Given below are a set of half-cell reactions (acidic medium) along with their `E_(@)` with respect to normal hydrogen electrode values. Using the data obtain the correct explanation to question given below. `{:(I_(2)+2e^(-)rarr2I^(-),E^(@)=0.54),(Cl_(2)+2e^(-)rarr2Cl^(-),E^(@)=1.36),(Mn^(2+)+e^(-)rarrMn^(2+),E^(@)=1.50),(Fe^(3+)+e^(-)rarrFe^(2+),E^(@)=0.77),(O_(2)+4H^(+)+4e^(-)rarr2H_(2)O,E^(@)=1.23):}`
While `Fe^(2+)` is stable, `Mn^(3+)` is not stable in acid solution because:
A. `O_(2)` oxidises `Mn^(2+)` to `Mn^(3+)` and `Fe^(2+)` to `Fe^(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)`
