A) \[Na\]
B) \[{{H}_{2}}\]
C) \[S{{O}_{2}}\]
D) \[S{{O}_{3}}\]
Correct Answer: B
Solution :
Water is reduced at the cathode and oxidized at the anode instead of \[N{{a}^{+}}\]and \[SO_{4}^{2-}\]. Cathode: \[2{{H}_{2}}O+2{{e}^{-}}\to {{H}_{2}}+2O{{H}^{-}}\] Anode : \[{{H}_{2}}O\to 2{{H}^{+}}+\frac{1}{2}{{O}_{2}}+2{{e}^{-}}\].
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