Redox reactions play a pivotal role in chemistry and biology. The values of standard redox potential `(E^(@))` of two half-cell reactions decide which way the reaction is expected to proceed. A simple example is Daniel Cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium) aong with their `E^(@)` (V with respect to normal hydrogen electrode) values. `{:(I_(2)+2e^(-) rarr 2I^(-),E^(@)=0.54),(Cl_(2) +2e^(-) rarr 2Cl^(-),E^(@)=1.36),(Mn^(3+)+e^(-)rarr Mn^(2+),E^(@)=1.50),(Fe^(3+)+e^(-) rarr Fe^(2+),E^(@)=0.77),(O_(2)+4H^(+)+4e^(-) rarr 2H_(2)O,E^(@)=1.23):}` Using these data, obtain the correct explanation for the following questions. Sodium fusion extract, obtained from anline, on treatment with iron (II) Sulphate and `H_(2)SO_(4)` in presence of air gives a prussian blue precipitate. Hence, 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 valu

1.	Redox reactions play a pivotal role in chemistry and biology. The values of standard redox potential `(E^(@))` of two half-cell reactions decide which way the reaction is expected to proceed. A simple example is Daniel Cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium) aong with their `E^(@)` (V with respect to normal hydrogen electrode) values. `{:(I_(2)+2e^(-) rarr 2I^(-),E^(@)=0.54),(Cl_(2) +2e^(-) rarr 2Cl^(-),E^(@)=1.36),(Mn^(3+)+e^(-)rarr Mn^(2+),E^(@)=1.50),(Fe^(3+)+e^(-) rarr Fe^(2+),E^(@)=0.77),(O_(2)+4H^(+)+4e^(-) rarr 2H_(2)O,E^(@)=1.23):}` Using these data, obtain the correct explanation for the following questions. Sodium fusion extract, obtained from anline, on treatment with iron (II) Sulphate and `H_(2)SO_(4)` in presence of air gives a prussian blue precipitate. Hence, 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+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))`

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