What mass of copper in gram would be deposited by a current of 1.00ampere passing for 965 seconds through…

Analyze the question: We need the mass of copper deposited during electrolysis of CuSO₄ solution, given the current (1.00 A), the time (965 s), the molar mass of Cu (63.5) and the Faraday constant (96500 C).

Concept: Faraday's laws of electrolysis.

• Quantity of charge: Q = I × t
• Cu²⁺ + 2e⁻ → Cu, so 2 Faradays (2 × 96500 C) deposit 1 mole (63.5 g) of copper.

Solution:

1. Charge passed: Q = I × t = 1.00 × 965 = 965 C

2. Mass deposited: If 193000 C deposits 63.5 g, then 965 C deposits

mass = (965 / 193000) × 63.5 mass = 0.005 × 63.5 mass = 0.3175 g ≈ 0.32 g

Answer: The mass of copper deposited is about 0.32 g, which is option A.

Tip: Notice that 965 is exactly 1/100 of 96500, so the charge is 0.01 F. Because copper(II) needs 2 electrons, the moles of Cu = 0.01 / 2 = 0.005 mol. Multiply by 63.5 to get 0.3175 g. Always check the ion's charge (here 2+) before dividing the charge by the Faraday constant.