Aqueous CuSO₄ is subjected to electrolysis using copper electrodes by passing 0.1 F of electricity. Correct statement about the process is

Attackable electrode is 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.

0.1 F can deposit 0.1 gram equivalents (0.05 moles) of copper at cathode.

0.1 F can decompose 0.1 gram equivalents (0.05 moles) of copper at anode