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

Redox reaction play a pivotal role in chemistry and biology. The value

1.	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
Answer» `O_(2)` OXIDISES `Mn^(2+)` to `Mn^(3+)` `O_(2)` oxidises both `Mn^(2+)` to `Mn^(3+) and Fe^(2+)` to `Fe^(3+)` `Fe^(3+)` oxidises `H_(2)O` to `O_(2)` `Mn^(3+)` oxidises `H_(2)O` to `O_(2)` SOLUTION :`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`