Two statements are given below.Statement I : | vedclass

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(C) During the electrolysis of molten $NaCl$,$Na$ metal is deposited at the cathode,and $Cl_2$ gas is liberated at the anode.At cathode: $Na^+ + e^- \

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Statement $I$ is correct but statement $II$ is not correct

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(C) During the electrolysis of molten $NaCl$,$Na$ metal is deposited at the cathode,and $Cl_2$ gas is liberated at the anode.At cathode: $Na^+ + e^- \rightarrow Na$At anode: $2Cl^- \rightarrow Cl_2 + 2e^-$Thus,Statement $I$ is correct.When aqueous $CuSO_4$ is electrolyzed using $Pt$ electrodes,$Cu$ is deposited at the cathode and $O_2$ is liberated at the anode.At cathode: $Cu^{2+} + 2e^- \rightarrow Cu$At anode: $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$Since $O_2$ is liberated at the anode,not the cathode,Statement $II$ is incorrect.Therefore,Statement $I$ is correct but Statement $II$ is incorrect.

Two statements are given below.
Statement $I$ : Molten $NaCl$ is electrolysed using $Pt$ electrodes. $Cl_2$ is liberated at anode.
Statement $II$ : Aqueous $CuSO_4$ is electrolysed using $Pt$ electrodes. $O_2$ is liberated at cathode.
The correct answer is

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Two statements are given below.Statement $I$ : Molten $NaCl$ is electrolysed using $Pt$ electrodes. $Cl_2$ is liberated at anode.Statement $II$ : Aqueous $CuSO_4$ is electrolysed using $Pt$ electrodes. $O_2$ is liberated at cathode.The correct answer is