108 g silver (molar mass 108 g mol ) is deposited at cathode from AgNO (aq) solution by a certain qu — Electrochemistry Chemistry Question

💡 Solution & Explanation

**Step 1: Calculate moles of silver deposited** Mass of Ag = 108 g, Molar mass = 108 g/mol Moles of Ag = 108/108 = 1 mol **Step 2: Determine electrons required for silver deposition** Cathode reaction: Ag⁺ + e⁻ → Ag For 1 mol Ag, electrons required = 1 mol e⁻ **Step 3: Apply the same quantity of electricity to water oxidation** At anode, water oxidation occurs: 2H₂O → O₂ + 4H⁺ + 4e⁻ 4 moles of electrons produce 1 mol O₂ **Step 4: Calculate moles of O₂ produced** Since 1 mol e⁻ is available: Moles of O₂ = 1 mol e⁻ × (1 mol O₂/4 mol e⁻) = 0.25 mol **Step 5: Calculate volume using ideal gas law at 273 K and 1 bar** Using PV = nRT V = nRT/P = (0.25 mol × 8.314 J mol⁻¹ K⁻¹ × 273 K)/(100 Pa) V = (0.25 × 8.314 × 273)/100 = 5.68 L ≈ 5.60 L (Alternatively, at STP conditions: V = 0.25 mol × 22.4 L/mol = 5.60 L) Therefore, the answer is 5.60.