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+))` ?

Options

A. `+1.006V`
B. `-1.006V`
C. `+0.77V`
D. `-0.77V`

Answer

C

Explanation

`E^(@)= (0.059)/(n) "log"_(10)K =(0.059)/(2) Log_(10)10^(8)= 0.236`
`E_("cell")^(@)= E_("Reduced species")^(@)- E_("Oxidised species")^(@) , 0.236 = E_(Fe^(3+)//Fe^(2+))^(@)- 0.54, E_(Fe^(3+)//Fe^(2+))^(@)= 0.77V`