The products formed when an aqueous solution of NaBr is electrolysed in a cell having inert electrodes are

On electrolysis of NaBr _(aq) the reactions taking place are 

${\mathrm{NaBr}}_{\left(\mathrm{aq}\right)}\to {\mathrm{Na}}^{+}+\hspace{0.17em}{\mathrm{Br}}^{-}+\hspace{0.17em}{\mathrm{H}}^{+}+{\mathrm{OH}}^{-}$

H⁺ is preffered over Na⁺  at cathode and Br^- preffered over OH^- at anode.

$$\begin{gathered} \mathrm{Cathode} :2\hspace{0.17em}{\mathrm{H}}^{+}+\hspace{0.17em}2{\mathrm{e}}^{-}\to {\mathrm{H}}_2 \\ \mathrm{Anode} :\hspace{0.17em}2{\mathrm{Br}}^{-}\to {\mathrm{Br}}_2+\hspace{0.17em}2{\mathrm{e}}^{-} \\ \mathrm{remaining} \mathrm{ions} \\ {\mathrm{Na}}^{+}\hspace{0.17em}+ {\mathrm{OH}}^{-}\to \mathrm{NaOH}\hspace{0.17em} \end{gathered}$$