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 \[N{{a}^{+}}\,\,and\,\,SO{{_{4}^{2-}}^{~}}.\] \[Cathode:\,\,2{{H}_{2}}O+2{{e}^{-}}\,\,\to \,\,{{H}_{2}}+2O{{H}^{-}}\] \[Anode:\,\,{{H}_{2}}O\,\,\to \,\,2{{H}^{+}}\,\,+\frac{1}{2}\,\,{{O}_{2}},\,\,+\,\,2{{e}^{-}}~\]You need to login to perform this action.
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