At a constant current of 1.26 A, how many grams of copper are deposited in 50.0 minutes?

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Multiple Choice

At a constant current of 1.26 A, how many grams of copper are deposited in 50.0 minutes?

Explanation:
This uses Faraday’s law of electrolysis: the mass deposited is proportional to the total charge passed, with m = (M I t)/(n F). For copper, deposition involves Cu2+ gaining 2 electrons, so n = 2. With M = 63.55 g/mol and F = 96485 C/mol, use I = 1.26 A and t = 50.0 min = 3000 s. The charge is Q = I t = 1.26 × 3000 = 3780 C. The moles of Cu deposited are Q/(nF) = 3780/(2 × 96485) ≈ 0.0196 mol. Multiplying by the molar mass gives m ≈ 0.0196 × 63.55 ≈ 1.24 g. Therefore about 1.24 g of copper are deposited.

This uses Faraday’s law of electrolysis: the mass deposited is proportional to the total charge passed, with m = (M I t)/(n F). For copper, deposition involves Cu2+ gaining 2 electrons, so n = 2. With M = 63.55 g/mol and F = 96485 C/mol, use I = 1.26 A and t = 50.0 min = 3000 s. The charge is Q = I t = 1.26 × 3000 = 3780 C. The moles of Cu deposited are Q/(nF) = 3780/(2 × 96485) ≈ 0.0196 mol. Multiplying by the molar mass gives m ≈ 0.0196 × 63.55 ≈ 1.24 g. Therefore about 1.24 g of copper are deposited.

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