1.

GivenE_(Cr^(3+)//Cr)^(@)=-0.72 V, E_(Fe ^(2+)//Fe)^(@)=-0.42V. The potential for the cell Cr|Cr^(3l) (0.1M)||Fr ^(2l)(0.01M)|Fe is

Answer»

`0.26V`
`0.336V`
`-0.339V`
`0.26V`

Solution :From the given representiaon of the cell, `E_(cell)` can be found as FOLLOWS.
`E_(cell) =(E_(Fe^(@+)//Fe)^(@)-E_(Cr^(3+)//Cr)^(@))-(0.059)/(6) log ""([Cr^(3+)])/([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 xx0.1)/(0.01xx0.01xx0.01)`
`=0.3-(0.059)/(6) log ""(10^(-2))/(10^(-6)) =0.3 -(0.059)/(6) xx4`
`= 0.30-0.0393 =0.26V`
Hence OPTION (d) is CORRECT answer.


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