Question

Redox reactions play a pivotal role in chemistry and biology.The values of standard reduction potential `E^(@)` of two half cell reaction decide which way the reaction is expected to proceed. A simple exaple is Danie cell in which zinc goes in to solution and copper gets deposited Given below are set of half cell reaction (acidic medium) along with their `E^(@)` values.
`I_(2)+2e^(-)rarr2I^(-) " "E^(@)=0.54`
`CI_(2)+2e^(-)rarr2CI^(-) " "E^(@)=0.54`
`Mn^(3+)+e^(-)rarrMn^(2+) " "E^(@)=1.36`
`Fe^(3+)+e^(-)rarrMn^(2+)" "E^(@)=0.77`
`O_(2)+4H^(+)e^(-)rarr2h_(2)O" "E^(@)=1.23`
Using these data , obtain the correct explanation for the following question.
Sodium fusion extact obvtined from aniline. On treatment with iron (II suphate and `H_(2)SO_(4))` in presence of air givers a prussian blue precipitte.The blue colour is due to the formation of
A. `Fe_(4)[Fe(CN)_(6)]^_(3)`
B. `Fe_(3)[Fe(CN)_(6)]^_(2)`
C. `Fe_(4)[Fe(CN)_(6)]^_(2)`
D. `Fe_(3)[Fe(CN)_(6)]^_(3)`

Answer 1

Correct Answer - A
`Na+C+NrarrNaCN`
(Sodium fusion extract)
`Fe^(2+)+6CN^(-)rarr[Fe(CN)_(6)]^(4-)`
In presence of air, `Fe^(2+)` ions get oxidised to `Fe^(3+)` ions.
`(Fe^(2+)rarrFe^(3+)+e^(-)]xx4,E^(@)=-0.77V)`
`(O_(2)+4H^(+)+4e^(-)rarr2H_(2)O,E^(@)=+1.23 V)/(4Fe^(2+)+4H^(+)+O_(2)rarr4Fe^(3+)+2H_(2)O)`
`Fe^(3+)` ions then combine with `[Fe(CN)_(6)]^(4-)` ion to form ferric ferrocyanide which has prussian blue colour `4Fe^(3+)+3[Fe(CN)_(6)]^(4-)rarrFe_(4)[Fe(CN)_(6)]_(3)`