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 .

Options

A. 0.26 V
B. 0.399 V
C. `-0.399 V`
D. `-0.26 V`

Answer

A

Explanation

As `E_(Cr^(3+)//Cr)^(@)=0.72 V, E_(Fe^(2+)//Fe)^(@)=-0.42 V`
`2Cr+ 3Fe^(2+) top 3Fe^(2+) to 3Fe+2Cr^(3+)`
`E_("cell")=E_("cell")^(@)=(0.0591)/(6)log""((Cr^(3+))^(2))/((Fe^(2+))^(3))`
`=(-0.42+0.72)-(0.0591)/(6) log ""((0.1)^(2))/((0.01)^(2))`
`=0.30-(0.0591)/(6) log""((0.1)^(2))/((0.01)^(3))`
`=0.30-(0.591)/(6)"log"(10^(-2))/(10^(-6))`
`=0.30-(0.0591)/(6) log 10^(4)`
`E_("cell")=0.2606 V`