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
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
`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)`
(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 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. While Fe^(3+) is stable, Mn^(3+) is not stable in acid solution because
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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 of
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 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)`
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A
Chloride ion is oxidised by `O_(2)`
B
`Fe^(2+)` is oxidised by iodide
C
Iodide ion is oxidised by chlorine
D
`Mn^(2+)` is oxidised by chlorine
Redox reactions play a pivotyal role in chemistry and biology. The values of standard redox potential (E^(@)) of two half-cell reactions decide which way the reaciton is expected to proced. A simple exampler is a Daniel cell in which zinc goes into solution and copper gerts deposited. Given below are a set of half-cell reactions (acidic medium) along with their E^(@) ( V with respect to normal hydrofgen electrode) values. Using this data : I_(2) + 2e^(-) rarr 2I^(-)" " E^(@) = 0.54 , CI_(2) + 2e^(-) rarr 2CI^(-) " "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 , Soldium fusion extract, obtained from aniline, on treatment with iron (II) sulphatge and H_(2)SO_(4) in presence of air gives a Prussion bule precipitate. The blue colour is due to the formation of :
Redox reactions play a pivotyal role in chemistry and biology. The values of standard redox potential (E^(@)) of two half-cell reactions decide which way the reaciton is expected to proced. A simple exampler is a Daniel cell in which zinc goes into solution and copper gerts deposited. Given below are a set of half-cell reactions (acidic medium) along with their E^(@) ( V with respect to normal hydrofgen electrode) values. Using this data : I_(2) + 2e^(-) rarr 2I^(-)" " E^(@) = 0.54 , CI_(2) + 2e^(-) rarr 2CI^(-) " "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 , Soldium fusion extract, obtained from aniline, on treatment with iron (II) sulphatge and H_(2)SO_(4) in presence of air gives a Prussion bule precipitate. 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)`
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Redox reaction 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 a Daniel cell in which zinc goes into solution and copper gets deposited. give below are a set of half-cell reaction (acidic medium) along with their E^(@) (V with respect to normal hydrogen electrode) values. using this data obtain the correct explanation to folloing Questions. I_(2)+2e^(-)to2I^(-)" "E^(0)=0.54 Cl_(2)+2e^(-)to2Cl^(-)" "E^(0)=1.36 Mn^(3+)+e^(-)toMn^(2+)" "E^(0)=1.50 Fe^(3+)+e^(-)toFe^(2+)" "E^(0)=0.77 O_(2)+4H^(+)+4e^(-)to2H_(2)O" "E^(0)=1.23 Q. Sodium fusion extract, obtained from aniline, on treatment with iron (II) sulphate and H_(2)SO_(4) in presence of air gives a prussian blue precipitate. the blue colour is due to the formation of
Redox reactions play a pivotal role in chemistry and biology. The value of standard reduction potetials (E^(@)) of the two half cells reactions decide which way the reaction is expected to proceed. A simple example is of Daniell cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium) along with with E^(@) values. Using the data obtain the correct explanation to the questions that are mentioned. I_(2)+e^(-) to 2I^(-) , E^(@)=0.54" V " Cl_(2)+2e^(-) to 2Cl^(-) , E^(@)=1.36" V" Mn^(2+)+2e^(-) to Mn , E^(@)=1.50" V" Fe^(3+)+e^(-) to Fe^(2+) , E^(@)=0.77" V" O_(2)+4H^(+)+4e^(-) to 2H_(2)O , E^(@)=1.23" V" . Sodium fusion extract obtained from aniline, on treatment with iron (II) sulphate and H_(2)SO_(4) in the presence of air, gives a prussian blue precipitate. The blue colour is due to the formation of :
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
Redox reactions play a pivotal role in chemistry and biology. The value of standard reduction potetials (E^(@)) of the two half cells reactions decide which way the reaction is expected to proceed. A simple example is of Daniell cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium) along with with E^(@) values. Using the data obtain the correct explanation to the questions that are mentioned. I_(2)+e^(-) to 2I^(-) , E^(@)=0.54" V " Cl_(2)+2e^(-) to 2Cl^(-) , E^(@)=1.36" V" Mn^(2+)+2e^(-) to Mn , E^(@)=1.50" V" Fe^(3+)+e^(-) to Fe^(2+) , E^(@)=0.77" V" O_(2)+4H^(+)+4e^(-) to 2H_(2)O , E^(@)=1.23" V" . Among the following, identify the correct statement :