a pure metal strip as the cathode. The metal dissolves from the anode as ions and deposits cleanly onto the cathode, while less active impurities collect as anode mud.
In simple words: Zone refining works because impurities prefer staying in molten liquid. Vapor phase refining changes metal to gas, separates it, and then breaks it down back to solid. Electrolytic refining uses electricity to dissolve metal from a dirty anode and plate it cleanly on a cathode.

Exam Tip: For each process, be prepared to name at least one metal refined by that method (e.g., Ge/Si for zone refining, Ni for vapor phase, Cu/Zn for electrolytic).

Question. Complete the following reactions: (i)Al2O3+ NaOH + H2O (ii) Au + CN- + O2+H2O (iii) [Ni(CO)4] (at 450-470K)
Answer: The completed chemical reactions are:
(i) Leaching of alumina: \[ \text{Al}_2\text{O}_3(\text{s}) + 2\text{NaOH}(\text{aq}) + 3\text{H}_2\text{O}(\text{l}) \rightarrow 2\text{Na}[\text{Al}(\text{OH})_4](\text{aq}) \]
(ii) Cyanide leaching of gold: \[ 4\text{Au}(\text{s}) + 8\text{CN}^-(\text{aq}) + \text{O}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l}) \rightarrow 4[\text{Au}(\text{CN})_2]^-(\text{aq}) + 4\text{OH}^-(\text{aq}) \]
(iii) Thermal decomposition of nickel tetracarbonyl: \[ [\text{Ni}(\text{CO})_4] \xrightarrow{450\text{-}470\text{ K}} \text{Ni} + 4\text{CO} \]
In simple words: These balanced reactions represent the conversion of alumina into soluble sodium aluminate, the oxidation of gold into a cyanide complex, and the thermal decomposition of nickel tetracarbonyl gas into pure nickel metal.

Exam Tip: Pay special attention to balancing the stoichiometry, especially for the gold leaching reaction which is a common source of calculation errors.

5 marks questions

Question. Describe the principle behind each of the following process. (i) Vapour phase refining (ii) Electrolytic refining of the metal (iii) Recovery of silver after silver ore was leached with NaCN (iv) Preparation of cast iron from pig iron (v) Preparation of pure alumina from bauxite
Answer:
(i) **Vapour Phase Refining:** The impure metal is treated with a specific reagent at a lower temperature to form a volatile compound, which is separated from non-volatile impurities. This gas is then thermally decomposed at a higher temperature to yield the ultra-pure metal (e.g., Mond's process for Ni, Van Arkel for Zr/Ti).
(ii) **Electrolytic Refining:** The impure metal block forms the anode and a thin strip of the pure metal forms the cathode in a solution of its soluble salt. When electric current is applied, pure metal ions dissolve from the anode and deposit onto the cathode, while less active impurities drop as anode mud.
(iii) **Recovery of Silver:** Silver leached with \( \text{NaCN} \) forms the soluble complex \( [\text{Ag}(\text{CN})_2]^- \). Silver is recovered by treating this complex with zinc metal, which acts as a reducing agent and displaces the less electropositive silver: \[ 2[\text{Ag}(\text{CN})_2]^- + \text{Zn} \rightarrow [\text{Zn}(\text{CN})_4]^{2-} + 2\text{Ag} \]
(iv) **Preparation of Cast Iron from Pig Iron:** Pig iron is melted with scrap iron and coke in a cupola furnace using a hot air blast. This oxidizes and removes some impurities (like carbon, sulfur, and phosphorus) as \( \text{CO}_2 \), \( \text{SO}_2 \), and \( \text{P}_2\text{O}_5 \), reducing the carbon content to approximately 3%.
(v) **Preparation of Pure Alumina from Bauxite (Leaching):** Aluminum oxide (\( \text{Al}_2\text{O}_3 \)) in bauxite is amphoteric and dissolves in concentrated hot \( \text{NaOH} \) to form soluble sodium aluminate, whereas basic impurities (like \( \text{Fe}_2\text{O}_3 \)) remain insoluble and are filtered out: \[ \text{Al}_2\text{O}_3 + 2\text{NaOH} + 3\text{H}_2\text{O} \rightarrow 2\text{Na}[\text{Al}(\text{OH})_4] \]
In simple words: Vapor phase refining uses a temporary gas form to leave solid impurities behind. Electrolytic refining uses electricity to plate pure metal onto a cathode. Silver is pushed out of its cyanide solution using zinc. Cast iron is made by melting pig iron with scrap to burn off excess carbon. Alumina is dissolved selectively in hot caustic soda to separate it from red mud impurities.

Exam Tip: Ensure that all five parts are answered with distinct headings and appropriate chemical reactions where applicable.

Question. Explain the role of each of the following in the extraction of metals from their ores : (i) CO in the extraction of nickel (ii) Zinc in the extraction of silver (iii) Silica in the extraction of copper (iv) Iodine in the extraction of titanium (v) Cryolite in the extraction of aluminium
Answer:
(i) **Role of \( \text{CO} \) in Nickel Extraction:** It acts as a volatile complexing agent in Mond's process, selectively reacting with nickel at \( 330\text{ - }350\text{ K} \) to form volatile nickel tetracarbonyl \( [\text{Ni}(\text{CO})_4] \) which is then thermally decomposed to give pure nickel.
(ii) **Role of Zinc in Silver Extraction:** Zinc acts as a reducing agent. Being more electropositive than silver, it displaces silver from its soluble cyano complex \( [\text{Ag}(\text{CN})_2]^- \) and precipitates it: \[ 2[\text{Ag}(\text{CN})_2] ^- + \text{Zn} \rightarrow [\text{Zn}(\text{CN})_4]^{2-} + 2\text{Ag} \]
(iii) **Role of Silica (\( \text{SiO}_2 \)) in Copper Extraction:** It serves as an acidic flux in smelting. It reacts with basic iron oxide (\( \text{FeO} \)) impurities in the copper matte to form a fusible slag (\( \text{FeSiO}_3 \)) that floats and can be separated easily: \[ \text{FeO} + \text{SiO}_2 \rightarrow \text{FeSiO}_3 \text{ (Slag)} \]
(iv) **Role of Iodine in Titanium Extraction:** It reacts with impure titanium at around \( 523\text{ K} \) to form volatile titanium tetraiodide (\( \text{TiI}_4 \network \)), separating it from non-volatile impurities. The vapor is then decomposed on a hot tungsten filament to obtain pure titanium metal.
(v) **Role of Cryolite (\( \text{Na}_3\text{AlF}_6 \)) in Aluminium Extraction:** It is added to alumina to lower the melting temperature of the mixture from \( 2323\text{ K} \) to \( 1173\text{ K} \) and to enhance the electrical conductivity of the electrolyte during Hall-Heroult electrolysis.
In simple words: Carbon monoxide turns nickel into a gas for purification. Zinc acts as a chemical pusher to force dissolved silver out of solution. Silica reacts with iron waste to form liquid slag. Iodine converts titanium into a gas for thermal refining. Cryolite lowers alumina's melting point and helps it conduct electricity.

Exam Tip: Remember to list both the chemical role (e.g., reducing agent, flux, complexing agent) and write a corresponding equation for each sub-question.

Question. Explain the following (a) Generally sulphide ores are converted into oxides before reduction (b) Carbon and hydrogen are not used as reducing agent at high temperature (c) Silica is added to sulphide ore of copper in the reverberatory furnace (d) NaCN acts as a depressant in preventing ZnS from forming the froth (e) Role of cryolite in the metallurgy of aluminium
Answer:
(a) Sulphides are thermodynamically very stable compared to oxides, and their direct reduction is highly non-spontaneous. Oxides have a much lower free energy of formation and are far easier to reduce using standard reducing agents like carbon.
(b) At very high temperatures, carbon and hydrogen can react directly with the liberated metals to form undesirable metal carbides and metal hydrides respectively, contaminating the product.
(c) Silica is added to copper ores to act as an acidic flux. It binds with the basic iron oxide (\( \text{FeO} \)) impurities formed during roasting, turning them into fusible iron silicate slag (\( \text{FeSiO}_3 \)) that is easily skimmed off.
(d) Sodium cyanide (\( \text{NaCN} \)) acts as a depressant by reacting with \( \text{ZnS} \) to form a soluble complex, \( \text{Na}_2[\text{Zn}(\text{CN})_4] \), on its surface. This makes the zinc ore hydrophilic (water-loving), preventing it from attaching to oil bubbles and floating, while galena (\( \text{PbS} \)) floats unhindered.
(e) Cryolite lowers the melting point of the alumina electrolyte from \( 2323\text{ K} \) to \( 1173\text{ K} \), thereby saving energy, and also significantly increases the electrical conductivity of the molten bath.
In simple words: Oxides are much easier to reduce than sulfides. High temperatures cause carbon/hydrogen to react with metals to make unwanted carbides/hydrides. Silica removes iron waste from copper. Cyanide selectively blocks zinc from floating. Cryolite lowers the melting point of alumina and increases conductivity.

Exam Tip: Be sure to write the slag-formation and complex-formation chemical equations to support parts (c) and (d).

Question. (a) Describe the principle of froth floatation process. What is the role of depressant? Give an example. (b) Define leaching. How is this process used in the benefaction of silver and gold ores?
Answer:
(a) **Principle of Froth Flotation:** This process is based on the differential wetting characteristics of ore and gangue particles. Sulphide ore particles are preferentially wetted by pine oil (making them hydrophobic), while gangue particles are wetted by water. When compressed air is bubbled through the mixture, the oil-wetted ore particles attach to air bubbles and rise to the surface as froth, which is skimmed off.
**Role of Depressant:** A depressant is added to selectively prevent one sulphide mineral from floating while allowing another to rise with the froth. For example, sodium cyanide (\( \text{NaCN} \)) is used to separate \( \text{PbS} \) from \( \text{ZnS} \). It forms a soluble complex \( \text{Na}_2[\text{Zn}(\text{CN})_4] \) with zinc, keeping it in the aqueous layer, while \( \text{PbS} \) floats.
(b) **Leaching:** Leaching is a chemical concentration method where the powdered ore is treated with a suitable chemical reagent that selectively dissolves the valuable metal-bearing mineral, leaving insoluble impurities behind.
**Benefaction of Silver and Gold:** The powdered silver or gold ore is treated with a dilute solution of sodium cyanide (\( \text{NaCN} \)) in the presence of oxygen. The metal dissolves as a soluble cyano complex: \[ 4\text{Au} + 8\text{CN}^- + \text{O}_2 + 2\text{H}_2\text{O} \rightarrow 4[\text{Au}(\text{CN})_2]^- + 4\text{OH}^- \] This solution is filtered to remove insoluble gangue, and zinc metal is added to displace and recover the precious metal: \[ 2[\text{Au}(\text{CN})_2]^- + \text{Zn} \rightarrow [\text{Zn}(\text{CN})_4]^{2-} + 2\text{Au} \]
In simple words: Froth flotation uses oil to float sulphide ores while waste sinks in water. Depressants like cyanide block zinc from floating so lead can be collected. Leaching is chemically dissolving the valuable metal out of the ore, which is how cyanide dissolves gold before zinc is added to recover it.

Exam Tip: Structure your answer clearly using subheadings for part (a) and (b), and provide the complete equations for leaching of gold or silver.

Question. Write the chemical reaction which takes place in the following operations: (a) Electrolytic reduction of Alumina (b) Mond’s process (c) Van Arkel method (d) Mac Arthur forest cynide process (e) Electrolysis of brine
Answer: The chemical reactions are:
(a) **Electrolytic Reduction of Alumina (Hall-Heroult Process):** Ionization: \[ \text{Al}_2\text{O}_3 \rightarrow 2\text{Al}^{3+} + 3\text{O}^{2-} \] At Cathode: \[ \text{Al}^{3+} + 3\text{e}^- \rightarrow \text{Al}(\text{l}) \] At Anode: \[ \text{C} + \text{O}^{2-} \rightarrow \text{CO} + 2\text{e}^- \] \[ \text{C} + 2\text{O}^{2-} \rightarrow \text{CO}_2 + 4\text{e}^- \] Overall reaction: \[ 2\text{Al}_2\text{O}_3 + 3\text{C} \rightarrow 4\text{Al} + 3\text{CO}_2 \]
(b) **Mond's Process (Nickel):** \[ \text{Ni} + 4\text{CO} \xrightarrow{330\text{-}350\text{ K}} [\text{Ni}(\text{CO})_4] \] \[ [\text{Ni}(\text{CO})_4] \xrightarrow{450\text{-}470\text{ K}} \text{Ni} + 4\text{CO} \]
(c) **Van Arkel Method (Titanium):** \[ \text{Ti} + 2\text{I}_2 \xrightarrow{523\text{ K}} \text{TiI}_4 \] \[ \text{TiI}_4 \xrightarrow{1700\text{ K}} \text{Ti} + 2\text{I}_2 \]
(d) **MacArthur-Forrest Cyanide Process (\( \text{M} = \text{Ag} \text{ or } \text{Au} \default \)):** \[ 4\text{M} + 8\text{CN}^- + \text{O}_2 + 2\text{H}_2\text{O} \rightarrow 4[\text{M}(\text{CN})_2]^- + 4\text{OH}^- \] \[ 2[\text{M}(\text{CN})_2]^- + \text{Zn} \rightarrow [\text{Zn}(\text{CN})_4]^{2-} + 2\text{M} \]
(e) **Electrolysis of Brine:** \[ 2\text{NaCl}(\text{aq}) + 2\text{H}_2\text{O}(\text{l}) \xrightarrow{\text{electrolysis}} 2\text{NaOH}(\text{aq}) + \text{H}_2(\text{g}) + \text{Cl}_2(\text{g}) \]
In simple words: These chemical equations represent Hall-Heroult electrolysis, Mond's gas purification, Van Arkel iodide refining, cyanide extraction of gold/silver, and electrolysis of salt water.

Exam Tip: Be sure to write both the anode and cathode half-reactions for processes like Hall-Heroult and brine electrolysis to secure full marks.