An aqueous solution of Na₂SO₄ in water is electrolyzed using Pt electrodes. The products at the cathode and anode respectively.

${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\to 2{\text{Na}}^{+}+{\mathrm{SO}}_4^{2-}$

$2{\text{H}}_{\text{2}}\text{O}\to 2{\text{H}}^{+}+2{\text{OH}}^{-}$

At cathode, both H⁺ and Na⁺ can reduced but hydrogen is preferentially reduced than Na, because the reduction potential of hydrogen is greater than sodium.

At cathode :  $4{\text{H}}^{+}+4e^{-}\to 2{\text{H}}_{\text{2}}$ 

At anode:       $$\begin{gathered} 4{\text{OH}}^{-}\to {\text{O}}_2+2{\text{H}}_{\text{2}}\text{O}+4{\text{e}}^{-} \\ [\mathrm{Discharge} \mathrm{potential} {\mathrm{SO}}_4^{2-}>{\text{OH}}^{-}] \end{gathered}$$