2024 H2 Chem P1 Q30 — Electrochemical cell Solution

The answer is B (+3).
Data Booklet values:
\(\text{SO}_4^{2-} + 4\text{H}^+ + 2e^- \rightleftharpoons \text{SO}_2 + 2\text{H}_2\text{O}\), \(E^\ominus = +0.17\) V
\(\text{VO}_2^+ + 2\text{H}^+ + e^- \rightleftharpoons \text{VO}^{2+} + \text{H}_2\text{O}\), \(E^\ominus = +1.00\) V
\(\text{VO}^{2+} + 2\text{H}^+ + e^- \rightleftharpoons \text{V}^{3+} + \text{H}_2\text{O}\), \(E^\ominus = +0.34\) V
\(\text{V}^{3+} + e^- \rightleftharpoons \text{V}^{2+}\), \(E^\ominus = -0.26\) V
\(\text{SO}_2\) is oxidised, so \(E^\ominus_\text{cell} = E^\ominus_\text{red} - E^\ominus_\text{ox} = E^\ominus(\text{V step}) - (+0.17)\). A step is feasible if \(E^\ominus_\text{cell} > 0\).
• \(\text{VO}_2^+\) (+5) to \(\text{VO}^{2+}\) (+4): \(E^\ominus_\text{cell} = +1.00 - 0.17 = +0.83\) V, feasible.
• \(\text{VO}^{2+}\) (+4) to \(\text{V}^{3+}\) (+3): \(E^\ominus_\text{cell} = +0.34 - 0.17 = +0.17\) V, feasible.
• \(\text{V}^{3+}\) (+3) to \(\text{V}^{2+}\) (+2): \(E^\ominus_\text{cell} = -0.26 - 0.17 = -0.43\) V, not feasible.
So excess \(\text{SO}_2\) reduces vanadium from +5 to +3 and stops there.
• A is wrong: the reduction of \(\text{V}^{3+}\) to \(\text{V}^{2+}\) is not feasible.
• C is wrong: \(\text{SO}_2\) is in excess and can also reduce \(\text{VO}^{2+}\) to \(\text{V}^{3+}\), so vanadium does not stop at +4.
• D is wrong: +5 is the starting oxidation state; \(\text{SO}_2\) does reduce \(\text{VO}_2^+\).