During electrolysis of an aqueous solution of sodium sulphate, 2.4L of oxygen at STP was liberated at anode. The volume of hydrogen at STP, liberated at cathode would be:
(a) 1.2L
(b) 2.4L
(c) 2.6L
(d) 4.8L

Hint: Electrolysis is a chemical process that uses DC current to drive an otherwise non-spontaneous chemical reaction. The anode is the positive terminal, while the cathode is the negative one. We equate the reactions in both to find the volume.

Complete step by step answer:
1: When aqueous solution of ${Na_2}S{O_4}$ is electrolyzed, the reaction occurring at the two electrodes are:
$
N{a_2}S{O_4} \to 2N{a^ + } + S{0_4}^{2 - } \\
{H_2}O \rightleftharpoons {H^ + } + O{H^ - } \\
$
2: Reaction at cathode:
$2{O^{2 - }} \to {O_2} + 4{e^ - }$
Reaction at anode:
$4{H^ + } + 2{e^ - } \to {H_2}$
3: So, as we can see in the anode, the charge released is double to that of cathode.
So, according to Faraday’s law:
$W = ZQ$
where, Q is the charge
W is the mass
Z is the constant of proportionality
4: Thus, it is evident that moles of ${H_2}$ and ${O_2}$ liberated in the negative electrodes are in the ratio 2:1; so will be the volume at STP.
5: Therefore, the hydrogen obtained at STP at cathode will be
$
 2.4 \times 2L \\
 = 4.8L \\
 $ at STP
Thus, option (d) is correct.

Note:
Students must first write down the equations at the cathode and anode and then check if the charges in both are equal or not. Also remember that in an electrolytic cell the anode is positive and the cathode is negative. Here, 2.4L is obtained for 2 ${e^2}$, so for 4${e^2}$ the hydrogen obtained at the cathode has to be twice this amount. The 4${e^2}$ when balanced in both the anode and cathode equations then get cancelled out and the final products at both the electrodes can be obtained. Electrolysis is commercially important as a stage in the separation of elements from naturally occurring sources such as ores using an electrolytic cell.