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During electrolysis of an AgNO3 solution, a charge of 9650 coulombs is passed through the cell. What mass of silver is deposited at the cathode?
- 21.6 g
- 108 g
- 1.08 g
- 10.8 g
Correct answer: 10.8 g
Solution
Moles of electrons = 9650/96500 = 0.1. Since Ag+ + e- -> Ag is a 1-electron reduction, 0.1 mol Ag is deposited = 0.1*108 = 10.8 g.
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