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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` While `Fe^(3+)` is stable, `Mn^(3+)` is not stable in acid solution becauseA. `O_(2)` oxidises `Mn^(2+)` to `Mn^(3+)`B. `O_(2)` oxidises both `Mn^(2+)` and `Fe^(2+)`C. `Fe^(3+)` oxidises `H_(2)O` to `O_(2)`D. `Mn^(3+)` oxidises `H_(2)O` to `O_(2)`

Answer» Correct Answer - d
`Mn^(3+)` oxidises `H_(2)O` to `O_(2)` because the standard reduction potential of `(Mn^(3+) rarr Mn^(2+))` is greater than that of `(O_(2) rarr H_(2)O)`.


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