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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

`-0.26V`

B

0.26V

C

0.339V

D

`-0.339V`

Text Solution

Verified by Experts

The correct Answer is:
B
`E_("cell")^(@)=-0.42 -(0.72)=+0.30V`
`2Cr(s)+3Fe^(2+) (0.01M) Leftrightarrow 2Cr^(3+) (0.1M) +3Fe(s)`
`Q=[Cr^(3+)]^(2))/([Fe^(2+)]^(3))=([0.1)^(2)])/([0.01])=10^(4)`
According to Nerst equation
`E=E^(@)-(0.059)/(n) log_(10)Q`
`=0.30-(0.059)/(n) log_(10)Q`
`=0.30 -(0.059)/(6) log 10^(4) (therefore n=6)`
`=0.261V`
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