Electrolysis of aqueous copper (II) sulphate | vedclass

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Question

(B) During the electrolysis of aqueous $CuSO_4$ using inert $Pt$ electrodes,the ions present in the solution are $Cu^{2+}$,$SO_4^{2-}$,$H^+$,and $OH^-

Vedclass · Accepted Answer

$O_2, Cu$

Vedclass · Answer

(B) During the electrolysis of aqueous $CuSO_4$ using inert $Pt$ electrodes,the ions present in the solution are $Cu^{2+}$,$SO_4^{2-}$,$H^+$,and $OH^-$. At the cathode,$Cu^{2+}$ ions have a higher reduction potential than $H^+$ ions,so $Cu^{2+}$ ions are reduced to $Cu$ metal: $Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$. At the anode,$OH^-$ ions are oxidized in preference to $SO_4^{2-}$ ions,leading to the evolution of oxygen gas: $4OH^-(aq) \rightarrow O_2(g) + 2H_2O(l) + 4e^-$. Therefore,'$X$' (anode product) is $O_2$ and '$Y$' (cathode product) is $Cu$.

Electrolysis of aqueous copper $(II)$ sulphate between $Pt$ electrodes gives '$X$' at anode and '$Y$' at cathode. $X$ and $Y$ are respectively

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