During electrolysis of an aqueous solution of sodium sulphate, 2.4 L of oxygen at STP was liberated at anode. The volume of hydrogen at STP, liberated at cathode would be

4.8 L

During the electrolysis of an aqueous solution of \(Na_2SO_4\), the following reactions occur at the electrodes.

At Anode: \(4OH^− \rightarrow 2H_2O  +  O_2  +  4e^−\)

At Cathode: \(2H^+  +  2e^− \rightarrow H_2\)

Number of Faradays needed to liberate 1 mole of \(O_2\) = \(4\)

∴ Charge =\(\frac{2.4 × 4}{22.4}F\)

Number of moles of \(H_2\) liberated by 1F of charge = \(\frac{1}{2}\)

Number of moles of \(H_2\) =  \(\frac{2.4 × 4}{2 ×22.4}\)

Volume of \(H_2\) at STP = \(\frac{2.4 × 4 × 22.4}{2 ×22.4} = 4.8 L \)