2021 H2 Chem P1 Q29 — Electrochemical cell Solution

The answer is C (\(\text{VO}_2^+\) is reduced all the way to \(\text{V}^{2+}\)).

Data Booklet values:
\(E^\circ(\text{Zn}^{2+}|\text{Zn}) = -0.76\) V
\(E^\circ(\text{VO}_2^+|\text{VO}^{2+}) = +1.00\) V
\(E^\circ(\text{VO}^{2+}|\text{V}^{3+}) = +0.34\) V
\(E^\circ(\text{V}^{3+}|\text{V}^{2+}) = -0.26\) V
\(E^\circ(\text{V}^{2+}|\text{V}) = -1.20\) V

With Zn as the reductant, \(E^\circ_\text{cell} = E^\circ(\text{V couple}) - (-0.76)\):
• \(\text{VO}_2^+ \rightarrow \text{VO}^{2+}\): \(+1.00 + 0.76 = +1.76\) V, feasible
• \(\text{VO}^{2+} \rightarrow \text{V}^{3+}\): \(+0.34 + 0.76 = +1.10\) V, feasible
• \(\text{V}^{3+} \rightarrow \text{V}^{2+}\): \(-0.26 + 0.76 = +0.50\) V, feasible
• \(\text{V}^{2+} \rightarrow \text{V}\): \(-1.20 + 0.76 = -0.44\) V, not feasible
With excess zinc and all feasible reactions occurring quickly, V is reduced from +5 to +2 and stops at \(\text{V}^{2+}\).

Half-equations: \(\text{VO}_2^+ + 4\text{H}^+ + 3\text{e}^- \rightarrow \text{V}^{2+} + 2\text{H}_2\text{O}\) (×2) and \(\text{Zn} \rightarrow \text{Zn}^{2+} + 2\text{e}^-\) (×3)
Overall: \(3\text{Zn} + 8\text{H}^+ + 2\text{VO}_2^+ \rightarrow 3\text{Zn}^{2+} + 2\text{V}^{2+} + 4\text{H}_2\text{O}\)

• A is wrong: it stops at \(\text{VO}^{2+}\) (+4), but excess Zn can also reduce \(\text{VO}^{2+}\) to \(\text{V}^{3+}\) (\(E^\circ_\text{cell}\) = +1.10 V) and on to \(\text{V}^{2+}\).
• B is wrong: it stops at \(\text{V}^{3+}\) (+3), but Zn can also reduce \(\text{V}^{3+}\) to \(\text{V}^{2+}\) (\(E^\circ_\text{cell}\) = +0.50 V).
• D is wrong: reducing \(\text{V}^{2+}\) to V by Zn has \(E^\circ_\text{cell}\) = \(-0.44\) V, so it is not feasible and no vanadium metal forms.