`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

Options

A. `10^(0.32//0.0591)`
B. `10^(0.32//0.0295)`
C. `10^(0.26//0.0295)`
D. `e^(0.32//0.295)`

Answer

B

Explanation

`Zn|Zn^(+2) (0.01M)||Fe^(+2) (0.001M)|Fe`
`E_(cell)^(@) = E_(cell)^(@) - (RT)/(nF) In K`
`E^(@) = E_(Cell)^(@) - (RT)/(2F) In K`
`E^(@) = 0.2905 - (0.0591)/(2) log.(10^(-3))/(10^(-2))`
`=0.2905 - (0.0591)/(2) log 10^(-1)`
`= 0.2905 +(0.0591)/(2) = 0.2905 +0.02950`
`= 0.32V`
`E^(@) = (RT)/(2F) In K`
`E^(@) = (0.0591)/(2) log K`
`log K = (0.32)/(0.0295) rArr K = 10^(0.32//0.0295)`