For a cell involving one electron `E_(cell)^(0)=0.59V` and 298K, the equilibrium constant for the cell reaction is:
[Given that `(2.303RT)/(F)=0.059V` at `T=298K`]

Options

A. `1.0 xx 10^(30)`
B. `1.0 xx 10^(2)`
C. `1.0 xx 10^(5)`
D. `1.0 xx 10^(10)`

Answer

D

Explanation

Nernst equation `E_("cell")^(Ɵ)= E_("cell")^(Ɵ)= (-0.059)/(n) log Q_(C)`
at equilibrium `E_("cell")= 0, Q_(C)= K_(C)`
`E_("cell")^(Ɵ)= (-0.059)/(n) log K_(C)` Value of `E_("cell")^(Ɵ)= 0.59V`
`0.59 = (0.059)/(1) log K_(C)` value of n=1
`K_(C)`= antilog 10
`K_(C)= 1 xx 10^(10)`