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 :
\[{{H}_{2}}S{{O}_{4}}\]can be prepared by electrolysis of\[50%\] \[{{H}_{2}}S{{O}_{4}}\]. In this method, hydrogen is liberated at cathode. \[{{H}_{2}}S{{O}_{4}}2{{H}^{+}}+2HSO_{4}^{-}\] 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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