Calculate the equilibrium constant for the reaction : `2Fe^(3+)+3I^(-)hArr 2Fe^(2+)+I_(3)^(-)`. Given: Standard reduction potential in acidic conditions is 0.78 V and 0.54 V, respectively, for `Fe^(3+)|Fe^(2+)` and `I_(3)^(-)|I^(-)` couples.

Explanation

Given, `E_(Fe^(3+)|Fe^(2+)^(0)=0.78V` and `E_(I_(3)^(0)|I^(-))^(0)=0.54V`
The half - cell reactions are as follows :
`{:("At anode: "3I^(-)" "rarr" "I_(3)^(-)+2e", "E^(0)=0.54V),(ul("At cathode: "2Fe^(3+)+2e" "rarr" "2Fe^(2+)", "E^(0)=+0.78V)),("Cell reaction: "2Fe^(3+)+3I^(-)hArr 2Fe^(2+)+I_(3)^(-)):}`
`therefore" "E_("cell")^(@)=(0.78-0.54)=0.24V`
We know, `E_("cell")^(@)=(0.059)/(2)logK_(eq)" or, "0.24 = 0.029log K_(eq)`
or,`" "logK_(eq)=(0.24)/(0.029)=8.27~=8.00 therefore K_(eq)=10^(8)`