Calculate the equilibrium constant for the reaction `:`
`2Fe^(3+)+3I^(Θ)hArr2Fe^(2+)+I_(3)^(Θ)`
The standard reduction potential in acidic conditions is `0.78V ` and `0.54V` , respectively, for `Fe^(3+)|Fe^(2+)` and `I_(3)^(c-)|I^(c-)` couples

Answer

`10^(8)`

Explanation

At anode `: 2I^(c-) rarr I_(3)^(c-)+2e^(-)`
At cathode `: 2Fe^(3+)+2e^(-) rarr 2Fe^(2+)`

At equlibrium, `E_(cell)=0`
`E^(c-)._(cell)=(E^(c-)._(red))_(c)-(E^(c-)._(red))_(a)`
`=E^(c-)._((Fe^(3+)|Fe^(2+)))-E^(c-)._((I_(3)^(c-)|3I^(c-)))`
`=0.78-0.54=0.24V`
`E_(cell)=E^(c-)._(cell)-(0.06)/(2)logK_(eq)(n_(cell)=2)`
`E^(c-)._(cell)=0.03logK_(eq)`
`logK_(eq)=(E^(c-)._(cell))/(0.03)=(0.24V)/(0.03V)=8.0`
`K_(eq)=10^(8)`