Given the overall formation constant of the `[Fe(CN)_(6)]^(4-)` ion as `10^(35)` and the standard potentials for the half reaction,
`Fe^(3+)+e^(-) hArr Fe^(2+) , E^(@)=0.77 V`
`[Fe(CN)_(6)]^(3-)+e^(-)hArr [Fe(CN)_(6)]^(4-) , E^(@)=0.36 V`
Calculate the overall formation constant of the`[Fe(CN)_(6)]^(3-)` ion.

Options

A. `8.59xx10^(41)`
B. `7.59xx10^(41)`
C. `5.59xx10^(41)`
D. `9.59xx10^(41)`

Answer

C

Explanation

(A) Let `K_(f)` be the formation constant of `[Fe(CN)_(6)]^(3-)` ion.
`Fe^(2+)+6CN^(-)hArr[Fe(CN)_(6)]^(4-),K_(f)=10^(35)`,
`Delta G_(1)^(@)=-2.303RT log K_(f)=-199704.69 J`
`Fe^(3+)+e^(-)hArr Fe^(2+), E^(@)=0.77V` ,
`Delta G_(2)^(@)=-96500xx0.77=-74305 J`
`Delta G_(3)^(@)=+96500xx0.36=34740 J`
`[Fe(CN)_(6)]^(4-) hArr [Fe(CN)_(6)]^(3-)+e^(-)`,
`E^(@)=-0.36 V`,
`bar(Fe^(3+)+6CN^(-)hArr [Fe(CN)_(6)]^(3-)`,
`Delta G_(4)^(@)=Delta G_(1)^(@)+Delta G_(2)^(@)+Delta G_(3)^(@)=-239269.69 J`
`Delta G_(4)^(@)=-2.303RT log K _(f) therefore K _(f)=8.59xx10^(41)`
`2Fe^(3+)+2e^(-)rarr 2Fe^(2+)` or