Question

Redox reactions 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 Daniel Cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium) aong with their `E^(@)` (V with respect to normal hydrogen electrode) values.
`{:(I_(2)+2e^(-) rarr 2I^(-),E^(@)=0.54),(Cl_(2) +2e^(-) rarr 2Cl^(-),E^(@)=1.36),(Mn^(3+)+e^(-)rarr Mn^(2+),E^(@)=1.50),(Fe^(3+)+e^(-) rarr Fe^(2+),E^(@)=0.77),(O_(2)+4H^(+)+4e^(-) rarr 2H_(2)O,E^(@)=1.23):}`
Using these data, obtain the correct explanation for the following questions.
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
Calculate the EMF of all the cell. Only the EMF of the cell involving oxidation of `H_(2)O` to `O_(2)` by `Mn^(3+)` is +ve
`{:(Mn^(3+)+e^(-) rarr Mn^(2+)"]"xx4, E^(@)=+1.50 V),(2H_(2)O rarr 4H^(+)+O_(2)+4e^(-), E^(@)=-1.23 V),(bar(4 Mn^(3+)+2 H_(2)O rarr 4 Mn^(2+)+O_(2)+4 H^(+)", "E_("cell")^(@)=+0.27 V)):}`
Thus, option (d) is corect.