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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

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

B

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

C

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

D

`(0.26)/(10^(0.0591))`

Text Solution

Verified by Experts

The correct Answer is:
B
`E_(cell)^(@)=(0.059)/(n)logK_(c),E_(cell)=0" at equilibrium"`
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