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Given standard electrode potentials: F...

Given standard electrode potentials:
`Fe^(3+) +3e^(-) rarr Fe, E^(@) =- 0.036` volt
`Fe^(2+) +2e^(-) rarr Fe, E^(@) =- 0.040` volt
The standard electrode potential `E^(@)` for
`Fe^(3+) +e^(-) rarr Fe^(2)` is:-

A

0.772 V

B

`-0.404` V

C

`+0.404 V`

D

`-0.772 V`

Text Solution

Verified by Experts

The correct Answer is:
A
`Delta G=-n FE^(@)`
`Fe^(+2)+2e^(-) to Fe " "…(i)`
`DeltaG=-2xxFxx(-0.440)=0.88 F`
`Fe^(+3)+3e^(-) to Fe" "…(ii)`
Substracting (i) from (ii) we get
`Fe^(+3)+-3e^(-) to Fe^(+2)`
`" "DeltaG=0.108 F-0.880 F =-0.772 F`
`E^(@)` for the reaction
`=-(DeltaG)/(nF)=(-(0.772 F))/(1 F)=+0.772 V`
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