Which of the following is not correct regarding electrolytic preparation of $H_{2} O_{2}$

Lead is used as cathode

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50%

H_{2} S O_{4}

is used

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Hydrogen is liberated at anode

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Sulphur acid undergoes oxidation

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Solution

The correct option is C Hydrogen is liberated at anode
$H_{2} O_{2}$ can be prepared by the electrolysis of 50% $H_{2} S O_{4}$ solution. In a cell, peroxy disulphuric acid is formed at the anode.

$2 H_{2} S O_{4} \to H_{2} S_{2} O_{8} (a q .) + H_{2} (g)$

Electrolysis Peroxy disulphuric acid

This is drawn off from the cell and hydrolysed with water to give $H_{2} O_{2}$ .

$H_{2} S_{2} O_{8} + 2 H_{2} O \to 2 H_{2} S O_{4} + H_{2} O_{2}$

The resulting solution is distilled under reduced pressure when $H_{2} O_{2}$ gets distilled while $H_{2} S O_{4}$ with high boiling point, remains undistilled.

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Which of the following is not correct regarding electrolytic preparation of H2O2

Which of the following is not correct regarding electrolytic preparation of $H_{2} O_{2}$