A current of 2 A flows for 30 min through a CuSO4 solution. Mass of Cu deposited (M_Cu = 63.5):
A2.36 g
B4.72 g
C0.59 g
D1.18 g
Answer & Solution
Correct answer: D. 1.18 g
Q = I × t = 2 × 30 × 60 = 3600 C. n = 2 (Cu²⁺ + 2e⁻ → Cu). m = QM/(nF) = (3600 × 63.5)/(2 × 96500) ≈ 1.184 g.
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