In 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

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2.4 L

2.6 L

4.8 L

1.2 L

answer is D.

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Detailed Solution

$\begin{array}{l} {Na}_{2} {SO}_{4} \to 2 {Na}^{+} + {SO}_{4}^{2 -} \\ {Anode: 2H}_{2} O \to 4 e^{-} + 4 H^{+} + O_{2} 2 .4 L \\ {Cathode: 2H}^{+} + 2 e^{-} \to H_{2} \\ 4 H^{+} + 4 e^{-} \to 2 H_{2} \\ {Total H}_{2} gas leberated = 2 \times 2.4=4.8 L \end{array}$

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