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A silver coulometer is connected in series with a water electrolysis cell. In one minute at constant current, 1.08 g of silver deposits on the cathode of the coulometer. If the anodic efficiency in the water cell is 90% and cathodic efficiency is 80%, find the total volume (in mL) of dry gases collected in the water cell at 1 atm and 273 K. (Atomic mass: Ag = 108)
- 123.2 mL
- 134.4 mL
- 187.6 mL
- 156.8 mL
Correct answer: 134.4 mL
Solution
Moles of Ag deposited = 1.08/108 = 0.01 mol. Since Ag⁺ + e⁻ -> Ag, charge passed = 0.01 * 96500 = 965 C = 0.01 F. Theoretical moles of H2 at cathode = 0.01/2 = 0.005 mol (since 2e- per H2). Actual H2 = 0.005 * 0.80 = 0.004 mol. Theoretical moles of O2 at anode = 0.01/4 = 0.0025 mol. Actual O2 = 0.0025 * 0.90 = 0.00225 mol. Total moles = 0.004 + 0.00225 = 0.00625 mol. Volume at STP = 0.00625 * 22400 mL = 140 mL. Hmm, not exactly matching. Let me try: 22400 * 0.006 = 134.4. Recalc: H2 = 0.005*0.8 = 0.004, O2 = 0.0025*0.9 = 0.00225. Total = 0.00625. V = 0.00625*22400 = 140 mL. Alternatively using 22.4 L/mol exactly: V = 6.25e-3 * 22400 = 140 mL. Still 140. If efficiency means differently or using 22.4 L for H2 and different: let me try V(H2) = 0.004*22400 = 89.6 mL, V(O2) = 0.00225*22400 = 50.4 mL, total = 140 mL.
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