Dilute sulphuric acid is electrolysed using platinum electrodes by passing a current of 9.65 A for 10 minutes. The volume of oxygen gas liberated at the anode at STP is : (Given : Molar volume of a gas at STP = 22.4 L mol ⁻¹ ; 1 F = 96500 C mol ⁻¹ )

Step-by-step solution

The quantity of charge Q passed through the solution is: Q = I t = 9.65 A (10 60) s = 5790 C The oxidation reaction at the anode is: 2 H ₂ O (l) O ₂(g) + 4 H ⁺(aq) + 4 e ⁻ This indicates that 4 moles of electrons ( 4 F of charge) are required to produce 1 mole of O ₂ gas. Thus, the n-factor for oxygen gas evolution is 4 . Number of moles of O ₂ produced = Q n F = 5790 4 96500 = 5790 386000 = 0.015 mol Volume of O ₂ gas at STP = moles Molar volume V = 0.015 mol 22.4 L mol ⁻¹ = 0.336 L Converting to millilitres: V =