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A current of 2 ampere was passed through...

A current of 2 ampere was passed through solutions of `CuSO_(4)` and `AgNO_(3)` in series . 0.635 g of copper was deposited . Then the weight of silver deposited will be

A

`0.59` g

B

`3.24` g

C

`1.08` g

D

`2.16` g

Text Solution

Verified by Experts

The correct Answer is:
D
A/c. to Faraday's `II^(nd)` law
or `(0.635)/(63.5//2) = (w_(Ag))/(108)`
`therefore W_(Ag) = 2.16 g `
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