Redox reactions play a pivolted role in chemistry and biology. The values of standard redox potential `(E^(@))` of two half-cell reactions decide which way be reaction is expected to proceed. A simple example is a Daniel cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reaction (acidic medium) along with their `E^(@)(V` with respect to normal hydrogen electrode) values. Using this data obtain the correct explanations to Question 14-16
`{:(I_(2)+2e^(-)rarr2I^(-),E^(@) =0.54),(CI_(2)+2e^(-)rarr2CI^(-),E^(@) =1.36),(Mn^(3+)+e^(-)rarrMn^(2+),E^(@) =1.50),(Fe^(3+)+e^(-)rarrFe^(2+),E^(@) =0.77),(O_(2)+4H^(+)+4e^(-)rarr2H_(2)O,E^(@) =1.23):}`
Among the following identify the correct statement.

Options

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)` to `O_(2)`
D. `Mn^(3+)` oxidises `H_(2)O` to `O_(2)`

Answer

D