2.68 xx10^(-3) moles of a solutioncontaing an ion A^(n+) require 1.61 xx10^(-3) moles MnO_(4)^(-) for the oxidation of A^(n+) to AO_(3)^(-) in acid medium what is the value of n ?

2.68 xx10^(-3) moles of a solutioncontaing an ion A^(n+) require 1.61

1.	2.68 xx10^(-3) moles of a solutioncontaing an ion A^(n+) require 1.61 xx10^(-3) moles MnO_(4)^(-) for the oxidation of A^(n+) to AO_(3)^(-) in acid medium what is the value of n ?
Answer» Solution :STEP 1 To write the reduction and oxidation half reaction Reducton : `Mno_(4)^(-) + 8H^(+)+5e^(-)rarrMn^(2+)+4H_(2)O` oxidation `A^(n+)+3H_(2)OrarrAO_(3)^(-)+6H^(+)+(5-n)E^(-)` Step 2 To FIND out the VALUE of n SINCE in a REDOX reaction number of electrons lost = number of electrons gained therefore multiple oxidant of Eq (i) i.e `MnO_(4)^(-)` by (5-n) and reductant of Eq (ii) i.e `A^(n+)` by 5 adn equate we have (5-n) `MnO_(4)^(-)=5 A^(n+) i.e (5-n)` moles of `MnO_(4)^(-)` will oxidise `An^(n+)` = moles Equationg the values of moles of `A^(n+)` actually oxidised =`2.68xx10^(-3)` moles Equating the value of Eq (iii) and (iv) we have `(5)/(5-n) xx1.61xx10^(-3)=2.68xx10^(-3) or5xxx1.61=(5-n)xx2.68 or 2.68 n = 5(2.68-1.61)=5xx1.07=5.35 or n =(5.35)/(2.68)=2`