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Given standard electode potenitals Fe...

Given standard electode potenitals
` Fe^(2+) + 2e^(-) rarr Fe, E^@ =- 0. 44 V ` ………………(1)
` Fe^(3+) + 2e^(-) rarr Fe, E^@ =- 0. 036 V` ………….(2)
The standard electrode pptential `E^@ ` for
`Fe^(2+) + e rarr Fe^(2+)` is.

A

`- 0.476V`

B

`-0.404V`

C

`+0.404V`

D

`+0.772V`

Text Solution

Verified by Experts

The correct Answer is:
D
`Fe^(2+)+ 2e^(-)rarrFe:E^(@)=-0.44V`
`Fe^(3+)+3e^(-)rarrFe:E^(@)=-0.036V`

Subtracting first from second
`Fe^(3+)+e^(-)rarrFe^(2+):E^(@)=0.108-( -0.88)`
=0.772 V mol
Since `n=1,E_(Fe^(3+)//Fe^(2+))^(@)=(0.772V"mol")/(1"mol")`
=0.722V
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