108 g of silver (molar mass 108 g mol –1 ) is deposited at cathode from AgNO₃(aq) solution by a certain quantity of electricity. The volume (in L) of oxygen gas produced at 273 K and 1 bar pressure from water by the same quantity of electricity is _______. (Round off to the nearest integer)

Solution

Chemistry ma’am · 5 steps

The same charge passes through both cells, so count it in faradays: one mole of silver needs exactly 1 F.

1. Charge from silver. Each Ag⁺ takes one electron: Ag⁺ + e⁻ → Ag. Here 108 g is 1 mol of Ag, so the charge passed is 1 mol of electrons, that is 1 F.
2. Oxygen at the anode. Water is oxidised at the anode, and four electrons are released for every O₂ formed.

   2H₂O → O₂ + 4H⁺ + 4e⁻

3. Moles of oxygen. 1 F gives 1/4 = 0.25 mol of O₂.
4. Volume at 1 bar. At 273 K and 1 bar, V = nRT/P = 0.25 × 0.08314 × 273 = 5.67 L. This uses the molar volume of 22.7 L at 1 bar, not 22.4 L at 1 atm.
5. So. 5.67 L rounds to 6.

The common slip: 11. Taking 2 electrons per O₂ instead of 4 gives 0.5 mol of oxygen and about 11 L. Each O atom loses 2 electrons, and an O₂ molecule has two O atoms.

The idea to keep. For two cells in series, convert everything to faradays first. Then divide by the electrons each product needs: 1 for Ag, 4 for O₂, 2 for H₂.

Answer from the NTA final answer key. Solution by PrepWiser.