A current of 1.25 A is passed through an aqueous solution of chromium(III) sulphate for 40 minutes . What is the amount of chromium metal deposited at the cathode? (Given: Molar mass of Cr = 52 g mol ⁻¹ ; 1 F = 96500 C mol ⁻¹ )

Step-by-step solution

The total charge passed through the solution is given by: Q = I t Q = 1.25 40 60 = 3000 C The reduction reaction for chromium(III) at the cathode is: Cr ³⁺ + 3 e ⁻ Cr Here, the number of moles of electrons required to deposit one mole of chromium is n = 3 . Using Faraday's first law of electrolysis, the mass of chromium deposited is: m = M Q n F m = 52 3000 3 96500 m = 520 965 0.539 g Answer: 0.539 g