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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

`0.399` V

C

`-0.339` V

D

`-0.26 V`

Text Solution

Verified by Experts

The correct Answer is:
A
As `E_(Cr^(3+)//Cr)^(@) = -0.72V and E_(Fe^(2+)//Fe)^(@)= -0.42V`
`2Cr +3Fe^(2+) to 3Fe +2Cr^(3+) `
`E_("cell") =E_("cell")^(@) - (0.0591)/(6)"log" ([0.1]^(2))/([Fe^(2+)]^(3)) = (-0.42 +0.72) -(0.0591)/(6)"log"([0.1]^(2))/([0.01]^(3))`
`=0.30 -(0.0951)/(6)"log"([0.1]^(2))/([0.01]^(3))=0.30 -(0.0591)/(6) log 10^(4)" " :. E_("cell")=0.2606V`
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