Redox reactions play a pivotal role in chemistry and biology. The value of standard reduction potetials `(E^(@))` of the two half cells reactions decide which way the reaction is expected to proceed. A simple example is of Daniell cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium) along with with `E^(@)` values. Using the data obtain the correct explanation to the questions that are mentioned. `I_(2)+e^(-) to 2I^(-) , E^(@)=0.54" V "` `Cl_(2)+2e^(-) to 2Cl^(-) , E^(@)=1.36" V"` `Mn^(2+)+2e^(-) to Mn , E^(@)=1.50" V"` `Fe^(3+)+e^(-) to Fe^(2+) , E^(@)=0.77" V"` `O_(2)+4H^(+)+4e^(-) to 2H_(2)O , E^(@)=1.23" V"`. While `Fe^(2+)` ion is stable, `Mn^(2+)` ion 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+)` to `Fe^(3+)`C. `Fe^(3+)` oxidises `H_(2)O` to `O_(2)`D. `Mn^(3+)` oxidises `H_(2)O` to `O_(2)`