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The equilibrium constant of the followin...

The equilibrium constant of the following redox reaction at 298 K is `1xx10^(8)`.
`2Fe^(3+)(aq)+2I^(-)(aq) hArr 2Fe^(2+)(aq)+I_(2)(s)`
If the standard reduction potential of iodine becoming iodide is `+0.54V`, what is the standard reduction potential of `Fe^(3+)//Fe^(2+)` ?

A

`+1.006V`

B

`-1.006 V`

C

`+0.77V`

D

`-0.77 V`

Text Solution

Verified by Experts

The correct Answer is:
C
`E_("cell")^(Ɵ)=(0.059)/(n)log K_(c) " " :. n=2`
`E_("cell")^(Ɵ)=(0.059)/(2)log 1xx10^(8)`
`=(0.059)/(2) xx 8 = 0.236V`
`E_("cell")^(Ɵ)=E^(Ɵ)(Fe^(3+)"|"Fe)-E^(@)(I_(2)"|"I^(-))`
`0.236=E^(Ɵ)(Fe^(3+)"|"Fe)-0.54`
`:.E^(Ɵ)(Fe^(3+)"|"Fe)=0.236+0.54`
`=0.776V`.
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