Question

On electrolysis of aqueous sodium sulphate using platinum electrodes, the product formed at cathode and anode are respectively:

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

Answer 1

The correct option is C $H_2,O_2$

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 $H_2$ and liberated at cathode.
$\text{At cathode :}2H_{2}O\left(l\right)+2e^{-}\rightleftharpoons H_2\left(g\right)+2O{H}^{-}\left(aq\right)$

At Anode:
Possible reactions:
$2S{O}_4^{2-}\left(aq\right)\rightleftharpoons S_2{O}_8^{2-}\left(aq\right)+2e^{-};E^0=-2.0V$

$2H_{2}O\left(l\right)\rightleftharpoons 4H^{+}\left(aq\right)+O_2\left(g\right)+4e^{-}$
$E_{H_{2}O/O_2}^0=-1.23V$

Hence, $H_{2}O$ has higher oxidation potential so it will undergo oxidation at anode and liberates $O_2$.