What current is required to plate 4.00 g copper in 60.0 minutes?

• A. 3.38 A
• B. 1.50 A
• C. 2.50 A
• D. 4.10 A

Answer: (A)

Electric plating follows Faraday’s law: the mass deposited is proportional to the charge passed, m = (M I t) / (n F). For copper, each Cu atom requires 2 electrons to plate, so n = 2. With a deposited mass m = 4.00 g, molar mass M = 63.55 g/mol, Faraday’s constant F = 96485 C/mol, and time t = 60.0 min = 3600 s, solve for the current: I = m n F / (M t). Plugging in gives I = (4.00 × 2 × 96485) / (63.55 × 3600) A ≈ 771,880 / 228,780 ≈ 3.38 A. So about 3.38 A is required.