Chemistry - Electrochemistry Question with Solution | TestHub

 ${\text{H}}_2\text{O}$  is more readily reduced at cathode than  $\text{N}{\text{a}}^{+}.$  It is also more readily oxidized at anode than  $\text{S}{\text{O}}_4^{2-}.$  Hence, the electrode reactions are $\begin{array}{l}2{\text{H}}_2\text{O}+2{\text{e}}^{-}\stackrel{}{\to }{\text{H}}_2\uparrow +2\text{O}{\text{H}}^{-}\left[\text{a}\text{t}\text{c}\text{a}\text{t}\text{h}\text{o}\text{d}\text{e}\right]\\ {\text{H}}_2\text{O}\stackrel{}{\to }\frac{1}{2}{\text{O}}_2\uparrow +2{\text{H}}^{+}+2{\text{e}}^{-}\left[\text{a}\text{t}\text{a}\text{n}\text{o}\text{d}\text{e}\right]\end{array}$