Question

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`]
A. `1.0xx10^(30)`
B. `1.0xx10^(2)`
C. `1.0xx10^(5)`
D. `1.0xx10^(10)`

Answer 1

Correct Answer - D
Nearest equation
`E_(cell)^(ɵ)=E_(cell)^(ɵ)(-0.059)/(n)logQ_(C)`
At equilibrium Ecell `=0,Q_(C)=K_(C)`
`E_(cell)^(ɵ)=(-0.059)/(n)logK_(C)" Value of "E_(cell)^(ɵ)=0.59V`
`0.59=(0.059)/(1)logK_(C)" value of "n=1`
`K_(C)="antilog "10`
`K_(C)=1xx10^(10)`