Redox and electrochemistry — Question 24

24a)

MnO₄ ⁻ (aq) (+1.51 V) is a stronger oxidising agent than Cu²⁺(aq) (+0.34 V) because it has a more positive standard potential. MnO₄ ⁻ (aq) is preferentially reduced and gains electrons from Cu(s), which is oxidised. Therefore, electrons flow from Cu electrode to Pt electrode.

• correctly identifies MnO₄ ⁻ as the stronger oxidising agent [1 mark] • correctly identifies that MnO₄ ⁻ gains electrons [1 mark] • determines that electrons flow from Cu to Pt electrode [1 mark]

Acceptable responses are: - MnO₄ ⁻ is reduced - Cu is oxidised - Cu loses electrons. Acceptable responses are that electrons flow from: - Cu electrode to Pt electrode - Cu(s) to MnO₄ ⁻ (aq).

24b)

$\mathrm{E}^{\mathrm{o}}_{\mathrm{cell}}\,=\,\mathrm{E}^{\mathrm{o}}_{\mathrm{red}}\,–\,\mathrm{E}^{\mathrm{o}}_{\mathrm{ox}}\,=\,+1.51\,–\,(+0.34)\,=\,1.17\,\mathrm{V}\,$

• determines that the standard reduction potential is 1.17 [1 mark]

Do not penalise for incorrect decimal places/significant figures.

24c)

Cu²⁺(aq) concentration will increase and the solution will become darker. MnO₄ ^–(aq) concentration will decrease and the solution will become lighter.

• predicts that the Cu²⁺(aq) solution will become darker [1 mark] • predicts that the MnO₄ ^–(aq) solution will become lighter [1 mark]

Acceptable responses are: - Cu electrode will become smaller - Cu²⁺ solution will become darker - MnO₄ ^– solution will become lighter - MnO₄ ^– solution will become colourless. Allow FT error from 24a).