If 48242 C of charge pass through a copper solution, how many grams of copper deposit?

• A. 63.6 g
• B. 15.9 g
• C. 47.7 g
• D. 31.8 g

Answer: (B)

When copper is deposited from a solution, the amount of metal formed is tied directly to the total electric charge that passes through the solution. Faraday’s law of electrolysis gives the relation m = (Q × M) / (n × F), where m is the mass of metal deposited, Q is the total charge, M is the molar mass of the metal, n is the number of electrons required to reduce one atom of metal, and F is Faraday’s constant (about 96485 C/mol e−).

For copper, each Cu atom gaining two electrons forms Cu metal, so n = 2. The molar mass of copper is about 63.55 g/mol. The charge passed is 48242 C. First find how many moles of copper are deposited: moles Cu = Q / (nF) = 48242 / (2 × 96485) ≈ 0.25 mol. Then convert to mass: mass = moles × M ≈ 0.25 mol × 63.55 g/mol ≈ 15.9 g.

So about 15.9 grams of copper would deposit.