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Redox reactions play a pivotal role in c...

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

`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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The correct Answer is:
A
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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 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 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

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