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Question
Two Faraday of electricity is passed through a solution of CuSO4. The mass of copper deposited at the cathode is : (at. mass of Cu = 63.5 amu)
Options
A.2g
B.127g
C.0 g
D.63.5 g
Solution
2F = 2 equivalents of charge
= 2 equivalents of Cu deposited
= 1 mol Cu will be deposited (for ′Cu′ n-factor = 2)
≡ 63.5 gm
= 2 equivalents of Cu deposited
= 1 mol Cu will be deposited (for ′Cu′ n-factor = 2)
≡ 63.5 gm
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