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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)`

Text Solution

Verified by Experts

The correct Answer is:
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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Redox reactions play a vital role in chemistry and biology. The values of standard redox potential (E^(@)) of two half-cells reactions decide which way the reaction is expected to proceed. A simple example is a Daniell cell in which zince goes into solution and copper gets deposited. Given below are set of half-cell reactions (acidic medium ) along with their E^(@) in V with respect to normal hydrogen electrode values. {:(l_(2)+2e^(-)rarr2l^(-),E^(@)=0.54),(Cl_(2)+2e^(-)rarr2Cl^(-)" ",E^(@)=1.36),(Mn^(3+)+e^(-)rarrMn^(2+),E^(2)=1.50),(Fe^(3+)+e^(-)rarrFe^(2+)" ",E^(@)=0.77),(O_(2)+4H^(+)+4e^(-)rarr2H_(2)O,E^(@)=1.23):} Among the following, identify the correct statement

Redox reactions play a vital role in chemistry and biology. The values of standard redox potential (E^(@)) of two half-cells reactions decide which way the reaction is expected to proceed. A simple example is a Daniell cell in which zince goes into solution and copper gets deposited. Given below are set of half-cell reactions (acidic medium ) along with their E^(@) in V with respect to normal hydrogen electrode values. {:(l_(2)+2e^(-)rarr2l^(-),E^(@)=0.54),(Cl_(2)+2e^(-)rarr2Cl^(-)" ",E^(@)=1.36),(Mn^(3+)+e^(-)rarrMn^(2+),E^(2)=1.50),(Fe^(3+)+e^(-)rarrFe^(2+)" ",E^(@)=0.77),(O_(2)+4H^(+)+4e^(-)rarr2H_(2)O,E^(@)=1.23):} while Fe^(3+) is stable, Mn^(3+) is not stable in acid solution because :

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

DINESH PUBLICATION-REDOX REACTIONS-JEE(Joint Entrance Examination) Comprehension 3
  1. Redox reactions play a pivotal role in chemistry and biology.The value...

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