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

Options

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)`

Answer

d

Explanation

`[Mn^(3+)+e^(-) rarr Mn^(2+)]xx4`
`2H_(2)Orarr H^(o+)+O_(2)+4e^(-)`
`ulbar(4Mn^(2+)+2H_(2)Orarr 4Mn^(2+)+4H^(o+)+O_(2))`
`(E^(c-)._(cell)`is `+ve)`