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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+) (0.1 M) || FE^(2+) (0.01 M) |` Fe is .

A

`-0.26` V

B

0.26 V

C

0.339 V

D

`-0.339 V`

Text Solution

Verified by Experts

The correct Answer is:
B
`E_(cell)^(@)=-0.42 -(-0.72)=+0.30V`
`2Cr(s)+3Fe^(2+)(0.01 M) hArr 2Cr^(3+) (0.1 M)+3Fe(s)`
`Q=([Cr^(3+)]^(2))/([Fe^(2+)]^(3))=([0.1]^(2))/([0.01]^(2))=10^(4)`
According to Nernst equation,
`E=E^(@)-0.059/n log_(10) Q`
`=0.30-0.059/6 log10^(4)" "( :' n=6)`
`=0.261 V`
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