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(A) solution of sodium sulphate $(Na_2SO_4)$ in water is electrolysed using inert platinum electrodes.The ions present in the solution are $Na^+$,$SO_

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$H_2, O_2$

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(A) solution of sodium sulphate $(Na_2SO_4)$ in water is electrolysed using inert platinum electrodes.The ions present in the solution are $Na^+$,$SO_4^{2-}$,$H^+$,and $OH^-$.At the cathode,the reduction reaction with a higher reduction potential occurs. Since $H^+$ ions have a higher reduction potential than $Na^+$ ions,$H_2$ gas is liberated:$2H^+ + 2e^- \rightarrow H_2 \uparrow$At the anode,the oxidation reaction with a lower discharge potential occurs. Since $OH^-$ ions have a lower discharge potential than $SO_4^{2-}$ ions,$O_2$ gas is liberated:$4OH^- \rightarrow 2H_2O + O_2 + 4e^-$Thus,the products at the cathode and anode are $H_2$ and $O_2$ respectively.

$A$ solution of sodium sulphate in water is electrolysed using platinum electrodes. The products at cathode and anode are respectively

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