A) \[{{\text{H}}_{\text{2}}}\]is liberated at cathode
B) \[{{\text{O}}_{\text{2}}}\]is produced at cathode
C) \[\text{C}{{\text{l}}_{\text{2}}}\] is obtained at cathode
D) \[\text{N}{{\text{H}}_{\text{3}}}\] is produced at anode
Correct Answer: A
Solution :
\[{{\text{H}}_{\text{2}}}\]is tibrated at cathode and\[{{\text{O}}_{\text{2}}}\] is liberated at anode. \[2{{H}^{+}}+2{{e}^{-}}\xrightarrow{{}}{{H}_{2}}\]
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