How many grams of copper are deposited in 60.0 minutes at 1.26 A?

• A. 0.75 g
• B. 1.49 g
• C. 2.10 g
• D. 1.00 g

Answer: (B)

This question uses Faraday’s laws of electrolysis: the mass of metal deposited is proportional to the total charge passed, and you need the right number of electrons for each atom of metal. Copper(II) reduction to solid copper requires 2 electrons per copper atom.

First find the total charge: current times time. 60.0 minutes is 3600 seconds, so Q = 1.26 A × 3600 s = 4536 C. The number of moles of electrons transferred is Q divided by the Faraday constant, F ≈ 96485 C/mol e−, giving 0.0470 mol e−. Since each copper atom needs 2 electrons, the moles of copper deposited are half that: 0.0470 / 2 = 0.0235 mol Cu.

Multiply by copper’s molar mass, 63.55 g/mol, to get the mass: 0.0235 mol × 63.55 g/mol ≈ 1.49 g.

So about 1.49 g of copper would be deposited.