Calculate the euilibrium constant for the reaction,
`2Fe^(3+) + 3I^(-) hArr 2Fe^(2+) + I_(3)^(-)`.
The standard reduction potential in acidic conditions are `0.77 V ` and `0.54 V` respectivelu for `Fe^(3+)//Fe^(2+)` and `I_(3)^(-)//I^(-)` couples.

Explanation

For the change
`2Fe^(3+) +3I^(-) hArr 2Fe^(2+) +I_(3)^(-)` ,
At equilibrium, `E = 0`
`E = E^(@)-(0.059)/(2)log_(10)K_(c)`
or `E^(@) = (0.059)/(2)log_(10)K_(c)`
Also `E_(cell)^(@) = E_(RP_(Fe^(3+)//Fe^(2+)))^(@)+E_(OP_(I^(-)//I_(3)^(-)))^(@)`
`= 0.77-0.54 = 0.23 V`
Thus, `0.23 = (0.059)/(2)log_(10)K_(c)`
`K_(c) = 6.26 xx 10^(7)`