One solution contains `Fe^(2+),Fe^(3+) and I^(-)` ions. Such solution is reacted with iodine solution at `35^(@)C`. If `E^(@)` of `Fe^(3+)//Fe^(2+)` is +0.77V and for `I_(2)//2I^(-),E^(@)=0.535V`. So what is probable redox reaction ?

Options

A. `I^(-)` will be reduced in `I_(2)`
B. There will be no redox reaction.
C. `I^(-)` will be oxidized in `I_(2)`
D. `Fe^(2+)` will be oxidized in `Fe^(3+)`.

Answer

C

Explanation

`Fe^(+3)|Fe^(2+)=0.77V and I_(2)|2I^(-)=0.536V` is given
`2(e^(-)+Fe^(+3)toFe^(+2))" "E_(red)^(@)=0.77V(Fe^(3+)|Fe^(2+))`
`underline(2I^(-) to I_(2)+2e^(-)" "E_(red)^(@)=0.536V(I_(2)|I^(-)))`
`2Fe^(+3)+2I^(-) to 2Fe^(+2)+I_(2)`
`E^(@)=E_(Fe^(3+)|Fe^(2+))^(@)-E_(I_(2)|I^(-))^(@)`
`=0.77-0.536=0.164V`
So reaction will occur.