Answer: C. For dilute NaCl(aq), \(\text{Cl}^-\) is too dilute to be preferentially discharged; instead \(\text{OH}^-\)/water is oxidised at the positive electrode to give oxygen.
Cathode: \(2\text{H}^+ + 2e^- \rightarrow \text{H}_2\)
Anode: \(4\text{OH}^- \rightarrow \text{O}_2 + 2\text{H}_2\text{O} + 4e^-\)
Electron balance: 4 mol electrons → 2 mol \(\text{H}_2\) and 1 mol \(\text{O}_2\). Mole ratio \(\text{H}_2 : \text{O}_2 = 2 : 1\). At the same T and p, equal moles of gas occupy equal volumes, so the volume ratio is also 2:1. \(V(\text{O}_2) = V(\text{H}_2)/2 = 25.0/2 = 12.5\) cm³.
So 12.5 cm³ of oxygen — option C. Options A and B name chlorine, which isn't produced from a dilute chloride solution. D has the right gas but wrong volume (it ignores the 2:1 stoichiometry).
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