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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

1.212V

B

0.111V

C

0.330V

D

1.653V

Text Solution

Verified by Experts

The correct Answer is:
A

The given cell is
`Fe|Fe^(2+)||Fe^(3+)|Fe^(2+)`
`E_("cell")^(@)=E_(Fe^(3+)//Fe^(2+))^(@)-E_(Fe^(2+)//Fe)^(@)`
`=0.771V-(-0.441V)=1.212V`
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