Which of the following is not correct regarding the electroplytic perparation of `H_(2)O_(2)`?

Options

A. Lead is used as cathode
B. 50% `H_(2)SO_(4)` is used.
C. Hydrogen is liberated at anode.
D. Sulphuric acid undergoes oxidation.

Answer

C

Explanation

`H_(2)O_(2)` is slightly stronger acid than water, it forms two series of salt,
`H_(2)O_(2) iff H^(+)+HO_(2)^(-)` (hydroperoxide ion)
`H_(2)O_(2) iff 2H^(+)+O_(2)^(2-)` (Peroxideion)
`H_(2)O_(2` can be prepared by electrolysis 50% of 50 `H_(2)SO_(4)`. In this method, hydrogen is liberated at cathode.
`H_(2)SO_(4) iff 2H^(+)+2HSO_(4)^(-)`
At anode `2HSO_(4)^(-) rarr H_(2)S_(2)O_(8)+2e^(-)`
`H_(2)S_(2)O_(8)+2H_(2)O rarr 2H_(2)SO_(4)+H_(2)O_(2)`
At cathode `2H^(+)+2e^(-) rarr H_(2)uarr`