The gas released at the cathode when acidulated water is electrolysed is :

O_{2}

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H_{2}

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both A and B

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none of these

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Solution

The correct option is B $H_{2}$
During electrolysis of acidified water,
At anode: $2 H_{2} O \to O_{2} + 4 H^{+} + 4 e^{-}$
At cathode: $4 H_{2} O + 4 e^{-} \to 2 H_{2} + 4 O H^{-}$

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