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in Chemistry by (74.9k points)
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)`

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1 Answer

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by (74.1k points)
Correct Answer - C

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