During the electrolysis of 500 ml of 2M CuSO₄ solution with a current of 9.65 A for 100 minutes, what is the final molar concentration of sulfuric acid in the solution?
- 0.5 M
- 0.6 M
- 1.0 M
- 1.2 M
📝 Batafsil tushuntirish
Total Charge Q = I × t = 9.65 × 6000 = 57900 C.Faradays F = 57900 / 96500 = 0.6 F.
CuSO₄ + H₂O → Cu + 0.5O₂ + H₂SO₄.
Initial moles CuSO₄ = 0.5 × 2 = 1 mol.
Reaction of 1 mol CuSO₄ requires 2 F.
Since we have 0.6 F, CuSO₄ is in excess.
Moles of H₂SO₄ formed = 0.6 / 2 = 0.3 mol.
Wait, 0.6 F deposits 0.3 mol Cu and forms 0.3 mol H₂SO₄.
Concentration = 0.3 / 0.5 = 0.6 M.
Checking the question: If the answer is 1.0, the current or time must be higher.
Based on 0.6 F: Concentration is 0.6 M.