`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//0.2995)`

Answer

b

Explanation

`Zn|Zn^(2+)(a=0.1M)||Fe^(2+)(a=0.01M)|Fe` the cell reaction
`i. Zn(s)rarr Zn^(2+)(aq)+2e^(-)`
`ii. Fe^(2+)(aq)+2e^(-) rarr Fe(s)`
`ulbar(Zn(s)+Fe^(2+)(aq) rarr Zn^(2+)+Fe(s))`
On applying Nernst equation,
`E_(cell)=E^(c-)._(cell)-(0.0591)/(n)log.([Zn^(2+)])/([Fe^(2+)])`
`0.2905=E^(c-)._(cell)-(0.0591)/(2)log.(0.1)/(0.01)`
`0.2905=E^(c-)._(cell)-0.02905xxlog10`
`0.295=E^(c-)._(cell0-0.2905xx1`
`:. E^(c-)._(cell)=0.2905+0.0295=0.32V`
At equilibrium `(E_(cell)=0)`,
`E_(cell)=E^(c-)._(cell)-(0.0591)/(n)log K_(c)`
`:. 0=E^(c-)._(cell)-(0.0591)/(n)logK_(c)`
or `E^(c-)._(cell)=(0.0591)/(2)logK_(c)`
`0.32=(0.0591)/(2)logK_(c)`
or `K_(c)=10^(0.32//0.295)`