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Given E((Cr^(3+)) / (Cr)=-0.72 V . E(Fe^...

Given `E_((Cr^(3+)) / (Cr)=-0.72 V . E_(Fe^(2+)//Fe)^(@)=-0.42 V` 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) hArr 2Cr^(3+) (0.1M) + 3Fe(s), Q= ([Cr^(3+)]^(2))/([Fe^(2=)]^(3))= ([0.1]^(2))/([0.01]^(3))= 10^(4)`
According to Nernst equation, `E= 0.30- (0.059)/(6) log 10^(4)= 0.261V (because n=6)`
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