A solution of sodium sulphate was electrolyzed using some inert electrode. The products at the electrodes are:

O_{2}, H_{2}

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O_{2}, N a

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O_{2}, S O_{2}

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O_{2}, S_{2} O_{8}^{2 -}

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Solution

The correct option is A $O_{2}, H_{2}$
Since the reduction potential of $H_{2} O$ is greater than reduction potential of $N a^{\oplus} s o H_{2} O$ undergoes reduction to give
$H_{2} (g)$ at cathode.
$H_{2} O + e^{-} \to ⊖ O H + \frac{1}{2} H_{2} (g)$
Similarly, the oxidation potential of $H_{2} O$ is greater than the oxidation potential of $S O_{4}^{2 -}$ ion. So, $H_{2} O$ undergoes oxidation to give $O_{2} (g)$ at anode.
$H_{2} O \to 2 H^{\oplus} + 2 e^{-} + \frac{1}{2} O_{2} (g)$

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