A solution of copper sulphate is electrolysed for 10 minutes with a current of 1.5 amperes. The mass of copper deposited at cathode is : (Given : Molar mass of Cu = 63 g mol ⁻¹ ; 1 F = 96487 C mol ⁻¹ )

Step-by-step solution

The reaction at the cathode is Cu ²⁺ + 2 e ⁻ Cu . The quantity of charge Q passed through the solution is given by Q = I t . Substituting the given values: Q = 1.5 A (10 60) s = 900 C From Faraday's first law of electrolysis, the mass of copper deposited is: m = M Q n F Here, M = 63 g mol ⁻¹ , n = 2 , and F = 96487 C mol ⁻¹ . m = 63 900 2 96487 m = 56700 192974 0.2938 g Answer: 0.2938 g