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A current of 2A was passed for 1.5 hours...

A current of 2A was passed for 1.5 hours through a solution of `CuSO_(4)` when 1.6g of copper was deposited. Calculate percentage current efficiency.

A

3.18g

B

0.318g

C

0.296g

D

0.150g

Text Solution

Verified by Experts

The correct Answer is:
C
`Q=1xxt=0.5xx30xx60=900C`
`Cu^(2+)+2e^(-)rarrCu`
`2xx96500C` deposit `Cu=63.5g`
`therefore 900C` will deposit Cu
`=(63.5)/(2xx96500)xx900=0.296g`
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