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How many coulombs of electric charge are required to deposit 25.0 g of Cu(s) at the cathode in the electrolysis of CuSO4(aq)?
25.0 g Cu(s) = 25.0/63.5 g/mol = 0.394 mol
Cu2+(aq) + 2 eCu(s)
so 0.394 mol Cu requires 2×0.394 = 0.788 mol electrons
coulombs = mols electrons × Faraday's constant = 0.788 × 96485 = 76000 coulombs.