The mass of aluminium deposited at the cathode when a current of 9.65 A is passed through molten aluminium chloride for 1000 seconds is: (Given: Molar mass of Al = 27 ~g ~mol ⁻¹, 1 ~F = 96500 ~C )

Step-by-step solution

The reduction reaction at the cathode for molten aluminium chloride is: Al ³⁺ + 3 e ⁻ Al ( s ) Here, the number of moles of electrons required per mole of Al is n = 3 . According to Faraday's First Law of Electrolysis: w = M I t n F Given: M = 27 ~g ~mol ⁻¹ I = 9.65 ~A t = 1000 ~s F = 96500 ~C Substituting the values: w = 27 9.65 1000 3 96500 w = 27 9650 289500 w = 27 30 = 0.9 ~g Thus, the mass of aluminium deposited is 0.9 ~g . Answer: 0.9 g