2020 H2 Chem P1 Q27 — Electrochemical cell Solution

The answer is A (X to Y; 0.40 V).

The electrolyte is 1 \(\text{mol dm}^{-3}\) NaOH(aq), so the alkaline half-equations from the Data Booklet are used:
\(2\text{H}_2\text{O} + 2\text{e}^- \rightleftharpoons \text{H}_2 + 2\text{OH}^-\) \(E^\ominus = -0.83\) V (hydrogen electrode, X)
\(\text{O}_2 + 2\text{H}_2\text{O} + 4\text{e}^- \rightleftharpoons 4\text{OH}^-\) \(E^\ominus = +0.40\) V (oxygen electrode, Y)

The half-cell with the more positive \(E^\ominus\) (oxygen) is reduced, and the other (hydrogen) is oxidised:
At X: \(\text{H}_2 + 2\text{OH}^- \rightarrow 2\text{H}_2\text{O} + 2\text{e}^-\) (anode, releases electrons)
At Y: \(\text{O}_2 + 2\text{H}_2\text{O} + 4\text{e}^- \rightarrow 4\text{OH}^-\) (cathode, takes in electrons)
So electrons flow in the external circuit from X to Y, and \(E^\ominus_\text{cell} = +0.40 - (-0.83) = +1.23\) V.

• B is wrong: the direction is correct, but +1.23 V is the \(E^\ominus\) for \(\text{O}_2 + 4\text{H}^+ + 4\text{e}^- \rightleftharpoons 2\text{H}_2\text{O}\) in acidic solution; in NaOH(aq) the oxygen electrode is \(\text{O}_2/\text{OH}^-\), +0.40 V.
• C is wrong: the \(E^\ominus\) value is correct, but electrons flow from the hydrogen electrode (where \(\text{H}_2\) is oxidised) to the oxygen electrode, that is X to Y, not Y to X.
• D is wrong: both the direction of electron flow and the \(E^\ominus\) value are wrong, for the reasons given in B and C.