A solution of Na_2SO_4 in water is electrolys | vedclass

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(C) During the electrolysis of an aqueous solution of $Na_2SO_4$,the ions present are $Na^+$,$SO_4^{2-}$,$H^+$,and $OH^-$.At the cathode,the reduction

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$H_2 ; O_2$

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(C) During the electrolysis of an aqueous solution of $Na_2SO_4$,the ions present are $Na^+$,$SO_4^{2-}$,$H^+$,and $OH^-$.At the cathode,the reduction of $H^+$ ions occurs because the reduction potential of $H^+ (0.00 \ V)$ is higher than that of $Na^+ (-2.71 \ V)$.Reaction at cathode: $2H_2O + 2e^- \rightarrow H_2(g) + 2OH^-(aq)$.At the anode,the oxidation of $OH^-$ ions (or $H_2O$) occurs because the oxidation potential of $OH^-$ is higher than that of $SO_4^{2-}$.Reaction at anode: $2H_2O \rightarrow O_2(g) + 4H^+(aq) + 4e^-$.Thus,the products at the cathode and anode are $H_2$ and $O_2$ respectively.

$A$ solution of $Na_2SO_4$ in water is electrolysed using inert electrodes. The products at cathode and anode are respectively:

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