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A solution contains Cr^(+2),Cr^(+3) and ...

A solution contains `Cr^(+2),Cr^(+3)` and `I^(-)` ions, This solution was treated with iodine at `35^(@)C.E^(@)` for `cr^(+3)//Cr^(+2)` is `-0.41V` and `E^(@)` for `I_(2)//2I^(-)=0.536V`, The favourable redox reaction is :-

A

`Cr^(+2)` will be oxidised to `Cr^(+3)`

B

`Cr^(+3)` will reduced to `Cr^(+2)`

C

There will be no redox reaction

D

`I^(-)` will oxidised to `I_(2)`

Text Solution

Verified by Experts

The correct Answer is:
A
`E^(@)(I_(2)//2I^(-))gtE^(@)(Cr^(+3)//Cr^(+2))`
So `I_(2)` will reduced to `I^(-)` and `Cr^(+2)` will oxidised to `Cr^(+3)`
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