A solution of CuSO_4 is electrolyzed for 1 minute with a current of 1…

The correct answer is Option (2) → (1890/F) g

By Faraday’s first law, the mass deposited is proportional to the charge passed. To deposit one mole of a metal from its ion \( \ce{M^{n+}} \), n moles of electrons, that is nF coulombs, are needed:

    \( \displaystyle m = \frac{M \times Q}{nF} \)

The charge passed in 1 minute at 1 ampere:

    \( \displaystyle Q = I \times t = 1\ \mathrm{A} \times 60\ \mathrm{s} = 60\ \mathrm{C} \)

At the cathode, copper is deposited from \( \ce{Cu^2+} \), which takes two electrons per atom, so n = 2:

    \( \displaystyle \ce{Cu^2+ + 2e- -> Cu(s)} \)

    \( \displaystyle m = \frac{63\ \mathrm{g\,mol^{-1}} \times 60\ \mathrm{C}}{2 \times F} = \frac{3780}{2F}\ \mathrm{g} = \frac{1890}{F}\ \mathrm{g} \)

So the mass of copper deposited is (1890/F) g, about 0.0196 g with F = 96487 C mol⁻¹.