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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

`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)`

Text Solution

Verified by Experts

The correct Answer is:
D
`4Mn^(3+)+2H_(2)Oto4Mn^(2+)+O_(2)+4H^(+)`
`E^(o)=E_(Mn^(3+)//Mn^(2+))^(o)+E_(H_(2)O//O_(2))^(o)=1.50+(-1.23)=0.27V`
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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.

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