During the electrolysis of a 0.2 M FeSO₄ solution, a current of 12 A flows for 1 hour. Find the mass of iron deposited at the cathode. (Use Faraday constant 96485 C, molar mass of Fe = 55.85 g mol⁻¹.)

Step-by-step solution

1. Total charge Q = I × t = 12 A × 3600 s = 43200 C.
2. Moles of electrons nₑ = Q/F = 43200/96485 ≈ 0.448 mol.
3. Fe²⁺ + 2e⁻ → Fe, so moles of Fe = nₑ/2 ≈ 0.224 mol.
4. Mass = moles × molar mass = 0.224 × 55.85 ≈ 12.5 g.

Concept tested: Mass deposition in electrolysis

Published 10 October 2026 · How this solution was prepared and checked