A solution of $CuSO_4$ is electrolyzed for 1 minute with a current of 1 ampere. Given the atomic mass of Cu as $63\, g\, mol^{-1}$, the mass of copper deposited at the cathode is

(1890/ F) g

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

We can calculate the mass of copper deposited using Faraday’s laws of electrolysis.

Given:

• Current, $I = 1 \, \text{A}$
• Time, $t = 1 \, \text{min} = 60 \, \text{s}$
• Atomic mass of Cu, $M = 63 \, \text{g/mol}$
• Cu²⁺ → Cu (2 electrons, $n = 2$)

Step 1: Charge passed

$Q = I \cdot t = 1 \times 60 = 60 \, \text{C}$

Step 2: Moles of electrons

$\text{Moles of electrons} = \frac{Q}{F} = \frac{60}{F}$

Step 3: Moles of Cu deposited

$\text{Cu}^{2+} + 2e^- \to \text{Cu}$

$\text{Moles of Cu} = \frac{\text{Moles of electrons}}{2} = \frac{60}{2F} = \frac{30}{F} \, \text{mol}$

Step 4: Mass of Cu deposited

$m = n \cdot M = \frac{30}{F} \times 63 = \frac{1890}{F} \, \text{g}$