During electrolysis of an aqueous copper(II) sulfate (CuSO4) solution, a constant electric current of 5.0 A is passed for 1,930 seconds. Given Faraday's constant F ≈ 96,500 C/mol and the molar mass of Cu = 63.5 g/mol, what mass of copper is deposited at the cathode (Cu2+ + 2 e- -> Cu)?

Answer and explanation

Correct answer: D. 3.18 g

Total charge Q = I t = 5.0 A 1,930 s = 9,650 Coulombs. Moles of electrons = Q / F = 9,650 C / 96,500 C/mol = 0.100 mol e-. Since reduction of Cu2+ requires 2 electrons per atom of copper (n = 2): moles of Cu = 0.100 / 2 = 0.050 mol Cu. Mass of copper = 0.050 mol * 63.5 g/mol = 3.175 g ≈ 3.18 g.

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