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During electrolysis of 270 g of water with a current efficiency of 75%, which of the following statements is correct?

1. 168 L of O2(g) will be produced at the anode at 1 atm and 273 K
2. Total 504 L of gases will be produced at 1 atm and 273 K
3. 336 L of H2(g) will be produced at the cathode at 1 atm and 273 K
4. 45 F of electricity will be consumed

Correct answer: 336 L of H2(g) will be produced at the cathode at 1 atm and 273 K

Solution

270 g of water is 15 moles. At 75% efficiency, 11.25 moles of H2O are actually electrolysed, producing 11.25 moles H2 (= 252 L) and 5.625 moles O2 (= 126 L). Checking option C: 336 L H2 would require 15 moles H2, which corresponds to 100% efficiency—so let me re-examine. At 100% efficiency: 15 mol H2O gives 15 mol H2 = 336 L and 7.5 mol O2 = 168 L. At 75% efficiency, 75% of these volumes are produced. However, option A says 168 L O2 (that is the 100%-efficiency value), option C says 336 L H2 (also 100%-efficiency value). The 75% efficiency most naturally means 75% of 60 F = 45 F are used productively. Option D says 45 F consumed—but that would mean efficiency * total F = effective F. 15 mol H2O needs 4 electrons per mole (2 for each H2 and O2 half-reaction combined for 2H2O -> 2H2 + O2 needs 4F per 2 mol, so 2F per mol H2O). Total F = 15 * 2 = 30 F effective. Total consumed = 30/0.75 = 40 F. None match exactly. Re-checking: electrolysis of water: at cathode 2H+ + 2e- -> H2 (2F per mol H2); at anode H2O -> 1/2 O2 + 2H+ + 2e- (2F per mol H2). For 15 mol H2O: 15 mol H2 needs 30 F, 7.5 mol O2 needs 15 F... This is getting complex. Most standard approach: moles H2O = 270/18 = 15 mol. With 75% efficiency the actual moles electrolysed = 0.75 * 15 = 11.25 mol. This gives 11.25 mol H2 = 252 L and 5.625 mol O2 = 126 L. Total = 378 L. None of the options match perfectly. Option C (336 L H2) corresponds to 15 mol H2 i.e. 100% efficiency of 15 mol water. Option A (168 L O2) corresponds to 7.5 mol O2 at 100%. The question likely intends: 270 g water -> 15 mol; effective moles electrolysed due to 75% efficiency; charge needed for 15 mol water = 2*15=30F, with 75% efficiency total charge passed = 30/0.75 = 40F. H2 at cathode: moles = 15 mol = 15*22.4 = 336 L. O2 at anode: 7.5 mol = 168 L. The 75% efficiency means 75% of charge does useful electrolysis, so all 15 mol water IS electrolysed but 40F is consumed, not 30F. Option C says 336 L H2 at cathode—correct!

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