1.

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 ?

Answer»

`I^(-)` will be REDUCED in `I_(2)`
There will be no redox reaction.
`I^(-)` will be oxidized in `I_(2)`
`Fe^(2+)` will be oxidized in `Fe^(3+)`.

Solution :`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.


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