A solution contains `Fe^(2+), Fe^(3+)` and `I^(-)` ions. The solution was treated with iodine at `35^(@)C. E^(@)` for `Fe^(3+)//Fe^(2+)` is `+0.77" V

Question

A solution contains `Fe^(2+), Fe^(3+)` and `I^(-)` ions. The solution was treated with iodine at `35^(@)C. E^(@)` for `Fe^(3+)//Fe^(2+)` is `+0.77" V

A solution contains `Fe^(2+), Fe^(3+)` and `I^(-)` ions. The solution was treated with iodine at `35^(@)C. E^(@)` for `Fe^(3+)//Fe^(2+)` is `+0.77" V "` and `E^(0)` for `I_(2)//2I^(-)` is `+0.536" V"`. The favourable redox reaction is :
A. `I_(2)` will be reduced to `I^(-)`
B. These will be no redox reaction
C. `I_(2)` will be oxidised to `I_(2)`
D. `Fe^(2+)` will be oxidised to `Fe^(3+)`

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - C
(c ) Since the reduction potential of `Fe^(2+)//Fe^(3+)` is more than that of `l_(2)//l^(-)`, `l^(-)` in the redox reaction. `l^(-)` ions will be oxidised to `l_(2)` at the cost of `Fe^(3+)` ions which will be reduced :
`2Fe^(3+)+2l^(-) to 2Fe^(2+)+l_(2)`

4 views Answered 2023-02-05 15:29