In the refining of silver by electrolytic method, what will be the mass of 72.8 g silver anode (60% — Electrochemistry Chemistry Question

💡 Solution & Explanation

Total charge $Q = 9.65 \text{ A} \times 3600 \text{ s} = 34740$ C. Faradays = $34740 / 96500 = 0.36$ F. Moles of Ag oxidized = 0.36 mol. Mass of Ag oxidized = $0.36 \times 108 = 38.88$ g. Because the anode is 60% pure silver, the oxidation of 38.88 g of silver releases accompanying impurities into the anode mud. Total mass lost by the anode = $38.88 / 0.60 = 64.8$ g. Final mass of the anode = $72.8 - 64.8 = 8.0$ g. Therefore, correct answer is 8.