Electrolysis of 500 ml of 1 M CuSO₄ solution was performed until 0.4 Faradays of electricity passed through the cell. Determine the final molar concentration of CuSO₄ in the solution.
- 0.6 M
- 0.8 M
- 0.2 M
- 0.5 M
📝 Detailed explanation
Initial moles CuSO₄ = 0.5 L * 1 M = 0.5 mol.Moles of Cu²⁺ reduced = 0.4 F / 2 = 0.2 mol.
Remaining moles CuSO₄ = 0.5 - 0.2 = 0.3 mol.
Final Molarity = 0.3 mol / 0.5 L = 0.6 M.