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If E(Fe^(2+))^(@)//Fe = -0.441 V and E...

If `E_(Fe^(2+))^(@)//Fe = -0.441 V`
and `E_(Fe^(3+))^(@)//Fe^(2+) = 0.771 V`
The standard `EMF` of the reaction
`Fe+2Fe^(3+) rarr 3Fe^(2+)`
will be:

A

`1.653" V "`

B

`1.212" V "`

C

`0.111" V "`

D

`0.330" V "`.

Text Solution

Verified by Experts

The correct Answer is:
B

(b) Anode :`Fe to Fe^(2+)+2e^(-)" " E^(@)=-0.441" V "`.
Cathode : `Fe^(3+)+e^(-) to Fe^(2+)]xx2E^(@)=+0.77" V "`.
`E_(cell)=E_(cathode)^(@)-E_(anode)^(@)`
`=+0.771-(-0.441)=1.212" V "`.
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