The products formed when an aqueous solution of NaBr is...

$: \mathrm{NaBr} \rightleftharpoons \mathrm{Na}^{+}+\mathrm{Br}^{-}$ $\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}^{+}+\mathrm{OH}^{-}$ At cathode : $2 \mathrm{H}^{+}+2 e^{-} \rightarrow \mathrm{H}_2$ At anode : $2 \mathrm{Br}^{-}-2 e^{-} \rightarrow \mathrm{Br}_2$ In solution : $\mathrm{Na}^{+}+\mathrm{OH}^{-} \rightarrow \mathrm{NaOH}$ As discharge potential of $\mathrm{H}^{+}$ions is lower than that of $\mathrm{Na}^{+}$ions and discharge potential of $\mathrm{Br}^{-}$ ions is lower than that of $\mathrm{OH}^{-}$ions, $\mathrm{H}_2$ is liberated at cathode and $\mathrm{Br}_2$ is liberated at anode.