Q: Question 2Extraction of Aluminium Step 1 : Dressing of the ore chemical method Baeyer's process - Step 1, Step 2, Step 3 14. Bauxite to - sodium aluminateAl2O3.2H2O+NaOH→150−200°C...............+...............\text{Al}_2\text{O}_3.2\text{H}_2\text{O} + \text{NaOH} \xrightarrow{150-200 \degree\text{C}} ............... + ...............Al2O3.2H2O+NaOH150−200°C...............+...............15. Sodium aluminate to - aluminium hydroxideNaAlO2+H2O→50−60°C...............+...............↓\text{NaAlO}_2 + \text{H}_2\text{O} \xrightarrow{50-60 \degree\text{C}} ............... + ............... \downarrowNaAlO2+H2O50−60°C...............+...............↓16. Aluminium hydroxide to pure aluminaAl(OH)3→1100°C...............+...............\text{Al(OH)}_3 \xrightarrow{1100 \degree\text{C}} ............... + ...............Al(OH)31100°C...............+...............Step 2: Conversion of conc. ore to its oxide Step not necessary since the ore is already an oxide Step 3: Reduction of metallic oxide to its metal Hall Heroult's Process - Reduction by electrolysis 17. Electrolytic reduction of pure aluminaAl2O3 ⇌ ............... + ...............18. At cathode2Al3+ + 6e- ⟶ ...............19. At anode3O2- - 6e- ⟶ 3[...............] ⟶ 3O2Step 4: Electrorefining of aluminium [reference]

Extraction of Aluminium Step 1 : Dressing of the ore chemical method Baeyer's process - Step 1, Step 2, Step 3 14. Bauxite to - sodium aluminateAl2O3.2H2O+2NaOH→150−200°C2NaAlO2+3H2O\text{Al}_2\text{O}_3.2\text{H}_2\text{O} + \text{2NaOH} \xrightarrow{150-200 \degree\text{C}} \bold{2NaAlO}_\bold{2} + \bold{3H}_\bold{2}\bold{{O}}Al2O3.2H2O+2NaOH150−200°C2NaAlO2+3H2O15. Sodium aluminate to - aluminium hydroxideNaAlO2+2H2O→50−60°CNaOH+Al(OH)3↓\text{NaAlO}_2 + \text{2H}_2\text{O} \xrightarrow{50-60 \degree\text{C}} \bold{NaOH} + \bold{Al(OH)}_\bold{3}\downarrowNaAlO2+2H2O50−60°CNaOH+Al(OH)3↓16. Aluminium hydroxide to pure alumina2Al(OH)3→1100°CAl2O3+3H2O\text{2Al(OH)}_3 \xrightarrow{1100 \degree\text{C}} \bold{Al}_\bold{2}\bold{O}_\bold{3} + \bold{3H}_\bold{2}\bold{{O}}2Al(OH)31100°CAl2O3+3H2OStep 2: Conversion of conc. ore to its oxide Step not necessary since the ore is already an oxide Step 3: Reduction of metallic oxide to its metal Hall Heroult's Process - Reduction by electrolysis 17. Electrolytic reduction of - pure aluminaAl2O3 ⇌ 2Al3+ + 3O2-18. At cathode2Al3+ + 6e- ⟶ 2Al19. At anode3O2- - 6e- ⟶ 3[O] ⟶ 3O2Step 4: Electrorefining of aluminium [reference]

Q: Question 6Compare roasting and calcination.

RoastingCalcinationThe ore is heated in the excess air.The ore is heated in the absence of airGenerally sulphide ores are roasted and hence sulphur dioxide is given out.Generally carbonate and hydrated ores are calcinated and hence carbon dioxide and water vapour is given off.2ZnS+3O2→800−900°C2ZnO+2SO2[g]2\text{ZnS} + 3\text{O}_2 \xrightarrow{800 - 900 \degree\text{C}} \text{2ZnO} + \text{2SO}_\text{2}[\text{g}]2ZnS+3O2800−900°C2ZnO+2SO2[g]ZnCO3→ZnO+CO2[g]\text{ZnCO}_3 \rightarrow \text{ZnO} + \text{CO}_\text{2} [\text{g}]ZnCO3→ZnO+CO2[g]Volatile impurities are removed as oxides (SO2, P2O5, As2O3) and the ore becomes porous and more reactive.Moisture and organic impurities are removed and the ore becomes porous and more reactive.

Q: Question 8(a) Some metallic oxides can be reduced by hydrogen, carbon, carbon monoxide and some cannot. Explain.(b) Write balanced equation for the reduction of copper (II) oxide by hydrogen.

(a) The metals in the middle of the activity series such as iron, lead, copper, etc., are moderately reactive and are found as sulphides or carbonates in nature. They are obtained by the reduction of their oxides with carbon, carbon monoxide and hydrogen.While, oxides of highly active metals like sodium and potassium have great affinity for oxygen and so cannot be reduced by common reducing agents like carbon, carbon monoxide or hydrogen.(b) CuOblack+H2→ΔCupink/brown+H2O\underset{\text{black}}{\text{CuO}} + \text{H}2 \xrightarrow{\Delta} \underset{\text{pink/brown}}{\text{Cu}} + \text{H}{2}\text{{O}}blackCuO+H2Δpink/brownCu+H2O

Q: Question 9How are the following metallic oxides reduced. Write equations :(a) Iron (II) oxide(b) Zinc oxide

(a) Iron (II) oxide is reduced by CO:FeO+CO→ΔFe+CO2[g]\text{FeO} + \text{CO} \xrightarrow{\Delta} \text{Fe} + \text{CO}_{2} [\text{g}]FeO+COΔFe+CO2[g](b) Zinc oxide is reduced by coke :ZnO+C→1400°CZn+CO[g]\text{ZnO} + \text{C} \xrightarrow{1400 \degree\text{C}} \text{Zn} + \text{CO} [\text{g}]ZnO+C1400°CZn+CO[g]

Q: Question 10State why aluminium is extracted from it's oxide by electrolysis while copper, lead, iron by reducing agents and mercury and silver by thermal decomposition.

Oxides of highly active metals like potassium, sodium, calcium, magnesium and aluminium have great affinity towards oxygen and so cannot be reduced by common reducing agents like carbon, carbon monoxide or hydrogen.They are obtained by electrolytic reduction of fused metallic salts (halides and oxides) using inert electrodes.The metals in the middle of the activity series such as iron, lead, copper, etc., are moderately reactive and are found as sulphides or carbonates in nature. They are obtained by the reduction of their oxides with carbon, carbon monoxide and hydrogen.Mercury and silver are very less reactive, placed in the end of the reactivity series. The oxides of these metals are reduced to metals by heating alone.

Question 1

Question 6In the extraction of aluminium from bauxite, the first step is the dressing of the ore by Baeyer's process. Give balanced equations for the conversion of impure bauxite to pure alumina using a concentrated solution of NaOH.

Accepted Answer

Conversion of impure bauxite to sodium aluminate:Al2O3.2H2O [Impure bauxite]Amphoteric oxide+2NaOH [conc. soln]Base→150−200°C2NaAlO2[sodium aluminate]Salt+3H2OWater\underset{	ext{Amphoteric oxide}}{\underset{	ext{[Impure bauxite]}}{	ext{Al}_2	ext{O}_3.2	ext{H}_2	ext{O }}} + \underset{	ext{Base}}{\underset{	ext{[conc. soln]}}{2	ext{NaOH }}} \xrightarrow{150 - 200 \degree	ext{C}} \underset{	ext{Salt}}{\underset{	ext{[sodium aluminate]}}{2	ext{NaAlO}_2}} + \underset{	ext{Water}}{ 	ext{3H}_2	ext{O}}Amphoteric oxide[Impure bauxite]Al2​O3​.2H2​O ​​+Base[conc. soln]2NaOH ​​150−200°C​Salt[sodium aluminate]2NaAlO2​​​+Water3H2​O​Conversion of sodium aluminate to aluminium hydroxide:NaAlO2[sodium aluminate]+2H2O→50−60°CNaOH[sodium hydroxide]+Al(OH)3[aluminium hydroxide]\underset{	ext{[sodium aluminate]}}{	ext{NaAlO}_2} + 2	ext{H}_2	ext{O} \xrightarrow{50 - 60 \degree	ext{C}} \underset{	ext{[sodium hydroxide]}}{	ext{NaOH}} + \underset{	ext{[aluminium hydroxide]}}{	ext{Al(OH)}_3}[sodium aluminate]NaAlO2​​+2H2​O50−60°C​[sodium hydroxide]NaOH​+[aluminium hydroxide]Al(OH)3​​Conversion of Al(OH)3 - to pure alumina - heat on Al(OH)3:2Al(OH)3[aluminium hydroxide]→1100°CAl2O3[pure alumina]+3H2O\underset{	ext{[aluminium hydroxide]}}{	ext{2Al(OH)}_3} \xrightarrow{1100\degree	ext{C}} \underset{	ext{[pure alumina]}}{	ext{Al}_2	ext{O}_3} + 	ext{3H}_2	ext{O}[aluminium hydroxide]2Al(OH)3​​1100°C​[pure alumina]Al2​O3​​+3H2​O

Question 2

Question 7In the electrolytic reduction of pure alumina to pure aluminium by Hall Herault's process, give the electrolytic reactions involved in the same, resulting in formation of aluminium at the cathode.

Accepted Answer

In the electrolytic reduction of pure alumina to pure aluminium by Hall Herault's process, the constitutents of electrolyte are cryolite (Na3AlF6) and fluorspar (CaF2) and alumina (Al2O3)The electrolytic reactions are as follows :Cryolite : Na3AlF6 ⇌ 3Na1+ + Al3+ + 6F1-Fluorspar : CaF2 ⇌ Ca2+ + 2F1-Alumina : Al2O3 ⇌ 2Al3+ + 3O2-At cathode : 2Al3+ + 6e- ⟶ 2AlAt anode : 3O2- - 6e- ⟶ 3[O] ⟶ 3O2Al3+ are discharged in preference to Na1+ and Ca2+ ions due to it's lower position in the electrochemical series.

Question 3

Question 8State the function of (i) Sodium hydroxide (ii) Cryolite (iii) Fluorspar in the metallurgy of aluminium in Hall Herault's process.

Accepted Answer

(i) Function of Sodium hydroxide — Sodium hydroxide is added to bauxite ore during purification of bauxite. Bauxite is reacted with a conc. solution of Sodium hydroxide under pressure for 2 hours as a first step in obtaining Al2O3.The impurities present in bauxite mainly Fe2O3 and SiO2 remain unaffected with conc. NaOH as impurities are not amphoteric.Bauxite, being amphoteric reacts with the base forming sodium salt [sodium aluminate] and water. Hence, impurities are separated out.Al2O3.2H2O [Impure bauxite]Amphoteric oxide+2NaOH [conc. soln]Base→150−200°C2NaAlO2[sodium aluminate]Salt+3H2OWater\underset{	ext{Amphoteric oxide}}{\underset{	ext{[Impure bauxite]}}{	ext{Al}_2	ext{O}_3.2	ext{H}_2	ext{O }}} + \underset{	ext{Base}}{\underset{	ext{[conc. soln]}}{2	ext{NaOH }}} \xrightarrow{150 - 200 \degree	ext{C}} \underset{	ext{Salt}}{\underset{	ext{[sodium aluminate]}}{2	ext{NaAlO}_2}} + \underset{	ext{Water}}{ 	ext{3H}_2	ext{O}}Amphoteric oxide[Impure bauxite]Al2​O3​.2H2​O ​​+Base[conc. soln]2NaOH ​​150−200°C​Salt[sodium aluminate]2NaAlO2​​​+Water3H2​O​(ii) Addition of Cryolite & FluorsparLowers the fusion point of the mixture i.e., mixture fuses around 950°C instead of 2050°C.Enhances the mobility of the fused mixture by acting as a solvent for the electrolytic mixture.Addition of Cryolite enhances the conductivity of the mixture since, pure alumina is almost a non-conductor of electricity.

Question 4

Question 9Give reasons for the following pertaining to Hall Herault's process.(i) The fusion temperature of the electrolyte has to be lowered before conducting the electrolytic reduction.(ii) The constituents of the electrolyte in addition to one part of fused alumina contains three parts of cryolite and one part of fluorspar.(iii) A layer of powdered coke sprinkled over the electrolytic mixture, protects the carbon electrodes.(iv) It is preferred to use a number of graphite electrodes as anode, instead of a single graphite electrode.

Accepted Answer

(i) Fused alumina has a melting point around 2050°C. Such high temperatures, create the below difficulties:A large amount of electrical energy is required to liberate Al2O3 and maintain the electrolyte in molten state at that temperature.The liberated Al metal [m.p. 660°C] may also tend to volatilize out and get wasted.Hence the fusion temperature of the electrolytic mixture has to be lowered.(ii) The reasons for addition of mainly Cryolite [in a higher ratio] and Fluorspar to the electrolytic mixture are:They lower the fusion point of the mixture i.e., the mixture fuses [melts] around 950°C instead of 2050°C.They enhance the mobility of the fused mixture by acting as a solvent for the electrolytic mixture. Thus cryolite in the molten state of subdivision dissolves aluminium oxide.Addition of cryolite enhances the conductivity of the mixture. Since pure alumina is almost a non-conductor of electricity.(iii) The layer of powdered coke is sprinkled over the electrolytic mixture as:It prevents burning of carbon electrodes in air at the emergence point from the bath.It minimizes or prevents heat loss by radiation.(iv) The oxygen evolved at the anode escapes as a gas or reacts with the carbon anode. The carbon anode is thus oxidized to carbon monoxide which either burns giving carbon dioxide or escapes out through an outlet. [2C + O2 ⟶ 2CO; 2CO + O2 ⟶ 2CO2] The carbon anode is hence consumed and renewed periodically after a certain period of usage. It is therefore preferable to use a number of graphite electrodes as anode instead of a single electrode.

Question 5

Question 11State (i) composition (ii) reason for alloying (iii) one use of each of the following alloys.(i) Brass(ii) Bronze(iii) Duralumin(iv) Solder [fuse metal](v) Stainless steel

Accepted Answer

AlloyCompositionReason for alloyingUse1.BrassCu [60 - 80%], Zn [40 - 20%]Hard, Lustrous, easily cast, malleable, ductileElectrical fittings, decorative articles2.BronzeCu [80%], Zn [1%], Sn [19%]Hard, brittle, takes up polishStatues, medals, coins3.DuraluminAl [95%], Mg [0.5%], Mn [0.5%], Cu [4%]Al - imparts lightness, Mg - imparts strengthAircrafts, light tools, cooker4.Solder [fuse metal]Pb [50%], Sn [50%]Sn - lowers melting point of alloyElectrical fuse, Soldering purposes5.Stainless steelFe [75%], Ni [10%], Cr [15%], C [.5%]Ni, Cr - imparts lustre, C - imparts hardnessUtensils, cutlery, surgical instruments

Question 6

Question 1Extraction of Metals — General details of stages involvedComplete and balance the equationStep 1 : Dressing of the ore   Step 2 : Conversion of conc. ore to its oxide   Roasting of sulphide ore   1. Roasting of zinc blendeZnS+O2→800°C...............+...............[g]	ext{ZnS} + 	ext{O}_2 \xrightarrow{800 \degree	ext{C}} ............... + ............... [	ext{g}]ZnS+O2​800°C​...............+...............[g]2. Roasting of iron pyriteFeS2+O2→...............+...............[g]	ext{FeS}_2 + 	ext{O}_2 ightarrow ............... + ............... [	ext{g}]FeS2​+O2​→...............+...............[g]Calcination of carbonate ore   3. Calcination of zinc carbonateZnCO3→<400°C...............+...............[g]	ext{ZnCO}_3 \xrightarrow{\lt 400 \degree	ext{C}} ............... + ............... [	ext{g}]ZnCO3​<400°C​...............+...............[g]4. Calcination of iron [II] carbonateFeCO3→...............+...............[g]	ext{FeCO}_3 ightarrow ............... + ............... [	ext{g}]FeCO3​→...............+...............[g]Step 3: Reduction of metallic oxide to its oxide Reduction by electrolysis 5. Electrolytic reduction of pure aluminaAl2O3 ⇌ ............... + ...............6. At cathode2Al3+ + 6e- ⟶ ...............Reduction by reducing agents [C, CO, H2] 7. Reduction of zinc oxide - by cokeZnO+C→Δ...............+...............[g]	ext{ZnO} + 	ext{C} \xrightarrow{\Delta} ............... + ............... [	ext{g}]ZnO+CΔ​...............+...............[g]8. Reduction of iron [III] oxide - by COFe2O3+CO→Δ...............+...............[g]	ext{Fe}_2	ext{O}_3 + 	ext{CO} \xrightarrow{\Delta} ............... + ............... [	ext{g}]Fe2​O3​+COΔ​...............+...............[g]9. Reduction of lead [II] oxide - by cokePbO+C→Δ...............+...............[g]	ext{PbO} + 	ext{C} \xrightarrow{\Delta} ............... + ............... [	ext{g}]PbO+CΔ​...............+...............[g]10. Reduction of copper [II] oxide - by cokeCuO+C→Δ...............+...............[g]	ext{CuO} + 	ext{C} \xrightarrow{\Delta} ............... + ............... [	ext{g}]CuO+CΔ​...............+...............[g]11. Reduction of copper [II] oxide - by hydrogenCuO+H2→Δ...............+...............[g]	ext{CuO} + 	ext{H}_2 \xrightarrow{\Delta} ............... + ............... [	ext{g}]CuO+H2​Δ​...............+...............[g]Reduction by thermal decomposition12. Thermal decomposition of mercury [II] oxideHgO→Δ...............+...............[g]	ext{HgO} \xrightarrow{\Delta} ............... + ............... [	ext{g}]HgOΔ​...............+...............[g]13. Thermal decomposition of silver [I] oxideAg2O→Δ...............+...............[g]	ext{Ag}_2	ext{O} \xrightarrow{\Delta} ............... + ............... [	ext{g}]Ag2​OΔ​...............+...............[g]Step 4: Electrorefining of impure metals

Accepted Answer

Extraction of Metals — General details of stages involvedComplete and balance the equationStep 1 : Dressing of the ore   Step 2 : Conversion of conc. ore to its oxide   Roasting of a sulphide ore   1. Roasting of zinc blende2ZnS+3O2→800°C2ZnO+2SO2[g]2	ext{ZnS} + 3	ext{O}_2 \xrightarrow{800 \degree	ext{C}} \bold{2ZnO} + \bold{2SO}_\bold{2}[	ext{g}]2ZnS+3O2​800°C​2ZnO+2SO2​[g]2. Roasting of iron pyrite4FeS2+11O2→2Fe2O3+8SO2[g]4	ext{FeS}_2 + 11	ext{O}_2 ightarrow \bold{2Fe}_\bold{2}\bold{O}_\bold{3} + \bold{8SO}_\bold{2} [	ext{g}]4FeS2​+11O2​→2Fe2​O3​+8SO2​[g]Calcination of a carbonate ore   3. Calcination of zinc carbonateZnCO3→<400°CZnO+CO2[g]	ext{ZnCO}_3 \xrightarrow{\lt 400 \degree	ext{C}} \bold{ZnO} + \bold{CO}_\bold{2} [	ext{g}]ZnCO3​<400°C​ZnO+CO2​[g]4. Calcination of iron [II] carbonateFeCO3→FeO+CO2[g]	ext{FeCO}_3 ightarrow \bold{FeO} + \bold{CO}_\bold{2} [	ext{g}]FeCO3​→FeO+CO2​[g]Step 3: Reduction of metallic oxide to its oxide Reduction by electrolysis 5. Electrolytic reduction of pure aluminaAl2O3 ⇌ 2Al3+ + 3O2-6. At cathode2Al3+ + 6e- ⟶ 2AlReduction by reducing agents [C, CO, H2] 7. Reduction of zinc oxide - by cokeZnO+C→ΔZn+CO[g]	ext{ZnO} + 	ext{C} \xrightarrow{\Delta} \bold{Zn} + \bold{CO} [	ext{g}]ZnO+CΔ​Zn+CO[g]8. Reduction of iron [III] oxide - by COFe2O3+3CO→Δ2Fe+3CO2[g]	ext{Fe}_2	ext{O}_3 + 3	ext{CO} \xrightarrow{\Delta} \bold{2Fe} + \bold{3CO}_\bold{2} [	ext{g}]Fe2​O3​+3COΔ​2Fe+3CO2​[g]9. Reduction of lead [II] oxide - by coke2PbO+C→Δ2Pb+CO2[g]2	ext{PbO} + 	ext{C} \xrightarrow{\Delta} \bold{2Pb} + \bold{CO}_\bold{2} [	ext{g}]2PbO+CΔ​2Pb+CO2​[g]10. Reduction of copper [II] oxide - by cokeCuO+C→ΔCu+CO[g]	ext{CuO} + 	ext{C}\xrightarrow{\Delta} \bold{Cu} + \bold{CO} [	ext{g}]CuO+CΔ​Cu+CO[g]11. Reduction of copper [II] oxide - by hydrogenCuO+H2→ΔCu+H2O[g]	ext{CuO} + 	ext{H}_2 \xrightarrow{\Delta} \bold{Cu} + \bold{H}_\bold{2}\bold{{O}} [	ext{g}]CuO+H2​Δ​Cu+H2​O[g]Reduction by thermal decomposition 12. Thermal decomposition of mercury [II] oxide2HgO→Δ2Hg+O2[g]2	ext{HgO} \xrightarrow{\Delta} \bold{2Hg} + \bold{O}_\bold{2} [	ext{g}]2HgOΔ​2Hg+O2​[g]13. Thermal decomposition of silver [I] oxide2Ag2O→Δ4Ag+O2[g]2	ext{Ag}_2	ext{O} \xrightarrow{\Delta} \bold{4Ag} + \bold{O}_\bold{2} [	ext{g}]2Ag2​OΔ​4Ag+O2​[g]Step 4: Electrorefining of impure metals

Question 7

Question 2Extraction of Aluminium Step 1 : Dressing of the ore chemical method Baeyer's process - Step 1, Step 2, Step 3 14. Bauxite to - sodium aluminateAl2O3.2H2O+NaOH→150−200°C...............+...............	ext{Al}_2	ext{O}_3.2	ext{H}_2	ext{O} + 	ext{NaOH} \xrightarrow{150-200 \degree	ext{C}} ............... + ...............Al2​O3​.2H2​O+NaOH150−200°C​...............+...............15. Sodium aluminate to - aluminium hydroxideNaAlO2+H2O→50−60°C...............+...............↓	ext{NaAlO}_2 + 	ext{H}_2	ext{O} \xrightarrow{50-60 \degree	ext{C}} ............... + ............... \downarrowNaAlO2​+H2​O50−60°C​...............+...............↓16. Aluminium hydroxide to pure aluminaAl(OH)3→1100°C...............+...............	ext{Al(OH)}_3 \xrightarrow{1100 \degree	ext{C}} ............... + ...............Al(OH)3​1100°C​...............+...............Step 2: Conversion of conc. ore to its oxide   Step not necessary since the ore is already an oxide   Step 3: Reduction of metallic oxide to its metal   Hall Heroult's Process - Reduction by electrolysis   17. Electrolytic reduction of pure aluminaAl2O3 ⇌ ............... + ...............18. At cathode2Al3+ + 6e- ⟶ ...............19. At anode3O2- - 6e- ⟶ 3[...............] ⟶ 3O2Step 4: Electrorefining of aluminium [reference]

Accepted Answer

Extraction of Aluminium Step 1 : Dressing of the ore chemical method Baeyer's process - Step 1, Step 2, Step 3 14. Bauxite to - sodium aluminateAl2O3.2H2O+2NaOH→150−200°C2NaAlO2+3H2O	ext{Al}_2	ext{O}_3.2	ext{H}_2	ext{O} + 	ext{2NaOH} \xrightarrow{150-200 \degree	ext{C}} \bold{2NaAlO}_\bold{2} + \bold{3H}_\bold{2}\bold{{O}}Al2​O3​.2H2​O+2NaOH150−200°C​2NaAlO2​+3H2​O15. Sodium aluminate to - aluminium hydroxideNaAlO2+2H2O→50−60°CNaOH+Al(OH)3↓	ext{NaAlO}_2 + 	ext{2H}_2	ext{O} \xrightarrow{50-60 \degree	ext{C}} \bold{NaOH} + \bold{Al(OH)}_\bold{3}\downarrowNaAlO2​+2H2​O50−60°C​NaOH+Al(OH)3​↓16. Aluminium hydroxide to pure alumina2Al(OH)3→1100°CAl2O3+3H2O	ext{2Al(OH)}_3 \xrightarrow{1100 \degree	ext{C}} \bold{Al}_\bold{2}\bold{O}_\bold{3} + \bold{3H}_\bold{2}\bold{{O}}2Al(OH)3​1100°C​Al2​O3​+3H2​OStep 2: Conversion of conc. ore to its oxide   Step not necessary since the ore is already an oxide   Step 3: Reduction of metallic oxide to its metal   Hall Heroult's Process - Reduction by electrolysis   17. Electrolytic reduction of - pure aluminaAl2O3 ⇌ 2Al3+ + 3O2-18. At cathode2Al3+ + 6e- ⟶ 2Al19. At anode3O2- - 6e- ⟶ 3[O] ⟶ 3O2Step 4: Electrorefining of aluminium [reference]

Question 8

Question 6Compare roasting and calcination.

Accepted Answer

RoastingCalcinationThe ore is heated in the excess air.The ore is heated in the absence of airGenerally sulphide ores are roasted and hence sulphur dioxide is given out.Generally carbonate and hydrated ores are calcinated and hence carbon dioxide and water vapour is given off.2ZnS+3O2→800−900°C2ZnO+2SO2[g]2	ext{ZnS} + 3	ext{O}_2 \xrightarrow{800 - 900 \degree	ext{C}} 	ext{2ZnO} + 	ext{2SO}_	ext{2}[	ext{g}]2ZnS+3O2​800−900°C​2ZnO+2SO2​[g]ZnCO3→ZnO+CO2[g]	ext{ZnCO}_3 ightarrow 	ext{ZnO} + 	ext{CO}_	ext{2} [	ext{g}]ZnCO3​→ZnO+CO2​[g]Volatile impurities are removed as oxides (SO2, P2O5, As2O3) and the ore becomes porous and more reactive.Moisture and organic impurities are removed and the ore becomes porous and more reactive.

Question 9

Question 8(a) Some metallic oxides can be reduced by hydrogen, carbon, carbon monoxide and some cannot. Explain.(b) Write balanced equation for the reduction of copper (II) oxide by hydrogen.

Accepted Answer

(a) The metals in the middle of the activity series such as iron, lead, copper, etc., are moderately reactive and are found as sulphides or carbonates in nature. They are obtained by the reduction of their oxides with carbon, carbon monoxide and hydrogen.While, oxides of highly active metals like sodium and potassium have great affinity for oxygen and so cannot be reduced by common reducing agents like carbon, carbon monoxide or hydrogen.(b) CuOblack+H2→ΔCupink/brown+H2O\underset{ ext{black}}{ ext{CuO}} + ext{H}2 \xrightarrow{\Delta} \underset{ ext{pink/brown}}{ ext{Cu}} + ext{H}{2} ext{{O}}blackCuO+H2Δpink/brownCu+H2O

Question 10

Question 9How are the following metallic oxides reduced. Write equations :(a) Iron (II) oxide(b) Zinc oxide

Accepted Answer

(a) Iron (II) oxide is reduced by CO:FeO+CO→ΔFe+CO2[g]	ext{FeO} + 	ext{CO} \xrightarrow{\Delta} 	ext{Fe} + 	ext{CO}_{2} [	ext{g}]FeO+COΔ​Fe+CO2​[g](b) Zinc oxide is reduced by coke :ZnO+C→1400°CZn+CO[g]	ext{ZnO} + 	ext{C} \xrightarrow{1400 \degree	ext{C}} 	ext{Zn} + 	ext{CO} [	ext{g}]ZnO+C1400°C​Zn+CO[g]

Question 11

Question 10State why aluminium is extracted from it's oxide by electrolysis while copper, lead, iron by reducing agents and mercury and silver by thermal decomposition.

Accepted Answer

Oxides of highly active metals like potassium, sodium, calcium, magnesium and aluminium have great affinity towards oxygen and so cannot be reduced by common reducing agents like carbon, carbon monoxide or hydrogen.They are obtained by electrolytic reduction of fused metallic salts (halides and oxides) using inert electrodes.The metals in the middle of the activity series such as iron, lead, copper, etc., are moderately reactive and are found as sulphides or carbonates in nature. They are obtained by the reduction of their oxides with carbon, carbon monoxide and hydrogen.Mercury and silver are very less reactive, placed in the end of the reactivity series. The oxides of these metals are reduced to metals by heating alone.

Question 12

Question 12(a) Define roasting. Name an ore on which roasting is done. Give balanced equation.(b) Define calcination. Give an example and equation for calcination.

Accepted Answer

(a) Roasting — It is the process of heating the concentrated ore to a high temperature in the presence of excess air.Example — Zinc blende (ZnS) is roasted to yield it's oxide, zinc oxide2ZnS+3O2→800°C2ZnO+2SO22 ext{ZnS} + 3 ext{O}2 \xrightarrow{800 \degree ext{C}} ext{2ZnO} + ext{2SO}{2}2ZnS+3O2800°C2ZnO+2SO2(b) Calcination — If an ore is carbonate or a hydrated oxide, it is heated in the absence of air to a temperature that is high but insufficient to melt the ore. The process is known as Calcination.Example — carbonate and hydrated ores are calcinated like ZnCO3ZnCO3→ZnO+CO2[g] ext{ZnCO}3 ightarrow ext{ZnO} + ext{CO}{2} [ ext{g}]ZnCO3→ZnO+CO2[g]

Question 13

Question 13How are the metals like sodium, potassium and calcium obtained. Give equations.

Accepted Answer

Oxides of highly active metals like sodium, potassium and calcium are obtained by electrolytic reduction of fused metallic salts (halides and oxides) using inert electrodes.SodiumElectrolyte : Fused sodium chlorideReaction : NaCl ⇌ Na+ + Cl-Reaction at cathode : Na+ + e- ⟶ NaReaction at anode : Cl- - e- ⟶ ClCl + Cl ⟶ Cl2PotassiumElectrolyte : Fused potassium bromideReaction : KBr ⇌ K+ + Br-Reaction at cathode : K+ + e- ⟶ KReaction at anode : Br- - e- ⟶ BrBr + Br ⟶ Br2CalciumElectrolyte : Fused calcium chlorideReaction : CaCl2 ⇌ Ca2+ + 2Cl-Reaction at cathode : Ca2+ + 2e- ⟶ CaReaction at anode : Cl- - e- ⟶ ClCl + Cl ⟶ Cl2

Question 14

Question 14Give equations for the reduction of :(a) Iron (II) oxide(b) Iron (III) oxide(c) Lead (II) oxide(d) Zinc oxide

Accepted Answer

(a) Reduction of Iron (II) oxideFeO+C→ΔFe+CO ext{FeO} + ext{C} \xrightarrow{\Delta} ext{Fe} + ext{CO}FeO+CΔFe+COFeO+CO→ΔFe+CO2 ext{FeO} + ext{CO} \xrightarrow{\Delta} ext{Fe} + ext{CO}_2FeO+COΔFe+CO2FeO+H2→ΔFe+H2O{ ext{FeO}} + ext{H}2 \xrightarrow{\Delta} { ext{Fe}} + ext{H}{2} ext{{O}}FeO+H2ΔFe+H2O(b) Reduction of Iron (III) oxideFe2O3+2Al→2Fe+Al2O3+heat ext{Fe}_2 ext{O}_3 + 2 ext{Al} ightarrow ext{2Fe} + ext{Al}_2 ext{O}_3 + ext{heat}Fe2O3+2Al→2Fe+Al2O3+heat(c) Reduction of Lead (II) oxidePbO+C→ΔPb+CO ext{PbO} + ext{C} \xrightarrow{\Delta} ext{Pb} + ext{CO}PbO+CΔPb+COPbO+CO→ΔPb+CO2 ext{PbO} + ext{CO} \xrightarrow{\Delta} ext{Pb} + ext{CO}_2PbO+COΔPb+CO2PbOyellow+H2→ΔPbgreenish white+H2O\underset{ ext{yellow}}{ ext{PbO}} + ext{H}2 \xrightarrow{\Delta} \underset{ ext{greenish white}}{ ext{Pb}} + ext{H}{2} ext{{O}}yellowPbO+H2Δgreenish whitePb+H2O(d) Reduction of Zinc oxideZnO+C→1400°CZn+CO ext{ZnO} + ext{C} \xrightarrow{1400\degree ext{C}} ext{Zn} + ext{CO}ZnO+C1400°CZn+CO

Question 15

Question 3In order to obtain 1 tonne of aluminium, the following inputs are required: 4 tonnes of bauxite, 150 kg of sodium hydroxide and 600 kg of graphite. The aluminium compound in bauxite is aluminium oxide and the main impurity is iron (III) oxide. Aluminium is obtained by the electrolysis of aluminium oxide dissolved in cryolite.(a) When bauxite is treated with sodium hydroxide solution, what happens to(i) the aluminium oxide(ii) the iron (III) oxide(b) (i) Name the process used for the purification of bauxite.(ii) Write the equation for the action of heat on aluminium hydroxide.(c) (i) Write the formula of cryolite.(ii) Write down the word which correctly completes the following sentence:By dissolving aluminium oxide in cryolite a ............... (conducting/non-conducting) solution is produced.(iii) Why is so much graphite required for the electrolytic process?(iv) Write the equation for the reaction which takes place at the cathode(v) What is the cathode made up of?

Accepted Answer

(a) Aluminium oxide dissolves and forms sodium meta aluminate. Iron (III) oxide is left behind as red mud and is removed by filtration.Al2O3.2H2O + 2NaOH ⟶ 2NaAlO2 + 3H2O(b) (i) Bayer's process(ii) The equation for the action of heat on aluminium hydroxide is:2Al(OH)3→1000°CAl2O3+3H2O2	ext{Al(OH)}_3 \xrightarrow{1000 \degree	ext{C}} 	ext{Al}_2	ext{O}_3 + 3	ext{H}_2	ext{O}2Al(OH)3​1000°C​Al2​O3​+3H2​O(c) (i) Na3AlF6(ii) By dissolving aluminium oxide in cryolite, a conducting solution is produced.(iii) Thick Graphite rods attached to copper clamps dipping into fused electrolyte are used as anode. The graphite (anode) is oxidized by oxygen to CO and further forms CO2, so it is consumed and has to be replaced from time to time. Hence, large amount of graphite is required.Equation:2C + O2 ⟶ 2CO2CO + O2 ⟶ 2CO2(iv) Reaction at the cathode:4Al3+ + 12e- ⟶ 4Al(v) Inner carbon lining of the electrolytic cell

Question 16

Question 4Aluminium is extracted from it's chief ore, bauxite. The ore is first purified and then the metal is extracted from it by electrolytic reduction.(a) Write three balanced equations for the purification of bauxite.(b) Name a chemical used for dissolving aluminium oxide. In which state is the chemical used?(c) Write an equation for the reaction which takes place at the anode during the extraction of aluminium by the electrolytic process.

Accepted Answer

(a) The three balanced equations for the purification of bauxite are:Al2O3.2H2O + 2NaOH ⟶ 2NaAlO2 + 3H2ONaAlO2 + 2H2O ⟶ NaOH + Al(OH)3↓2Al(OH)3 →1000°Cheat\xrightarrow[1000 \degree	ext{C}]{	ext{heat}}heat1000°C​ Al2O3 + 3H2O(b) The chemical used for dissolving aluminium oxide is cryolite. It is used in molten state.(c) At anode:6O2- - 12e- ⟶ 6[O]3[O] + 3[O] ⟶ 3O2Anode is oxidised to carbon monoxide which further forms carbon dioxide2C + O2 ⟶ 2CO2CO + O2 ⟶ 2CO2

Question 17

Question 11Assertion (A): Sodium and potassium are reactive elements.Reason (R): Sodium and potassium occur in free statesBoth A and R are true and R is the correct explanation of A.Both A and R are true but R is not the correct explanation of A.A is true but R is false.A is false but R is true.

Accepted Answer

A is true but R is false.Explanation — Sodium and potassium are highly reactive in nature. Hence, assertion (A) is true. Sodium and potassium are so reactive. So, they do not occur in free states. They quickly combine with other elements. Hence reason (R) is false.

Question 18

Question 19Assertion (A): For electrical heating devices like electric iron and heater, alloys are used.Reason (R): Alloys have a greater resistance and low melting point.Both A and R are true and R is the correct explanation of A.Both A and R are true but R is not the correct explanation of A.A is true but R is false.A is false but R is true.

Accepted Answer

Both A and R are true but R is not the correct explanation of A.Explanation — Alloys are used for heating devices like electric iron and heater because of its high resistance to electricity. Hence, assertion (A) is true. Generally alloys have great resistance to electricity. Thus, they can produce more heat and alloys also have lower melting point than its individual component. Hence, reason (R) is true.However, in case of electrical iron and heater, alloys have high melting point to withstand the heat, Hence, reason (R) is not correct explanation of assertion (A).

Question 19

Question 5The following questions are relevant to the extraction of Aluminium :(a) State the reason for addition of caustic alkali to bauxite ore during purification of bauxite.(b) Give a balanced chemical equation for the above reaction.(c) Along with cryolite and alumina, another substance is added to the electrolyte mixture. Name the substance and give one reason for the addition.

Accepted Answer

(a) Bauxite is reacted with a conc. solution of NaOH (caustic alkali) under pressure for 2 hrs as a first step in obtaining Al2O3.The impurities present in bauxite mainly Fe2O3 and SiO2 remain unaffected with conc. NaOH as impurities are not amphoteric.Bauxite, being amphoteric reacts with the base forming salt [sodium aluminate] and water. Hence, impurities are separated out. Therefore, caustic alkali is added to bauxite ore during purification of bauxite.(b) Balanced chemical equation for the conversion of impure bauxite to sodium aluminate is given below:Al2O3.2H2O [Impure bauxite]Amphoteric oxide+2NaOH [conc. soln]Base→150−200°C2NaAlO2[sodium aluminate]Salt+3H2OWater\underset{	ext{Amphoteric oxide}}{\underset{	ext{[Impure bauxite]}}{	ext{Al}_2	ext{O}_3.2	ext{H}_2	ext{O }}} + \underset{	ext{Base}}{\underset{	ext{[conc. soln]}}{2	ext{NaOH }}} \xrightarrow{150 - 200 \degree	ext{C}} \underset{	ext{Salt}}{\underset{	ext{[sodium aluminate]}}{2	ext{NaAlO}_2}} + \underset{	ext{Water}}{ 	ext{3H}_2	ext{O}}Amphoteric oxide[Impure bauxite]Al2​O3​.2H2​O ​​+Base[conc. soln]2NaOH ​​150−200°C​Salt[sodium aluminate]2NaAlO2​​​+Water3H2​O​(c) Fluorspar is added to the electrolyte mixture as it lowers the fusion point of the mixture i.e., the mixture fuses around 950°C instead of 2050°C.

Question 20

Question 7Write the equation for the reaction where the aluminium oxide for the electrolytic extraction of aluminium is obtained by heating aluminium hydroxide.

Accepted Answer

Conversion of Al(OH)3 - to pure alumina - heat on Al(OH)32Al(OH)3[aluminium hydroxide]→1100°CAl2O3[pure alumina]+3H2O\underset{	ext{[aluminium hydroxide]}}{	ext{2Al(OH)}_3} \xrightarrow{1100\degree	ext{C}} \underset{	ext{[pure alumina]}}{	ext{Al}_2	ext{O}_3} + 	ext{3H}_2	ext{O}[aluminium hydroxide]2Al(OH)3​​1100°C​[pure alumina]Al2​O3​​+3H2​O

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