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Using the data given below, E(Cr(2)O(7...

Using the data given below,
`E_(Cr_(2)O_(7)^(2-)//Cr^(3+))^(Ɵ)=1.33V " " E_(Cl_(2)//Cl^(-))^(Ɵ)=1.36V`
`E_(MnO_(4)^(-)//Mn^(2+))^(Ɵ)=1.51V" " E_(Cr^(3+)//Cl)^(Ɵ)=0.74V`
the order of reducing power is

A

`Cr^(3+) lt Cl^(-) lt Mn^(2+) lt Cr`

B

`Mn^(2+) lt Cl^(-) lt Cr^(3+) lt Cr`

C

`Cr^(3+) lt Cl^(-) lt Cr_(2)O_(7)^(2-) lt MnO_(4)^(-)`

D

`Mn^(2+) lt Cr^(3+) lt Cl^(-) lt Cr`

Text Solution

Verified by Experts

The correct Answer is:
B
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Given E_(Cr^(3+)//Cr)^(@)=-0.74V,E_(MnO_(4)^(-)//Mn^(2+))^(@) =1.51V E_(Cr_(2)O_(7)^(2-)//Cr^(3+))^(@)=1.33V,E_(Cl//C//l^(-))^(@)=1.36V Based on the data given above , strongest oxidising agent will be :

E_(Mn)^(@)+7|""_(Mn)+2 = 1.5 V E_(Mn)^(@)+4|""_(Mn)+2 = 1.2 V, then E_(Mn)^(@)+7|""_(Mn)+4 is

Calculate the potential of the following cell at 298 K Zn//Zn^(2+)(a=0.1)//Cu^(2+)(a=0.01)//Cu E_(Zn^(2+)//Zn)^(@)=-0.762V E_(Cu^(2+)//Cu)^(@)=+0.337V Compare the free energy change for this cell with the free enegy of the cell in the standard state.

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