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