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The equilibrium constant for the reactio...

The equilibrium constant for the reaction,
`Cu(s)+Cu^(2+) (aq.) hArr 2Cu^(+) (aq.)`
`E_(Cu^(2+)//Cu)^(@)=0.34 V" "E_(Cu^(2+)//Cu)^(@)=0.15 V`
(Given : log 3.72 = 0.571)

A

`3.72 xx 10^(-6)`

B

`3.72 xx 10^(-5)`

C

`3.72 xx 10^(-7)`

D

`3.72 xx 10^(-8)`

Text Solution

Verified by Experts

The correct Answer is:
C
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