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The equilibrium constant (K) for the reaction
`Cu(s)+2Ag^(+) (aq) rarr Cu^(2+) (aq)+2Ag(s)`, will be
[Given, `E_(cell)^(@)=0.46 V`]
A. `K_(c)=` antilog 15.6
B. `K_(c)=` antilog 2.5
C. `K_(c)=` antilog 1.5
D. `K_(c)=` antilog 12.2

1 Answer

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Best answer
Correct Answer - A
`e^(@)=0.059/n log K_(C)`
`0.46=0.059/2 log K_(C)`
`log K_(C)=(0.46xx2)/0.059=15.6`
`K_(C)=` antilog 15.6

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