2024 H2 Chem P1 Q28 — Electrochemical cell Solution

The answer is B (1 and 2 only).
Data Booklet values:
\(\text{O}_2 + 4\text{H}^+ + 4e^- \rightleftharpoons 2\text{H}_2\text{O}\), \(E^\ominus = +1.23\) V
\(\text{Br}_2 + 2e^- \rightleftharpoons 2\text{Br}^-\), \(E^\ominus = +1.07\) V
\(E^\ominus_\text{cell} = +1.23 - (+1.07) = +0.16\) V, so the oxidation of \(\text{Br}^-\) by \(\text{O}_2\) is feasible under standard conditions (1 bar gas, 1 mol \(\text{dm}^{-3}\) solutes, 298 K). The question asks what could stop it being observed.
• Statement 1 is a possible reason: air contains only about 21% oxygen, so the oxygen pressure is well below 1 atm. A lower [\(\text{O}_2\)] shifts the position of equilibrium of \(\text{O}_2 + 4\text{H}^+ + 4e^- \rightleftharpoons 2\text{H}_2\text{O}\) to the left, so \(E(\text{O}_2/\text{H}_2\text{O})\) becomes less positive than +1.23 V and \(E_\text{cell}\) becomes smaller than +0.16 V, making the reaction less feasible.
• Statement 2 is a possible reason: \(E^\ominus\) values show only whether a reaction is energetically feasible, not how fast it goes. With a high activation energy the reaction can be too slow to observe.
• Statement 3 is not a reason: a \([\text{Br}^-]\) much larger than 1 mol \(\text{dm}^{-3}\) shifts the position of equilibrium of \(\text{Br}_2 + 2e^- \rightleftharpoons 2\text{Br}^-\) to the left, making \(E(\text{Br}_2/\text{Br}^-)\) less positive and \(E_\text{cell}\) more positive, so the reaction becomes more feasible, not less.
• A is wrong: it includes statement 3, which would favour the reaction.
• C is wrong: it leaves out statement 1, which is a valid reason.
• D is wrong: statement 3 is not a reason, and statements 1 and 2 are left out.