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Zn|Zn^(2+)(a=0.1M)||Fe^(2+)(a=0.01M)Fe. ...

`Zn|Zn^(2+)(a=0.1M)||Fe^(2+)(a=0.01M)Fe.`
The `EMF` of the above cell is `0.2905`. The equilibrium constant for the cell reaction is

A

`10^(0.32//0.059)`

B

`10^(0.32//0.0295)`

C

`10^(0.26//0.0295)`

D

`10^(0.32//0.295)`

Text Solution

Verified by Experts

Cell reaction is, `zn+Fe^(2+)toZn^(2+)+Fe`
Using, `E=E^(@)-0.059/1 log. (10^(-2))/(10^(-3))`
`(n=2,[zn^(2+)]=10^(-2),[Fe_(2+)]=10^(-3))`
Since, `E=0.2905 therefore 0.2905-E^(@)-0.0295`
or `E^(@)=0.2905+0.0295=0.3200V`
Again, `E^(@)=0.059/2 log.k_(eq) therefore 0.32=0.059/2 log K_(eq) or K_(eq) =10.^(0.32)/(0.0296)`
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