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