`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//0.0991)`
B. `10^(0.32//0.0295)`
C. `10^(0.26//0.0295)`
D. `e^(0.32//0.295)`

Answer

B

Explanation

The half-cell reactions are `{:((i),Zn(s) rarr Zn^(2+) (aq) + 2e^(-)),((ii),ul(fe^(2+) (aq) + 2e^(-) rarr Fe(s))),(,ul(Zn(s) + Fe^(2+)(aq) rarr Zn^(2+)(aq) + Fe(s))
):}`
On applying the Nernst equation
`E_("cell")= E_("cell")^(0)- 0.0591//n log 10 [Zn^(2+)]//[Fe^(2+)]`
`0.2905= E_("cell")^(0)- 0.0591//n log 10 0.1//0.01`
`E_("cell")^(0)= 0.32V`
`E_("celll")= E_("cell")^(0) - 0.0591//n log 10 K_(c )`
`log_(10)K_(c )= 0.32//0.02955`