During the electrolysis of 200 ml of 0.5 M CuSO₄ solution, a current of 2 A was passed for 48.25 minutes. Determine the final molar concentration of copper ions in the solution.
- 0.2 M
- 0.35 M
- 0.15 M
- 0.5 M
📝 Detailed explanation
Total charge Q = 2 × (48.25 × 60) = 5790 C.Faradays = 5790 / 96500 = 0.06 F.
Cu²⁺ + 2e⁻ → Cu.
Moles of Cu²⁺ removed = 0.06 / 2 = 0.03 mol.
Initial moles Cu²⁺ = 0.2 × 0.5 = 0.1 mol.
Remaining moles = 0.1 - 0.03 = 0.07 mol.
Molarity = 0.07 / 0.2 = 0.35 M.