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Two Faraday of electricity is passed thr...

Two Faraday of electricity is passed through a solution of `CuSO_(4)`. The mass of copper deposited at the cathode is : (at. mass of Cu = 63.5 amu)

A

63.5g

B

127.0g

C

31.75g

D

250g

Text Solution

Verified by Experts

The correct Answer is:
A
In `CuSO_(4).Cu^(2+)+2e^(-)rarrCu`
`therefore Eq. wt. "of" Cu=(63.2)/(2)=31.75`
Now `1F=31.75g "of" Cu`
`2F=31.75xx2=63.5g`
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