2024 H2 Chem P1 Q27 — Electrolysis Solution

The answer is C (1.97 g).
\(Q = It = 5.00 \times (20.0 \times 60) = 6000\) C
\(n(e^-) = \frac{Q}{F} = \frac{6000}{96500} = 0.06218\) mol
At the cathode: \(\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)\), so 2 mol of electrons deposit 1 mol of Cu.
\(n(\text{Cu}) = \frac{0.06218}{2} = 0.03109\) mol
Mass of Cu \(= 0.03109 \times 63.5 = 1.97\) g
• A is wrong: 0.0329 g comes from using the time in minutes (20.0) instead of seconds.
• B is wrong: 0.902 g is only \(\frac{0.902}{63.5} = 0.0142\) mol of Cu, which would need just \(0.0142 \times 2 \times 96500 = 2740\) C, less than half of the 6000 C passed.
• D is wrong: 3.95 g comes from forgetting that each \(\text{Cu}^{2+}\) needs 2 electrons (not dividing by 2).