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The equilibrium constant of the reaction...

The equilibrium constant of the reaction :
`A_((s))+2B_((aq))^(+)hArr A_((aq))^(2+)+2B_((s)): E_(cell)^(@)=0.0295V` is :

A

10

B

`2xx10^(2)`

C

`3xx10^(2)`

D

`2xx10^(5)`

Text Solution

Verified by Experts

The correct Answer is:
A
`E^(@)=(2.303 RT)/(nF)log K`
At 298 K, `E^(@)=(0.0591)/(n)log K`
`0.0295 = (0.0591)/(2)log K`
or `log K = (2xx0.0295)/(0.0591)=1`
K = Antilog (1) = 10.
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