(a)Give two reasons why aluminium is preferred to copper for making overhead electric cables.
(ii) Describe briefly the electrolytic extraction of aluminium from purified bauxite.
(b) The diagram below represents an electrolytic cell used for the purification of copper.
(i) Which of the electrodes I and II increases in mass during the electrolysis? Give reasons for your answers
(ii) State with reason the site of oxidation
(iii) Identify III and explain why its colour does not change in intensity during the electrolysis.
(c) Calculate the current in amperes that will deposit 8.00 g of calcium from used CaCl\(_2\) in 1 hour 15 minutes. [Ca = 40.0; 1 Faraday = 96500C]
(a)(i) Two reasons aluminium is preferred to copper for overhead cables: aluminium is much lighter (lower density), so the cables sag less and need fewer supports; aluminium is cheaper and more abundant than copper (while still being a good conductor).
(ii) Electrolytic extraction of aluminium: Purified alumina (Al2O3) is dissolved in molten cryolite (Na3AlF6) to lower the melting point and improve conductivity. The molten mixture is electrolyzed in a steel tank lined with graphite (the cathode), with carbon anodes.
Cathode: Al3+ + 3e- → Al (molten aluminium collects at the bottom and is tapped off).
Anode: 2O2- → O2 + 4e-; the oxygen burns the carbon anodes to CO2, so they are replaced from time to time.
(b) In copper purification the impure copper is the anode and pure copper is the cathode, in copper(II) tetraoxosulphate(VI) solution.
(i) The cathode (pure copper) increases in mass, because Cu2+ ions are discharged and deposited on it (Cu2+ + 2e- → Cu).
(ii) Oxidation occurs at the anode, because there copper atoms lose electrons to form ions (Cu → Cu2+ + 2e-).
(iii) III is the electrolyte, copper(II) sulphate solution. Its blue colour does not fade because the rate at which Cu2+ ions leave the solution at the cathode equals the rate at which Cu2+ ions enter the solution from the dissolving anode, so [Cu2+] stays constant.
(c) Deposit 8.00 g Ca in 1 h 15 min = 4500 s, Ca2+ + 2e- → Ca.
\[ n(\text{Ca})=\frac{8.00}{40}=0.200\ \text{mol};\quad n(e^-)=0.400\ \text{mol} \]
\[ Q=0.400\times96500=38600\ \text{C};\quad I=\frac{Q}{t}=\frac{38600}{4500}=8.58\ \text{A} \]
(a)(i) Two reasons aluminium is preferred to copper for overhead cables: aluminium is much lighter (lower density), so the cables sag less and need fewer supports; aluminium is cheaper and more abundant than copper (while still being a good conductor).
(ii) Electrolytic extraction of aluminium: Purified alumina (Al2O3) is dissolved in molten cryolite (Na3AlF6) to lower the melting point and improve conductivity. The molten mixture is electrolyzed in a steel tank lined with graphite (the cathode), with carbon anodes.
Cathode: Al3+ + 3e- → Al (molten aluminium collects at the bottom and is tapped off).
Anode: 2O2- → O2 + 4e-; the oxygen burns the carbon anodes to CO2, so they are replaced from time to time.
(b) In copper purification the impure copper is the anode and pure copper is the cathode, in copper(II) tetraoxosulphate(VI) solution.
(i) The cathode (pure copper) increases in mass, because Cu2+ ions are discharged and deposited on it (Cu2+ + 2e- → Cu).
(ii) Oxidation occurs at the anode, because there copper atoms lose electrons to form ions (Cu → Cu2+ + 2e-).
(iii) III is the electrolyte, copper(II) sulphate solution. Its blue colour does not fade because the rate at which Cu2+ ions leave the solution at the cathode equals the rate at which Cu2+ ions enter the solution from the dissolving anode, so [Cu2+] stays constant.
(c) Deposit 8.00 g Ca in 1 h 15 min = 4500 s, Ca2+ + 2e- → Ca.
\[ n(\text{Ca})=\frac{8.00}{40}=0.200\ \text{mol};\quad n(e^-)=0.400\ \text{mol} \]
\[ Q=0.400\times96500=38600\ \text{C};\quad I=\frac{Q}{t}=\frac{38600}{4500}=8.58\ \text{A} \]
