2017 H2 Chem P1 Q17 — Electrolysis Solution

The answer is A (60 \(\text{cm}^3\)).
Cathode: \(2\text{CH}_2{=}\text{CHCN} + 2\text{H}_2\text{O} + 2\text{e}^- \rightarrow \text{NC(CH}_2)_4\text{CN} + 2\text{OH}^-\), so 1 mol of electrons is used per mol of acrylonitrile.
Anode (oxygen is produced from water): \(2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4\text{e}^-\), so 1 mol of \(\text{O}_2\) releases 4 mol of electrons.
\(n(\text{e}^-) = n(\text{acrylonitrile}) = 0.01\) mol
\(n(\text{O}_2) = \dfrac{0.01}{4} = 0.0025\) mol
\(V(\text{O}_2) = 0.0025 \times 24\,000 = 60 \text{ cm}^3\) (molar volume at r.t.p. = 24.0 \(\text{dm}^3 \text{ mol}^{-1}\))
The overall equation is \(4\text{CH}_2{=}\text{CHCN} + 2\text{H}_2\text{O} \rightarrow 2\text{NC(CH}_2)_4\text{CN} + \text{O}_2\), which also gives 1 mol \(\text{O}_2\) per 4 mol acrylonitrile.
• B is wrong: 120 \(\text{cm}^3\) is 0.005 mol \(\text{O}_2\), which would mean only 2 mol of electrons per mol of \(\text{O}_2\). Forming one \(\text{O}_2\) from water releases 4 electrons.
• C is wrong: 240 \(\text{cm}^3\) is 0.01 mol \(\text{O}_2\), one \(\text{O}_2\) per acrylonitrile, which ignores the 4 : 1 electron ratio.
• D is wrong: 480 \(\text{cm}^3\) is 0.02 mol \(\text{O}_2\), which would need 0.08 mol of electrons, eight times the 0.01 mol actually transferred.