During the electrolysis of 500 ml of a 0.2 M Copper(II) chloride solution, a current of 1.93 A was passed for 1000 seconds. Determine the mass of the gas released at the anode.
- 0.71 g
- 0.355 g
- 1.42 g
- 0.16 g
📝 Detailed explanation
Total charge Q = 1.93 * 1000 = 1930 C.Faradays F = 1930 / 96500 = 0.02 F.
At the anode: 2Cl⁻ → Cl₂ + 2e⁻.
Moles of Cl₂ = 0.02 / 2 = 0.01 mol.
Mass Cl₂ = 0.01 * 71 = 0.71 g.