During electrolysis of an aqueous solution of sodium sulphate, 2.4 L…

The correct answer is Option (4) → 4.8 L

In the electrolysis of aqueous sodium sulphate, water is both reduced and oxidised. At the cathode, water is reduced in preference to \( \ce{Na^+} \), whose reduction potential is far more negative. At the anode, water is oxidised in preference to \( \ce{SO4^2-} \).

Written for the same 4 electrons:

    Cathode: \( \displaystyle \ce{4H2O(l) + 4e^- -> 2H2(g) + 4OH^-(aq)} \)
    Anode: \( \displaystyle \ce{2H2O(l) -> O2(g) + 4H^+(aq) + 4e^-} \)
    Net reaction: \( \displaystyle \ce{2H2O(l) -> 2H2(g) + O2(g)} \)

The same charge passes through both electrodes, so 2 mol of \( \ce{H2} \) form for every 1 mol of \( \ce{O2} \). At the same temperature and pressure, volume is proportional to moles:

    \( \displaystyle V_{\ce{H2}}=2\times V_{\ce{O2}} \)

    \( \displaystyle V_{\ce{H2}}=2\times2.4=4.8\,\mathrm{L} \)

The volume of hydrogen liberated at the cathode is 4.8 L.