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

Given
`E_(cr^(3+)//Cr)^(0)=-0.74 V, E_(MnO_(4)//Mn^(2+))^(0)=1.51 cm`
`E_(Cr_(2)O_(7)^(2-)//Cr^(3+))^(0)=1.33 V, E_(Cl//Cl^(-))^(0)=1.36 V`
Based on the data given above, strongest oxidising agent will be:

A

`MnO_(4)^(-)`

B

`CI^(-)`

C

`Cr^(3+)`

D

`Mn^(2+)`

Text Solution

Verified by Experts

The correct Answer is:
a
Higher the `E^(@)` stronger is the oxidising agent thus `MnO_(4)^(-)` with `E^(@)=1.51` V is the stongest oxidising agent
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