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

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Solution

The ions present in aqueous sodium suphate are $N a^{+}, S O_{4}^{2 -}, H^{+} and O H^{-}$

$N a^{+} and H^{+}$ migrates to the cathode $S O_{4}^{2 -} and O H^{-}$ migrates to the anode.
$E_{H_{2} O / H_{2}}^{0} = - 0.83 V$
$E_{N a^{+} / N a}^{0} = - 2.71 V$

Reduction potential of $H^{+}$ is more than $N a^{+}$ so $H^{+}$ get reduced to $H_{2}$ which is liberated at cathode.
$At cathode : 2 H_{2} O (l) + 2 e^{-} ⇌ H_{2} (g) + 2 O H^{-} (a q)$

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