Question

When dil. $H_{2}S{O}_4$ is electrolysed by using platinum electrodes the gas evolved at cathode is

• A. $O_2$
• B. $S{O}_2$
• C. $S{O}_3$
• D. $H_2$

Answer 1

The correct option is D $H_2$

When $\text{dil.}{H}_{2}S{O}_4$ is electrolysed, the possible ions in the solution were $H^{+},O{H}^{-}\text{and}S{O}_4^{2-}.$
From the electrochemical series we can find that the reduction potential $H^{+}$ is higher than other ions in the solution so it will get reduce to $H_2$ molecule and liberated at the cathode. Also the oxidation potential of $O{H}^{-}$ is higher than $S{O}_4^2-$ so it will oxidise to $O_2$ and liberate at the anode.
Reaction at Cathode:
$2H^{+}+2e^{-}\to H_2$
Reaction at Anode:
$2O{H}^{-}\to H_{2}O+\frac{1}{2}{O}_2+2e^{-}$