A current of 9.65 ~A is passed through dilute sulphuric acid for 10 minutes. The volume of oxygen gas evolved at the anode at STP is: (Given: 1 ~F = 96500 ~C , Molar volume of gas at STP = 22400 ~mL ~mol ⁻¹ )

Step-by-step solution

According to Faraday's First Law of Electrolysis, the total charge passed is: Q = I t Q = 9.65 ~A (10 60) ~s = 5790 ~C Number of moles of electrons passed = Q F = 5790 96500 = 0.06 ~mol The reaction at the anode for the evolution of oxygen is: 2 H ₂ O ( l ) O ₂( g ) + 4 H ⁺( aq ) + 4 e ⁻ From the stoichiometry, 4 moles of electrons produce 1 mole of O ₂ gas. Moles of O ₂ produced = 0.06 4 = 0.015 ~mol Volume of O ₂ at STP = 0.015 ~mol 22400 ~mL ~mol ⁻¹ = 336 ~mL Answer: 336 ~mL