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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

1 Answer

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Best answer
Correct Answer - 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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