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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

`0.336` V

C

`-0.339` V

D

`0.26 V `

Text Solution

Verified by Experts

The correct Answer is:
D
From the given representation of the cell `E_(cell)` can be found as follows .
`E_(cell) = E_(-Fe^(2+)// Fe)^(@) - E_(Cr^(3+) // Cr)^(@) - (0.059)/(6) "log" ([Cr^(3+) ]^(2))/([Fe^(2+)]^(3) ` [Nernst - Equ.]
`=-0.42 - (0.72) - (0.059)/(6) "log" ((0.1)^(2))/((0.01)^(3))`
`=-0.42 + 0.72 - (0.059)/(6) "log" (0.1 xx 0.1)/(0.01 xx 0.01 xx 0.01)`
`= 0.3 - (0.059)/(6) "log" (10^(-2))/(10^(-6)) = 0.3 - (0.059)/(6) xx 4`
`=0.30 - 0.0393 = 0.26 V`
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DISHA PUBLICATION-ELECTROCHEMISTRY -Exercise
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