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Given, E(Cr^(3+)//Cr)^(0)=-0.74V,E(MnO(4...

Given, `E_(Cr^(3+)//Cr)^(0)=-0.74V,E_(MnO_(4)^(-)//Mn^(2+))^(0)=1.51V`,
`E_(Cr_(2)O_(7)^(2-)//Cr^(3+))^(0)=1.33V,E_(Cl//Cl^(-))^(0)=1.36V,` from above given information decide which is strong oxidizing agent ?

A

`Cl^(-)`

B

`Cr^(3+)`

C

`Mn^(2+)`

D

`MnO_(4)^(-)`

Text Solution

Verified by Experts

The correct Answer is:
D
From, given data, `E^(@)(MnO_(4)^(-)//Mn^(2+))=1.51V`, which is a maximum value. So reduction of `MnO_(4)^(-)` is easy among given all electrodes. So `MnO_(4)^(-)` is strongest oxidizing agent.
`MnO_(4)^(-)+5e^(-)+8H^(+)toMn^(2+)+4H_(2)O`
`Cl+e^(-) to Cl^(-)`
`Cr_(2)O_(7)^(2-)+14H^(+)+6e^(-) to 2Cr^(3+)+7H_(2)O`
`Cr^(3+)+3e to Cr`.
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