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 ofA. `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)`

Redox reactions play a pivotal role in chemistry and biology.The value

1.	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 ofA. `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» 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)`

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