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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

0.339 V

B

`-0.339V`

C

`-0.26V`

D

0.26 V

Text Solution

Verified by Experts

The correct Answer is:
D
`E_(cell)=E_(cell)^(@)-(0.059)/(6)log(([Cr^(+3)]^(2))/([Fe^(+2)]^(3))`
`=0.3-(0.056)/(6)log(((0.1)^(2))/((0.01)^(3)))=0.3-0.04`
`=0.26V`
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