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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

`0.26V`

B

`0.339V`

C

`-0.339V`

D

`-0.26V`

Text Solution

Verified by Experts

The correct Answer is:
A
Anode
`2Cr - 6e rarr 2Cr^(3+)`
Cathode `3Fe^(2+) +6e rarr 3Fe`
Net cell reaction
`2Cr +3Fe^(2+) rarr 2Cr^(3+) + 3Fe`
Applying Nernst equation
`E = E_(cell)^(@)-(0.059)/(6)log.([Cr^(3+)]^(2))/([Fe^(2+)]^(3))`
`= 0.30 - (0.059)/(6)log.((01)^(2))/((0.01)^(3))`
`= 0.3 - 0.01 xx log 10^(4)`
`= 0.3 - 0.04 = 0.26 V`
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