2021 H2 Chem P1 Q26 — Electrochemical cell Solution

The answer is D (using 1.0 mol dm\(^{-3}\) ethanoic acid instead of hydrochloric acid).

The copper electrode is the cathode (reduction) and the hydrogen electrode is the anode (oxidation):
\(E_\text{cell} = E_\text{reduction} - E_\text{oxidation} = E(\text{Cu}^{2+}|\text{Cu}) - E(\text{H}^+|\text{H}_2)\)
For \(E_\text{cell}\) to become more positive, either \(E(\text{Cu}^{2+}|\text{Cu})\) becomes more positive, or \(E(\text{H}^+|\text{H}_2)\) becomes more negative.

D: ethanoic acid is a weak acid, so 1.0 mol dm\(^{-3}\) ethanoic acid has a much lower [\(\text{H}^+\)] than 1.0 mol dm\(^{-3}\) HCl. The equilibrium \(2\text{H}^+ + 2\text{e}^- \rightleftharpoons \text{H}_2\) shifts left, \(E(\text{H}^+|\text{H}_2)\) becomes more negative and \(E_\text{cell}\) becomes more positive than +0.34 V.

• A is wrong: lowering \(p_{\text{H}_2}\) shifts \(2\text{H}^+ + 2\text{e}^- \rightleftharpoons \text{H}_2\) to the right, so \(E(\text{H}^+|\text{H}_2)\) becomes more positive and \(E_\text{cell}\) decreases.
• B is wrong: increasing [\(\text{H}^+\)] also shifts \(2\text{H}^+ + 2\text{e}^- \rightleftharpoons \text{H}_2\) to the right, so \(E(\text{H}^+|\text{H}_2)\) becomes more positive and \(E_\text{cell}\) decreases.
• C is wrong: 1.0 mol dm\(^{-3}\) copper(II) nitrate has the same [\(\text{Cu}^{2+}\)] as the standard copper(II) sulfate solution, and the anion takes no part in the electrode reaction, so \(E_\text{cell}\) does not change.