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 )

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Correct answer: B. 0.2938 g

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

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