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The equilibrium constant of the followin...

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

A

`+1.006V`

B

`-1.006V`

C

`+0.77V`

D

`-0.77V`

Text Solution

Verified by Experts

The correct Answer is:
C
`E^(@)=(0.059)/(n) log_(10)K`
`=(0.059)/(2) log_(10) 10^(8)`
=0.236
`E_("cell")^(@)=E_("Reduced species")^(@)-E_("Oxidised species")^(0)`
`0.236 =E_(Fe^(3+)//Fe^(2+)^@-0.54`
`E_(Fe^(3+)//Fe^(2+))^(@)=0.77V`
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