If the plating current is doubled from 1.26 A to 2.52 A while depositing 2.08 g of copper, approximately how many minutes are required?

• A. 83.6 min
• B. 5016 min
• C. 133 min
• D. 41.8 min

Answer: (D)

Plating copper follows Faraday’s law: the mass deposited is proportional to the total electric charge passed. This gives m = (M I t) / (n F), where M is the molar mass, I is the current, t is time, n is the number of electrons per atom deposited, and F is Faraday’s constant.

Rearrange to t = m n F / (M I). Use copper values: M = 63.55 g/mol, n = 2, F = 96485 C/mol, m = 2.08 g, I = 2.52 A.

Compute approximate numbers: m/M ≈ 2.08 / 63.55 ≈ 0.0327 mol. Then (m n F) ≈ 0.0327 × 2 × 96485 ≈ 6.31 × 10^3 C. Divide by the current: t ≈ 6310 C / 2.52 A ≈ 2.5 × 10^3 s. Convert to minutes: 2500 s ÷ 60 ≈ 41.7 min, which rounds to 41.8 min.

So the required time is about 41.8 minutes.