2021 H2 Chem P1 Q28 — Electrolysis Solution

The answer is C (0.89 g).

At the anode, both Cu and Ni in the alloy are oxidised to \(\text{Cu}^{2+}\) and \(\text{Ni}^{2+}\).
At the cathode, the solution contains \(\text{Cu}^{2+}\) and \(\text{Ni}^{2+}\) (1.0 mol dm\(^{-3}\) each). \(E^\circ(\text{Cu}^{2+}|\text{Cu}) = +0.34\) V is more positive than \(E^\circ(\text{Ni}^{2+}|\text{Ni}) = -0.25\) V, so \(\text{Cu}^{2+}\) is preferentially reduced and only Cu is deposited; \(\text{Ni}^{2+}\) stays in solution.

\(Q = I \times t = 0.50 \times 1.5 \times 60 \times 60 = 2700\) C
\(n_{\text{e}^-} = \frac{Q}{F} = \frac{2700}{96\,500} = 0.027979\) mol
\(\text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu}\), so \(n_{\text{Cu}}\) deposited = \(\frac{1}{2} \times 0.027979 = 0.013990\) mol
Increase in mass of cathode = \(0.013990 \times 63.5 = 0.888\) g = 0.89 g

• A is wrong: 0.82 g = \(0.013990 \times 58.7\), the mass if Ni were deposited instead of Cu. \(\text{Ni}^{2+}\) is harder to reduce than \(\text{Cu}^{2+}\), so no Ni is deposited.
• B is wrong: 0.87 g = \(0.75 \times 0.888 + 0.25 \times 0.821\), which assumes the cathode deposit has the same 75% Cu, 25% Ni composition as the anode. The composition of the anode does not decide what is deposited; only Cu is deposited.
• D is wrong: 1.8 g = \(0.027979 \times 63.5\), which uses the moles of electrons as the moles of Cu. Each \(\text{Cu}^{2+}\) needs 2 electrons, so the moles of Cu are half the moles of electrons.