A current of 2.00 A passes through a solution of copper(II) sulfate for 30.0 minutes with copper electrodes. What mass of copper deposits on the cathode? (F = 96 500 C/mol, M(Cu) = 63.5 g/mol)
Answer and explanation
Correct answer: B. 1.18 g
- Step 1: Q = It = 2.00 x 1800 = 3600 C.
- Step 2: n(e-) = 3600 / 96 500 = 0.0373 mol.
- Step 3: Cu2+ + 2e- -> Cu, so n(Cu) = 0.01865 mol and m = 0.01865 x 63.5 = 1.18 g.
Answer
1.18 g.
Why not the others: 2.37 g uses one electron per copper atom; 0.59 g halves the answer; 0.0197 g uses 30 as the time in seconds.
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