Question

Redox reaction play a pivotal role in chemistry and biology. The values of standard redox potential `(E^(@))` of two half-cell reactions decide which way the reaction is expected to proceed. A simple example is a Daniel cell in which zinc goes into solution and copper gets deposited. give 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 explanation to folloing Questions.
`I_(2)+2e^(-)to2I^(-)" "E^(0)=0.54`
`Cl_(2)+2e^(-)to2Cl^(-)" "E^(0)=1.36`
`Mn^(3+)+e^(-)toMn^(2+)" "E^(0)=1.50`
`Fe^(3+)+e^(-)toFe^(2+)" "E^(0)=0.77`
`O_(2)+4H^(+)+4e^(-)to2H_(2)O" "E^(0)=1.23`
Q. While `Fe^(3+)` is stable, `Mn^(3+)` 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+) 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)`

Answer 1

Correct Answer - D
`4Mn^(3+)+2H_(2)Oto4Mn^(2+)+O_(2)+4H^(+)`
`E^(o)=E_(Mn^(3+)//Mn^(2+))^(o)+E_(H_(2)O//O_(2))^(o)=1.50+(-1.23)=0.27V`