A) lead is used as cathode
B) 50% \[{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\] is used
C) hydrogen is liberated at anode
D) sulphuric acid undergoes oxidation
Correct Answer: C
Solution :
\[{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\]can be prepared by electrolysis of 50% \[{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{.}\]In this method, hydrogen is liberated at cathode. \[{{H}_{2}}S{{O}_{4}}2{{H}^{+}}+2HSO_{4}^{-}\] At anode: \[2HSO_{4}^{-}\to {{H}_{2}}{{S}_{2}}{{O}_{8}}+2{{e}^{-}}\] \[{{H}_{2}}{{S}_{2}}{{O}_{8}}+2{{H}_{2}}O\to 2{{H}_{2}}S{{O}_{4}}+{{H}_{2}}{{O}_{2}}\] At cathode: \[2{{H}^{+}}+2{{e}^{-}}\to {{H}_{2}}\uparrow \]
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