Question

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

• 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 1

Answer: (C)

Hydrogen is liberated at anode

Answer 2

$H _{2} O _{2}$ can be prepared by electrolysis of $50 \%$ $H _{2} SO _{4}$. In this method, hydrogen is liberated at cathode. $H _{2} SO _{4} \rightleftharpoons 2 H ^{+}+2 HSO _{4}^{-}$ At anode : $2 HSO _{4}^{-} \longrightarrow H _{2} S _{2} O _{8}+2 e^{-}$ $H _{2} S _{2} O _{8}+2 H _{2} O \longrightarrow 2 H _{2} SO _{4}+ H _{2} O _{2}$ At cathode $: 2 H ^{+}+2 e^{-} \longrightarrow H _{2} \uparrow$