A solution of sodium sulphate in water is electrolysed using platinum electrodes. The products at cathode and anode are respectively:

H_{2}, O_{2}

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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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Solution

The correct option is A $H_{2}, O_{2}$

At cathode (negative electrode) reduction occurs,
$2 H^{+} + 2 e^{-} ⟶ H_{2} ↑$
At anode (positive electrode) oxidation occurs.
$2 O^{2 -} ⟶ O_{2} + 4 e^{-}$
(Refer to Image)

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