Question

Which gas is evolved at cathode on electrolysis of aqueous sodium sulphate using platinum electrodes?

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

Answer 1

The correct option is A $H_2$

The ions present in aqueous sodium suphate are $N{a}^{+},S{O}_4^{2-},H^{+}\text{and}O{H}^{-}$

$N{a}^{+}\text{and}{H}^{+}$ migrates to the cathode $S{O}_4^{2-}\text{and}O{H}^{-}$ migrates to the anode.
$E_{H_{2}O/H_2}^0=-0.83V$
$E_{N{a}^{+}/Na}^0=-2.71V$

Reduction potential of $H^{+}$ is more than $N{a}^{+}$ so $H^{+}$ get reduced to $H_2$ which is liberated at cathode.
$\text{At cathode :}2H_{2}O\left(l\right)+2e^{-}\rightleftharpoons H_2\left(g\right)+2O{H}^{-}\left(aq\right)$