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Given E(Cr^(3+)//cr)^@ =- 0.72 V, E(Fe^(...

Given `E_(Cr^(3+)//cr)^@ =- 0.72 V, E_(Fe^(2+)//Fe)^@ =- 0.42 V`. The potential for the cell
`Cr | Cr^(3+) (0.1 M) || FE^(2+) (0.01 M) |` Fe is .

A

`-0.26" V"`

B

`0.26" V "`

C

`0.339" V "`

D

`-0.339" V "`.

Text Solution

Verified by Experts

The correct Answer is:
B
(b) `Cr to Cr^(3+)+3e^(-),E_(red)^(@)=-0.72" V"]xx2`
`Fe^(2+)+2e^(-) to Fe,E_(red)^(@)=-0.42" V"]xx3`
`2Cr+3Fe^(2+) to 2Cr^(3+)+3Fe`
`E_(cell)^(@)=E_(cathode)^(@)-E_(anode)^(@)`
=-0.42-(-0.72)=0.3
According to Nernst equation,
`E_(cell)=E_(cell)^(@)-(0.059)/(n)"log"([Cr^(3+)]^(2))/([Fe^(2+)]^(3))`
`=0.3-(0.059)/(6)"log"((0.1)^(2))/((0.01)^(3))`
`=0.3-(0.059)/(6)"log"10^(4)`
`=0.3-0.39=0.261" V "`
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