$\left. \begin{gathered}
2{H_2}O\, + \,2e\, \to {H_2}\, + O{H^ - } \hfill \\
\hfill \\
N{a^ + }\, + \,O{H^ - }\, \to \,NaOH \hfill \\
\end{gathered} \right]$ 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)