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

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

A

`0.26V`

B

`0.339V`

C

`-0.399V`

D

`-0.26V`

Text Solution

Verified by Experts

The correct Answer is:
A
As `E_(cr^(3+)//Cr)^(0)=-0.72V and E_(Fe^(2+)//Fe)^(0)=-0.42V`
`2Cr+3F^(2+)rarr3Fe+2Cr^(3+)`
Six electrons (n = 6) are used in redox change.
`E_(cell)=E_(cell)^(0)-(0.0591)/(n)log.([Cr^(3+)]^(2))/([Fe^(2+)]^(3))`
`=(-0.42+0.72)-(0.059)/(6)log.((0.1)^(2))/((0.01)^(3))`
`=0.30-(0.0591)/(6)log.((0.1)^(2))/((0.01)^(3))`
`=0.30-(0.0591)/(6)log10^(4)`
`=0.2606V`
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