Chemistry - Electrochemistry Question with Solution | TestHub

$:\mathrm{NaBr}\rightleftharpoons {\mathrm{Na}}^{+}+{\mathrm{Br}}^{-}$ ${\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{H}}^{+}+{\mathrm{OH}}^{-}$ At cathode : $2{\mathrm{H}}^{+}+2e^{-}\to {\mathrm{H}}_2$ At anode : $2{\mathrm{Br}}^{-}-2e^{-}\to {\mathrm{Br}}_2$ In solution : ${\mathrm{Na}}^{+}+{\mathrm{OH}}^{-}\to \mathrm{NaOH}$ 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.