2025 H2 Chem P1 Q28 — Electrochemical cell Solution

The answer is C (\(\text{Sn}^{2+}\)).
Ni is oxidised to \(\text{Ni}^{2+}\) and the metal ion is reduced, so \(E^\ominus_\text{cell} = E^\ominus_\text{reduction} - E^\ominus(\text{Ni}^{2+}/\text{Ni})\), with \(E^\ominus(\text{Ni}^{2+}/\text{Ni}) = -0.25\) V. The reaction is feasible only if \(E^\ominus_\text{cell}\) is positive, i.e. the metal ion has an \(E^\ominus\) more positive than \(-0.25\) V.
• \(\text{Sn}^{2+}/\text{Sn}\): \(E^\ominus = -0.14\) V, so \(E^\ominus_\text{cell} = -0.14 - (-0.25) = +0.11\) V. Positive, so Ni reduces \(\text{Sn}^{2+}\) to Sn.
• A is wrong: \(E^\ominus(\text{Cr}^{2+}/\text{Cr}) = -0.91\) V, so \(E^\ominus_\text{cell} = -0.91 - (-0.25) = -0.66\) V, not feasible.
• B is wrong: \(E^\ominus(\text{Mn}^{2+}/\text{Mn}) = -1.18\) V, so \(E^\ominus_\text{cell} = -1.18 - (-0.25) = -0.93\) V, not feasible.
• D is wrong: \(E^\ominus(\text{V}^{2+}/\text{V}) = -1.20\) V, so \(E^\ominus_\text{cell} = -1.20 - (-0.25) = -0.95\) V, not feasible.