Aqueous CuSO₄ is subjected to electrolysis using copper electrodes by passing 0.2 F of electricity. Correct statement about the process is

Attackable electrodes are used.

At cathode :

${\mathrm{E}}_{{\mathrm{Cu}}^{+2}/\mathrm{Cu}}^{\mathrm{o}}> {\mathrm{E}}_{{\mathrm{H}}^{+}/\frac{1}{2}{\mathrm{H}}_2,\mathrm{Pt}}^{\mathrm{o}}$

    ${\mathrm{Cu}}^{+2}+2\stackrel{-}{\mathrm{e}}\to \mathrm{Cu}$

At anode :

 ${\mathrm{E}}_{\mathrm{Cu}/{\mathrm{Cu}}^{+2}}^{\mathrm{o}}> {\mathrm{E}}_{{\mathrm{H}}_2\mathrm{O}/{\mathrm{O}}_2}^{\mathrm{o}}$

 $\mathrm{Cu}\to {\mathrm{Cu}}^{+2}+2\stackrel{-}{\mathrm{e}}$

Thus 2 F can deposit or decompose 2 gram equivalents (1 mole) of copper.

0.2 F can deposit 0.2 gram equivalents (0.1 moles) of copper at cathode.

0.2 F can decompose 0.2 gram equivalents (0.1 moles) of copper at anode.