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+)` ?

Options

A. `+1.006 V `
B. `-1.006 V`
C. `+0.77V`
D. `-0.652` V

Answer

D

Explanation

`E_(cell)^(Theta) = (0.059)/(n) log K_(c) because n = 2 `
`E_(cell)^(Theta) = (0.059)/(2) log 1 xx 10^(8)`
`= (0.059)/(2) xx 8 = 0.236 V`
`E_(cell)^(Theta) = E^(Theta) (Fe^(3+) |Fe) - E^(@) (I_(2) |I^(-))`
`0.236 = E^(Theta) (Fe^(3+) |Fe) -0.54`
`therefore E^(Theta) (Fe^(3+) |Fe) = 0.236 + 0.54`
`=0.776 V `