\(\left. \begin{gathered}
2{H_2}O\, + \,2e\, \to {H_2}\, + O{H^ - } \hfill \\
\hfill \\
N{a^ + }\, + \,O{H^ - }\, \to \,NaOH \hfill \\
\end{gathered} \right]\) At cathode
\(B{r^ - }\, \to \,Br\, + \,{e^ - }\)
\(Br\, + \,Br\, \to \,B{r_2}\) At anode
So the products are \(H_2\) and \(NaOH\) (at cathode) and \(Br_2\) (at anode)
$(A)$ $Sn^{+4}+ 2e^{-} \rightarrow Sn^{2+}$, $E^o= + 0.15\,V$
$(B)$ $2Hg^{+2} + 2e^{-} \rightarrow Hg_{2}^{+2}$, $E^o = + 0.92\,V$
$(C)$ $PbO_2 + 4H^{+} + 2e^{-} \rightarrow Pb^{+2} + 2H_2O$, $E^o = + 1.45\,V$
${{\text{E}}^o }{\text{C}}{{\text{u}}^{{\text{2}} + }}{\text{/Cu = + 0}}{\text{.34 V, E}}_{{\text{F}}{{\text{e}}^{ + {\text{2}}}}/Fe}^o = \,\,{\text{ - 0}}{\text{.44 V}}$