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Lalit Rao Sharaf
Asked: 2022-11-09T17:29:41+05:30 In:

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.
Among the following, identify the correct statement
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 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.
Among the following, identify the correct statement
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
Parabola
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  1. Correct Answer – C
    Calculate the EMF of all the cells. Only the EMF of the cell involving the oxidation of `I^(-)` ion by `Cl_(2)` is +ve.
    `{:(2I^(-) rarr I_(2)+2e^(-)”, “E^(@)=-0.54 V),(Cl_(2)+2e^(-) rarr 2Cl^(-)”, “E^(@)=+1.36 V),(bar(Cl_(2)+2I^(-) rarr 2Cl^(-)+I_(2), E_(“cell”)^(@)+0.82 V)):}`
    Thus, option (c) is correct.

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Geetanjali Feroz Trivedi
Asked: 2022-11-02T06:57:08+05:30 In:

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)`

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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  1. 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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Rehman Misra
Asked: 2022-11-01T15:17:27+05:30 In:

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)`

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)`
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  1. 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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Nakul Kannan
Asked: 2022-10-30T14:44:11+05:30 In:

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
A. `O_(2)` oxidises `Mn^(2+)` to `Mn^(3+)`
B. `O_(2)` oxidises both `Mn^(2+)` to `Mn^(3+)` and `Fe^(2+)` to `Fe^(3+)`
C. `Fe^(3+)` oxidises `H_(2)O` to `O_(2)`
D. `Mn^(3+)` oxidises `H_(2)O` to `O_(2)`

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
A. `O_(2)` oxidises `Mn^(2+)` to `Mn^(3+)`
B. `O_(2)` oxidises both `Mn^(2+)` to `Mn^(3+)` and `Fe^(2+)` to `Fe^(3+)`
C. `Fe^(3+)` oxidises `H_(2)O` to `O_(2)`
D. `Mn^(3+)` oxidises `H_(2)O` to `O_(2)`
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  1. Correct Answer – D
    Calculate the EMF of all the cell. Only the EMF of the cell involving oxidation of `H_(2)O` to `O_(2)` by `Mn^(3+)` is +ve
    `{:(Mn^(3+)+e^(-) rarr Mn^(2+)”]”xx4, E^(@)=+1.50 V),(2H_(2)O rarr 4H^(+)+O_(2)+4e^(-), E^(@)=-1.23 V),(bar(4 Mn^(3+)+2 H_(2)O rarr 4 Mn^(2+)+O_(2)+4 H^(+)”, “E_(“cell”)^(@)=+0.27 V)):}`
    Thus, option (d) is corect.

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