Redox reactions play a pivotal role in chemistry and biology. The values standard redox potential `(E^(c-))` of two half cell reactions decided which way the reaction is expected to preceed. A simple example is a Daniell cell in which zinc goes into solution and copper sets deposited. Given below are a set of half cell reactions `(` acidic medium `)` along with their `E^(c-)(V` with respect to normal hydrogen electrode `)` values. Using this data, obtain correct explanations for Question.
`I_(2)+2e^(-) rarr 2I^(c-)," "E^(c-)=0.54`
`Cl_(2)+2e^(-) rarr 2Cl^(c-), " "E^(c-)=1.36`
`Mn^(3+)+e^(-) rarr Mn^(2+), " "E^(c-)=1.50`
`Fe^(3+)+e^(-) rarr Fe^(2+)," "E^(c-)=0.77`
`O_(2)+4H^(o+)+4e^(-) rarr 2H_(2)O, " "E^(c-)=1.23`
Sodium fusion extract obtained from aniline on treatment with iron `(II)` sulphate and `H_(2)SO_(4)` in the presence of air gives a Prussion blue precipitate. The blue colour is due to the formation of

Options

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

A

Explanation

`Na+C+N rarr NaCN`
`Fe^(2+)+6 CN^(-) rarr [Fe(CN)_(6)]^(4-)`
In presence of air, `Fe^(2+)` ions get oxidised to `Fe^(3+)` ions.
`{:(Fe^(2+) rarr Fe^(3+) +e^(-) "]"xx4, E^(@)=-0.77 V),(O_(2)+4 H^(+)+ 4e^(-) + 4e^(-) rarr 2H_(2)O, E^(@)=+1.23 V),(bar(4 Fe^(2+)+4 H^(+)+O_(2) rarr 4 Fe^(3+) +2 H_(2)O", "E_("cell")^(@)= +0.46 V)):}`
`Fe^(3+)` ions then combine with `[Fe(CN)_(6)]^(4-)` ion to form ferric ferrocyanide which has prussian blu colour.
`4 Fe^(3+)+3[ Fe(CN)_(6)]^(4-) rarr underset("Prussian blue")(Fe_(4)[Fe(CN)_(6)]_(3))`