2017 H2 Chem P1 Q29 — Electrolysis Solution

The answer is C (current and time only).
At the cathode \(\text{Cu}^{2+}(\text{aq}) + 2\text{e}^- \rightarrow \text{Cu(s)}\), so the amount of copper deposited is set by the amount of electrons (charge) passed.
\(Q = It\) and \(n(\text{e}^-) = \dfrac{Q}{F}\), so \(n(\text{Cu}) = \dfrac{It}{2F}\) and the number of Cu atoms \(= \dfrac{It}{2F} \times L\).
F, L and the charge on the ion (2) are constants, so the number of atoms depends only on the current \(I\) and the time \(t\). The concentration of \(\text{Cu}^{2+}\) does not appear: as long as \(\text{Cu}^{2+}\) ions are present at the cathode, each 2 mol of electrons deposits 1 mol of Cu.
• A is wrong: it ticks \([\text{Cu}^{2+}]\), which does not affect the amount deposited, and crosses time, which does (a longer time passes more charge).
• B is wrong: it ticks \([\text{Cu}^{2+}]\) and crosses both current and time, which are the two factors that set the charge passed.
• D is wrong: it crosses current. A larger current passes more charge in the same time, so more Cu is deposited.