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Given E(Cr^(3+)//Cr)^(@)= 0.72V, E(Fe^(2...

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+) (0.01M)`| Fe is :

A

0.339V

B

`-0.339V`

C

`-0.26V`

D

`0.26V`

Text Solution

Verified by Experts

The correct Answer is:
D
`E_(cell)=E_(cell)^(o)-(0.059)/(n)log" "Q`
`=(-0.42+0.72)-(0.059)/(6)"log"((0.1)^(2))/((0.01)^(3))([CrtoCr^(3+)+3overline(e)]xx2,[Fe^(2+)+2overline(e)toFe]xx3)/(2Cr+3Fe^(++)to2Cr^(3+)+3Fe`
`E_(cell)=0.30-(0.059)/(6)"log"(10^(-2))/(10^(-6))=0.30-(4xx0.059)/(6)log10`
`=0.30-0.0393=0.2607`volt.
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