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Two faradays of electricity are passed through a solution of `CuSO_(4)`.The mass of copper deposited at the cathode is (atomic mass of Cu=63.5 g)
A. 2 g
B. 127 g
C. 0 g
D. 63.5 g

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Correct Answer - D
(d) The reduction reaction is :
`Cu^(2+)(aq)+underset(2F)2e^-) to underset(=63.5 g)underset(1 g" atom")(Cu(s))`

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