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 , Among the following , identify the correct statement :

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 , Among the following , identify the correct statement :
Answer» CHLORIDE ion is oxidised by `O_2` `Fe^(2+)` is oxidised by iodine Iodide ion is oxidised by chlorine `Mn^(2+)` is oxidised by chlorine SOLUTION :` I_2 // 2I^(-)(0.54V) , Fe^(+3)//Fe^(+2)(0.77V)` `O_2 , 4H^(oplus) // 2H_2O (1.23V) , Cl_2 // 2Cl^(Ɵ)(1.36V) , Mn^(+3)//Mn^(+2)(1.50V)`