A-LevelChemistryElectrochemistry9701/41 · May/June 2018 · Q311 Marks

Complete the table, identifying the substance liberated at each electrode during electrolysis with inert electrodes.

(a)Complete the table, identifying the substance liberated at each electrode during electrolysis with inert electrodes. electrolyte | substance liberated at the anode | substance liberated at the cathode --- | --- | --- AgNO3(aq) | | concentrated NaCl(aq) | | CuSO4(aq) | | [3]
(b)(i)Molten calcium iodide, CaI2, is electrolysed in an inert atmosphere with inert electrodes. Write ionic equations for the reactions occurring at the electrodes.[2]
(b)(ii)The electrolysis of molten CaI2 is a redox process. Identify the ion that is oxidised and the ion that is reduced, explaining your answer by reference to oxidation numbers.[2]
(b)(iii)Describe two visual observations that would be made during this electrolysis.[1]
(c)An oxide of iron dissolved in an inert solvent is electrolysed for 2.00 hours using a current of 0.800A. The electrolysis products are iron and oxygen. The mass of iron produced is 1.11g. Calculate the oxidation number of Fe in the oxide of iron. Show all your working.[3]

Mark scheme

Marking points from the official mark scheme (May/June 2018). The mark or mark type is shown in brackets.

[1] AgNO3(aq) anode: oxygen / O2, cathode: silver / Ag
[1] saturated NaCl(aq) anode: chlorine / Cl2, cathode: hydrogen / H2
[1] CuSO4(aq) anode: oxygen / O2, cathode: copper / Cu
[1] 2I− → I2 + 2e−
[1] Ca2+ + 2e− → Ca
[1] Ca / Calcium reduced and I / iodine oxidised
[1] Oxidation number of calcium decreases from 2 to 0 Oxidation number of iodine increases from −1 to 0
[1] metal / grey / silvery
[1] purple AND vapour / gas / fumes
[1] amount of melt decreases
[1] 2 × 60 × 60 × 0.8 = 5760 C AND 5760 / 96500 = 0.060 (0.0597) F
[1] 1.11 / 55.8 = 0.020 (0.0199) mol of Fe
[1] 0.06 / 0.02 = 3 ∴ Fe3+ or +3 or 3

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