Calculate the mass of copper deposited at the cathode when an electric current of 2.0 A is passed through a copper(II) sulfate solution for 1,930 seconds. (Faraday constant F = 96,500 C/mol; Atomic mass of Cu = 63.5 g/mol)

Answer and explanation

Correct answer: B. 1.27 g

  1. Step 1: Total electric charge passed: Q = I × t = 2.0 A × 1930 s = 3,860 C. Step 2. Reduction half-reaction: Cu²⁺ + 2e⁻ -> Cu. It requires 2 Faradays per mole of Cu (z = 2). Step 3. Moles of electrons = 3,860 / 96,500 = 0.040 mol. Step 4. Moles of Cu deposited = 0.040 / 2 = 0.020 mol. Step 5. Mass = 0.020 × 63.5 = 1.27 g. Answer: 1.27 g. Why not the others: 2.54 g forgets z = 2; 0.64 g halves the mass; 3.81 g assumes 3 electrons.

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