JEE Main 2015 · 4 Apr · Shift 1 · Chemistry Q38

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ChemistrySingle correct+4 / -1Question 38 of 90

Two Faraday of electricity is passed through a solution of $\ce{CuSO4}$ . The mass of copper deposited at the cathode is : (at. mass of Cu = 63.5 amu)

0 g

63.5 g

2 g

127 g

Answer and worked solution

Official answer: B

$\ce{Cu^2+ + 2e- -> Cu}$ : 2 F deposit 1 mol, i.e. 63.5 g.

Topics:Electrolysis and Faraday's laws

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JEE Main 2015 · 4 Apr · Shift 1, Chemistry question 38