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^(2)`
B. `1.0 xx 10^(5)`
C. `1.0 xx 10^(10)`
D. `1.0 xx 10^(20)`

Answer

C

Explanation

`E_("cell") = E_("cell")^(@) -0.059/n log Q`............(i)
(At equilibrium, `Q = K_(eq)` and `E_("cell) =0`)
`0 = E_("cell")^(@) - 0.059/1 log K_(eq)` (from equation (i))
`log K_(eq)= E_("cell")^(@)/0.059 = 0.59/0.059 = 10`
`K_(eq) = 10^(10) = 1 xx 10^(10)`