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Given below are a set of half-cell react...

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

Text Solution

Verified by Experts

The correct Answer is:
C
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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:

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 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):} Among the following, identify the correct statement

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 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

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