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Question

If 96500 coulombs of electricity liberates one gram equivalent of any substance, the time taken for a current of 0.15 amperes to deposite 20 mg of copper from a solution of copper sulphate is
Given:
(Equivalent weight of copper = 32 g)

A
4 minutes 2 secconds
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B
3 minutes 20 secconds
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C
5 minutes 50 secconds
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D
6 minutes 42 secconds
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Solution

The correct option is D 6 minutes 42 secconds
Given :
Mass deposited , W=20 mg
Current, I =0.15 A

By Faraday's First law,
The mass deposited/released of any substance during electrolysis is proportional to the amount of charge passed into the electrolyte.

WQ
W=ZQ
where,

W: Mass deposited or liberated
Q: Amount of charge passed
Z: Electrochemical equivalent of the substance
Since,
Z=E96500

W=E96500×Q
where,
E is Equivalent weight
Charge, Q =I×t

Hence,
W=Z×i×t

20×103=(3296500)×0.15×t
t=402 s
t=6 minutes 42 secconds

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