Q37. A solution of CuSO4 is electrolysed for 10 minutes using a current of 1.5 amperes. (1 mol of Cu = 63 g). How much copper is deposited at cathode. 1F = 96487 Cmol-1
(a) 0.2938 g
(b) 2.938 g
(c) 29

Question

Q37. A solution of CuSO4 is electrolysed for 10 minutes using a current of 1.5 amperes. (1 mol of Cu = 63 g). How much copper is deposited at cathode. 1F = 96487 Cmol-1
(a) 0.2938 g
(b) 2.938 g
(c) 29.38 g
(d) 0.02938 g

FirstInTest · Accepted Answer

Correct answer: a. The reaction is, Cu2+ + 2e- → CuI = 1.5 amperet = 10 x 60 = 600sQ = I x t = 1.5 x 600 = 900 CThe reaction states that, 2 x 96487 C required to deposit 63.5 g Cu at cathode.So, 900 C will deposit = (63.5 g / (2 x 96487 C)) x 900 C = 0.296 g Cu at cathode.The calculated answer 0.296 g matches (a) 0.2938 g approximately.

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