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The emf of the cell, Zn|Zn^(2+)(0.01M)...

The emf of the cell,
`Zn|Zn^(2+)(0.01M)|| Fe^(2+)(0.001M) | Fe`
at 298 K is 0.2905 then the value of equilibrium constant for the cell reaction is:

A

`e^((0.32)/(0.0295))`

B

`10^((0.32)/(0.0295))`

C

`10^((0.26)/(0.0295))`

D

`10^((0.32)/(0.059))`

Text Solution

Verified by Experts

The correct Answer is:
B
`Zn|Zn^(2+)(0.01M)||Fe^(2+)(0.001M)|Fe," "E=0.2905`
cell reaction, `Zn+Fe^(2+)hArrZn^(2+)+Fe`.
`0.2905=E^(@)-(0.0591)/(2)log((0.01)/(0.001)`
`E^(@)=0.32"volt"`
At equilibrium `E_(cell)=0`
`0=0.32-(0.0591)/(2)logK_(eq)`
`K_(eq)=10^(0.32//0.0295)`
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