A) \[C{{l}_{2}}\]
B) \[{{H}_{2}}\]
C) \[{{O}_{2}}\]
D) \[{{H}_{2}}O\]
Correct Answer: A
Solution :
\[2NaCl+2{{H}_{2}}O\to 2NaOH+\underset{\text{Anode}}{\mathop{C{{l}_{2}}}}\,+\underset{\text{Cathode}}{\mathop{{{H}_{2}}}}\,\]
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