On electrolysis of aqueous Na₂SO₄ , the reactions taking place are 

${\mathrm{Na}}_2\mathrm{S}{\mathrm{O}}_4\stackrel{\mathrm{aqueous} }{\to }2{\mathrm{Na}}^{{ }_{+}}\hspace{0.17em}\hspace{0.17em}+\hspace{0.17em}\mathrm{S}{{\mathrm{O}}_4}^{2-}+\hspace{0.17em}{\mathrm{H}}^{+\hspace{0.17em}}+\hspace{0.17em}{\mathrm{OH}}^{-}$

at cathode H+ and Na+ are competerors , but only H+ is deposited at cathode . 

At Cathode : $2{\mathrm{H}}^{+}\hspace{0.17em}\hspace{0.17em}+\hspace{0.17em}\hspace{0.17em}2{\mathrm{e}}^{-}\stackrel{ }{\to }{{\mathrm{H}}_2}_{\left(\mathrm{gas}\right)}$

Therefore H₂ gas is liberated at cathode .