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The mass of copper deposited when a current of 5.0 A flows for 30 minutes through a solution of CuSO₄ (M_Cu = 63.5 g/mol, F = 96500 C) is approximately

A2.96 g
B1.98 g
C5.92 g
D11.85 g
Answer & Solution
Correct answer: A. 2.96 g
1. Charge Q = I × t = 5.0 × 30 × 60 = 9000 C. 2. Moles of electrons = 9000 / 96500 ≈ 0.0933. 3. Cu²⁺ needs 2 electrons per Cu atom, so moles of Cu = 0.0933 / 2 ≈ 0.0466. 4. Mass = 0.0466 × 63.5 ≈ 2.96 g of copper deposited. _Source: syllabus Chemistry, Unit 3 "Electrochemistry", §3.4_
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