Electrolysis of 250 ml of a 0.4 M copper(II) sulfate solution was conducted with a current of 5 A for 965 seconds. Determine the mass of the copper deposited and the volume of the gas released at the anode (STP).
- 1.5875 g Cu and 0.28 L O₂
- 3.175 g Cu and 0.56 L O₂
- 1.5875 g Cu and 0.56 L O₂
- 0.79 g Cu and 0.14 L O₂
📝 Detailed explanation
Total charge Q = 5 * 965 = 4825 C.Faradays F = 4825 / 96500 = 0.05 F.
Cu²⁺ + 2e⁻ → Cu.
n(Cu) = 0.05 / 2 = 0.025 mol.
Mass Cu = 0.025 * 63.5 = 1.5875 g.
Anode: 2H₂O → O₂ + 4H⁺ + 4e⁻.
n(O₂) = 0.05 / 4 = 0.0125 mol.
Volume O₂ = 0.0125 * 22.4 = 0.28 L.