A solution contains `Fe^(2+), Fe^(3+)` and `T^(-)` ions. This solution was treated with iodine at `35^(@)C. E^(@)` for `Fe^(3+), Fe^(2+)` is `0.77 V` and `E^(@)` for `I_(2)//2I^(-)` = 0.536 V. The favourable redox reaction is:

Options

A. `I_2` will be reduced to `I^(-)`
B. There will be no redox reaction.
C. ( `I^(-)` will be oxidised to `I_2`.
D. `Fe^(2+)` will be oxidised to `Fe^(3+)`.

Answer

C

Explanation

(c) Given `Fe^(3+)//Fe^(2+)` = +0.77 V
and `I_(2)//2I^(-)` = 0.536V
`2(Fe^(3+) +e^(-) rarr Fe^(2+))" "E^(circ)` = 0.77V

`underline(2I^(-) rarr I_2 + 2e^(-)" "E^(circ)` =-0.536V