Statement: During electrolysis of acidified water, at the cathode, $H_{2}$ is evolved and at the anode, $O_{2}$ is evolved.
Enter 1 if true else 0.

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Solution

During electrolysis of acidified water using platinum electrodes, hydrogen is liberated at the cathode.
At the cathode, $H^{+} + e^{-} \to \frac{1}{2} H_{2}$

At the anode, $O H^{-}$ is oxidized to oxygen.

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

$2 O \to O_{2}$
Hence the given statement is true.

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