Redox reactions play a pivotal role I chemistry and bilogy . The values of standard redox potential (E^(@)) of two half - cell reaction decided which way the reaction is expected to proceed. A simple example is a Daniel cell in which zinc goes into solution and copper get deposited. Given below are a set of half 0 cell reactions ( acidic medium ) along with their E^(@) values ( with respectto normal hydrogen electrode ) Using this data : I_2 + 2e^(-) to 2I^(-) E^(@) = 0.54 , Cl_2 + 2e^(-) to 2Cl^(-) ""E=1.36V Mn^(3+) + e^(-) to Mn^(2+) E^(@)= 1.50 , Fe^(3+) + e^(-) to Fe^(2+) E = 0.77V O_2 +4H^(+) + 4e^(-) to 2H_2O"" E^(@) = 1.23 , While Fe^(3+) is stable , Mn^(3+) is not stable in acid solution because :

Redox reactions play a pivotal role I chemistry and bilogy . The value

1.	Redox reactions play a pivotal role I chemistry and bilogy . The values of standard redox potential (E^(@)) of two half - cell reaction decided which way the reaction is expected to proceed. A simple example is a Daniel cell in which zinc goes into solution and copper get deposited. Given below are a set of half 0 cell reactions ( acidic medium ) along with their E^(@) values ( with respectto normal hydrogen electrode ) Using this data : I_2 + 2e^(-) to 2I^(-) E^(@) = 0.54 , Cl_2 + 2e^(-) to 2Cl^(-) ""E=1.36V Mn^(3+) + e^(-) to Mn^(2+) E^(@)= 1.50 , Fe^(3+) + e^(-) to Fe^(2+) E = 0.77V O_2 +4H^(+) + 4e^(-) to 2H_2O"" E^(@) = 1.23 , 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_2O " to " O_2` `Mn^(3+)` oxidises ` H_2O " to " O_2 ` SOLUTION :`Mn^(+3)` oxidises `H_2O ` to `O_2`