2020 H2 Chem P1 Q28 — Electrolysis Solution

The answer is C (\(9.33 \times 10^{-3}\) mol).

At the anode, water is oxidised: \(2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4\text{e}^-\), so 4 mol of electrons are needed for each mol of \(\text{O}_2\).
\(Q = It = 20.0 \times (3.00 \times 60) = 3600\) C
\(n(\text{e}^-) = \frac{Q}{F} = \frac{3600}{96500} = 0.0373\) mol
\(n(\text{O}_2) = \frac{0.0373}{4} = 9.33 \times 10^{-3}\) mol

• A is wrong: \(1.55 \times 10^{-4}\) mol comes from leaving the time in minutes: \(\frac{20.0 \times 3.00}{96500} = 6.22 \times 10^{-4}\) mol of electrons, then dividing by 4.
• B is wrong: \(3.11 \times 10^{-4}\) mol comes from leaving the time in minutes and also dividing by 2 instead of 4: \(6.22 \times 10^{-4} \div 2\).
• D is wrong: \(1.87 \times 10^{-2}\) mol comes from dividing by 2 instead of 4 (\(0.0373 \div 2\)); this is the amount of \(\text{H}_2\) formed at the cathode (\(2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2\)), not the amount of \(\text{O}_2\) at the anode.