A) by the action of\[{{H}_{2}}{{O}_{2}}\]on\[Ba{{O}_{2}}\]
B) by the action of\[{{H}_{2}}S{{O}_{4}}\]on \[N{{a}_{2}}{{O}_{2}}\]
C) by electrolysis of 50% \[{{H}_{2}}S{{O}_{4}}\]
D) by burning hydrogen in excess of oxygen
Correct Answer: C
Solution :
Electrolysis of 50% sulphuric acid is the commercial method for the preparation of hydrogen peroxide. \[{{H}_{2}}S{{O}_{4}}={{H}^{+}}+HSO_{4}^{-}\] At anode: \[2HS{{O}_{4}}\xrightarrow[{}]{{}}{{H}_{2}}{{S}_{2}}{{O}_{8}}+2{{e}^{-}}\] \[{{H}_{2}}{{S}_{2}}{{O}_{8}}+2{{H}_{2}}\text{O}\xrightarrow[{}]{{}}2{{H}_{2}}S{{O}_{4}}+{{H}_{2}}{{O}_{2}}\] At cathode: \[2H+2{{e}^{-}}\xrightarrow[{}]{{}}{{H}_{2}}\]
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