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Electrolysis and Faraday's Law

AP Chemistry· difficulty 3/5

Cu²⁺ + 2e⁻ → Cu(s); F = 96485 C/mol e⁻.

Cu cathode CuSO₄ solution; I = 2.00 A; t = 1930 s

What mass of Cu deposits? (Cu = 63.5 g/mol)

  • A

    ~2.54 g

  • B

    ~3.86 g

  • C

    ~0.635 g

  • D

    ~1.27 g

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Explanation

Q = It = (2.00)(1930) = 3860 C; mol e⁻ = 3860/96485 ≈ 0.0400; mol Cu = 0.0200; mass = 0.0200(63.5) ≈ 1.27 g.

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