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The copper anode of a cell containing silver nitrate solution weighs 60.0 g. After passing current fElectrochemistry Chemistry Question

Question

The copper anode of a cell containing silver nitrate solution weighs 60.0 g. After passing current for some time, it is found that 3.24 g of silver if deposited on the platinum cathode. What is the final weight of the anode? (Ag = 108, Cu = 64)

Answer: C

💡 Solution & Explanation

Equivalents of Ag deposited = $3.24 / 108 = 0.03$ eq. This equals equivalents of Cu oxidized at the anode. Moles of Cu = $0.03 / 2 = 0.015$ mol. Mass of Cu oxidized = $0.015 \times 64 = 0.96$ g. Final anode weight = $60.0 - 0.96 = 59.04$ g. Therefore, correct answer is C.

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