The ratio of volumes of $H_{2}$ and $O_{2}$ liberated on electrolysis of water is

1 : 2

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1 : 3

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2 : 1

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3 : 1

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Solution

The correct option is B $2 : 1$
$H_{2} O \to H^{+} + O H^{-}$
Cathode : 4 $O H^{-} \to 2 H_{2} O + O_{2} + 4 e^{-}$
Anode : 4 $H^{+} + 4 e^{-} \to 2 H_{2}$
The ratio of volumes of H $_{2}$ and O $_{2}$ is $2 : 1$ .

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