Given `E_("Cr_(2)O_(7)^(2-)//Cr^(3+))^(@)=1.33V,E_(MnO_(4)^(-)//Mn^(2+))^(@)=1.51V`
Among the following, the strongest reducing agent is `E_(Cr^(3+)//Cr)^(@)=-0.74V^(x),E_(MnO_(4)^(-)//Mn^(2+))^(@)=1.51V`
`E_(Cr_(2)O_(7)^(2-)//Cr^(3+))^(@)=1.33V,E_(Cl//Cl^(-))^(@)=1.36V`
Based on the data given above strongest oxidising agent will be

Options

A. `MnO_(4)^(-)`
B. `Cl^(-)`
C. `Cr^(2+)`
D. `Mn^(2+)`

Answer

A

Explanation

In the given reaction, i has been oxidised to 1 and `Cr_(2)O_(7)^(2-)` has been reduced to `Cr^(3+)`
`therefore E_("cell")^(@)=E_(Cr_(2)O_(7)^(2-))^(@)-E_(I_(2))^(@)`
`therefore 0.79=1.33-E_(I_(2))^(@)` or `E_(I_(2))^(@)=1.33-0.79=0.54`