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The equilibrium constant of the followin...

The equilibrium constant of the following redox reaction at 298 K is `1 xx 10^(8) 2Fe^(3+) (aq.)+ 2I^(-) (aq.) rarr 2Fe^(2+) (aq)+I_(2) (s)` If the standard reduction potential of iodine becoming iodide is +0.54 V. What is the standard reduction potential of `(Fe^(3+))/(Fe^(2+))` ?

A

+ 1.006 V

B

-1.006 V

C

+0.77 V

D

-0.77 V

Text Solution

Verified by Experts

The correct Answer is:
B
`E^(@)=(0.059)/(n) log_(10) k=(0.059)/(2)log_(10)^(8) =0.236`
`E_(fe^(3+))^(@)//fe^(2+^(@))=0.77V`
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