A solution of $N a_{2} S O_{4}$ in water is electrolysed using inert electrodes. The products at the cathode and anode respectively, are:

O_{2}, H_{2}

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O_{2}, N a

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H_{2}, O_{2}

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O_{2}, S O_{2}

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Solution

The correct option is C $H_{2}, O_{2}$
When a solution of $N a_{2} S O_{4}$ in water is electrolysed then reactions take place at cathode and anode are

At anode (oxidation):

$2 H_{2} O \to O_{2} + 4 e^{-} + 4 H^{+}$

At cathode ( reduction):

$2 H^{+} + 2 e^{-} \to H_{2}$ .

So, the gas librated at cathode $= H_{2}$

and gas librated at anode $= O_{2}$ .

Option C is correct.

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A solution of Na2SO4 in water is electrolysed using inert electrodes. The products at the cathode and anode respectively, are:

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A solution of $N a_{2} S O_{4}$ in water is electrolysed using inert electrodes. The products at the cathode and anode respectively, are: