Given below are a set of half-cell reactions (acidic medium) along with their `E_(@)` with respect to normal hydrogen electrode values. Using the data obtain the correct explanation to question given below. `{:(I_(2)+2e^(-)rarr2I^(-),E^(@)=0.54),(Cl_(2)+2e^(-)rarr2Cl^(-),E^(@)=1.36),(Mn^(2+)+e^(-)rarrMn^(2+),E^(@)=1.50),(Fe^(3+)+e^(-)rarrFe^(2+),E^(@)=0.77),(O_(2)+4H^(+)+4e^(-)rarr2H_(2)O,E^(@)=1.23):}`
While `Fe^(2+)` is stable, `Mn^(3+)` is not stable in acid solution because:
Given below are a set of half-cell reactions (acidic medium) along with their `E_(@)` with respect to normal hydrogen electrode values. Using the data obtain the correct explanation to question given below. `{:(I_(2)+2e^(-)rarr2I^(-),E^(@)=0.54),(Cl_(2)+2e^(-)rarr2Cl^(-),E^(@)=1.36),(Mn^(2+)+e^(-)rarrMn^(2+),E^(@)=1.50),(Fe^(3+)+e^(-)rarrFe^(2+),E^(@)=0.77),(O_(2)+4H^(+)+4e^(-)rarr2H_(2)O,E^(@)=1.23):}`
While `Fe^(2+)` is stable, `Mn^(3+)` is not stable in acid solution because:
While `Fe^(2+)` is stable, `Mn^(3+)` is not stable in acid solution because:
A
`O_(2) "oxidises lt" n^(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)`
Text Solution
Verified by Experts
The correct Answer is:
D
In this case also, only the e.m.f of the cell involving the oxidation of `H_(2)O to O_(2)` by `Mn^(3+)` ion is positive
`{:(Mn^(3+)+e^(-)rarrMn^(2+)xx4,E^(@)=+1.50 V),(2H_(2)Orarr4H^(+)+O_(2)+4e^(-),E^(@)=-1.23V),(Mn^(3+)+2h_(2)Orarr4Mn^(2+)+O_(2)+4H^(+),E^(@)=+0.27V):}`
`{:(Mn^(3+)+e^(-)rarrMn^(2+)xx4,E^(@)=+1.50 V),(2H_(2)Orarr4H^(+)+O_(2)+4e^(-),E^(@)=-1.23V),(Mn^(3+)+2h_(2)Orarr4Mn^(2+)+O_(2)+4H^(+),E^(@)=+0.27V):}`
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Redox reactions play a pivoted role in chemistry and biology. The values of standard redox potential and biology. The values of standard redox potential (E^0) of two half cell reactions decide which way the raction is expected to proceed. A simple example is a Daniel cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium) along with their E^0 (V with respect to normal hydrogen eletrode ) values. Using this dat obtain the correct explanations to Questions I_2+2e^_ rarr2I^- E^0=0.54 Cl_2+2e^_ rarr2Cl^- E^0=1.36 Mn^(3+)+e^_ rarrMn^(2+) E^0=1.50 Fe^(3+)+e^_ rarrFe^(2+) E^0=0.77 O_2+4H^++4e^(-) rarr 2H_2O E^0=1.23 While Fe^(3+) is stable, Mn^(3+) is not stable in acid soltuion because
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 :
Knowledge Check
Given below are a set of half-cell reactions (acidic medium) along with their E_(@) with respect to normal hydrogen electrode values. Using the data obtain the correct explanation to question given below. {:(I_(2)+2e^(-)rarr2I^(-),E^(@)=0.54),(Cl_(2)+2e^(-)rarr2Cl^(-),E^(@)=1.36),(Mn^(2+)+e^(-)rarrMn^(2+),E^(@)=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:
Given below are a set of half-cell reactions (acidic medium) along with their E_(@) with respect to normal hydrogen electrode values. Using the data obtain the correct explanation to question given below. {:(I_(2)+2e^(-)rarr2I^(-),E^(@)=0.54),(Cl_(2)+2e^(-)rarr2Cl^(-),E^(@)=1.36),(Mn^(2+)+e^(-)rarrMn^(2+),E^(@)=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:
A
Chloride ion is oxidised by `O_(2)`
B
`Fe^(2+)` is oxidised by iodine.
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 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 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 While Fe^(3+) is stable, Mn^(3+) is not stable in acid solution because
A
`O_(2)` oxidizes `Mn^(2+)` to `Mn^(3+)`
B
`O_(2)` oxidizes both `Mn^(2+)` to `Mn^(3+)` and `Fe^(2+)` to `Fe^(3+)`
C
`Fe^(3+)` oxidizes `H_(2)O` to `O_(2)`
D
`Mn^(3+)` oxidized `H_(2)O` to `O_(2)`
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. While Fe^(3+) is stable, Mn^(3+) is not stable in acid solution because
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. 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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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 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" . While Fe^(2+) ion is stable, Mn^(2+) ion is not stable in acid solution because :
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