(a) (i) Name the ore mostly used in the extraction of aluminium.
(ii) Name two major impurities in the ore named in (a)(i).
(iii) Name the material used in making the electrodes in the extraction of aluminium.
(iv) Give two reasons why aluminium is commonly recycled.
(v) Explain briefly why the anode has to be replaced at regular intervals during the extraction of aluminium.
(b) A current of 0.75 amperes was passed through an electrolysis containing chromium ions for one hour and four minutes. If the mass of chromium deposited was 0.52 g, calculate the:
(i) quantity of electricity passed;
(ii) moles of chromium deposited;
(iii) quantity of electricity required to deposit one mole of chromium;
(iv) charge on the chromium ion.
[Cr = 52.0, 1 F = 96500 C]
(c) In the contact process for the manufacture of tetraoxosulphate(VI) acid, the following reaction occurs:
\[2\mathrm{SO}_{2(g)} + \mathrm{O}_{2(g)} \qquad 2\mathrm{SO}_{3(g)} \qquad H = -197\ \mathrm{kJ\ mol}^{-1}\]
(i) Name the catalyst used in the reaction;
(ii) State the optimum temperature for this reaction;
(iii) What would be the effect on the yield of \(\mathrm{SO}_3\) if a temperature higher than the optimum is used?
(d)(i) State two chemical methods by which temporary hardness of water can be removed.
(ii) Write a balanced chemical equation for each of the methods stated in (d)(i).
(a)(i) Bauxite (Al2O3·2H2O).
(a)(ii) Iron(III) oxide (Fe2O3) and silica (SiO2).
(a)(iii) Graphite (carbon).
(a)(iv) Recycling saves the large amount of electrical energy needed for extraction and conserves the limited bauxite reserves (also reduces waste/pollution).
(a)(v) The carbon anode reacts with the oxygen liberated at it, burning away as CO2, so it is gradually consumed and must be replaced.
(b) \(t = 1\,\text{h}\,4\,\text{min} = 3840\ \text{s}\).
(i) \( Q = It = 0.75 \times 3840 = 2880\ \text{C} \)
(ii) \( n(\text{Cr}) = \dfrac{0.52}{52} = 0.01\ \text{mol} \)
(iii) Quantity to deposit 1 mol = \( \dfrac{2880}{0.01} = 288000\ \text{C} \)
(iv) Number of Faradays per mole = \( \dfrac{288000}{96500} \approx 3 \), so the charge on the chromium ion is +3 (Cr3+).
(c)(i) Vanadium(V) oxide, V2O5.
(c)(ii) About 450 °C (400-450 °C).
(c)(iii) Since the forward reaction is exothermic, a temperature higher than the optimum decreases the yield of SO3 (equilibrium shifts backward).
(d)(i) & (ii) Removal of temporary hardness
- Boiling: \( \text{Ca(HCO}_3)_2 \to \text{CaCO}_3 + \text{H}_2\text{O} + \text{CO}_2 \)
- Adding slaked lime (Clark's method): \( \text{Ca(HCO}_3)_2 + \text{Ca(OH)}_2 \to 2\text{CaCO}_3 + 2\text{H}_2\text{O} \)
(a)(i) Bauxite (Al2O3·2H2O).
(a)(ii) Iron(III) oxide (Fe2O3) and silica (SiO2).
(a)(iii) Graphite (carbon).
(a)(iv) Recycling saves the large amount of electrical energy needed for extraction and conserves the limited bauxite reserves (also reduces waste/pollution).
(a)(v) The carbon anode reacts with the oxygen liberated at it, burning away as CO2, so it is gradually consumed and must be replaced.
(b) \(t = 1\,\text{h}\,4\,\text{min} = 3840\ \text{s}\).
(i) \( Q = It = 0.75 \times 3840 = 2880\ \text{C} \)
(ii) \( n(\text{Cr}) = \dfrac{0.52}{52} = 0.01\ \text{mol} \)
(iii) Quantity to deposit 1 mol = \( \dfrac{2880}{0.01} = 288000\ \text{C} \)
(iv) Number of Faradays per mole = \( \dfrac{288000}{96500} \approx 3 \), so the charge on the chromium ion is +3 (Cr3+).
(c)(i) Vanadium(V) oxide, V2O5.
(c)(ii) About 450 °C (400-450 °C).
(c)(iii) Since the forward reaction is exothermic, a temperature higher than the optimum decreases the yield of SO3 (equilibrium shifts backward).
(d)(i) & (ii) Removal of temporary hardness
- Boiling: \( \text{Ca(HCO}_3)_2 \to \text{CaCO}_3 + \text{H}_2\text{O} + \text{CO}_2 \)
- Adding slaked lime (Clark's method): \( \text{Ca(HCO}_3)_2 + \text{Ca(OH)}_2 \to 2\text{CaCO}_3 + 2\text{H}_2\text{O} \)
