A) \[\operatorname{Na}\,and\,B{{r}_{2}}\]
B) \[\operatorname{Na}\,\,and\,\,{{O}_{2}}\]
C) \[{{H}_{2}},\,B{{r}_{2}}\,and\,\,NaOH\]
D) \[{{\operatorname{H}}_{2}}\,and\,\,{{O}_{2}}\]
Correct Answer: C
Solution :
At cathode \[{{H}^{+}}\] reduces to \[{{H}_{2}}\] and at anode \[B{{r}^{-}}\] oxidises to\[B{{r}_{2}}\], leaving NaOH in the solution.
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