In the following reaction,
xMn2O7⟶yMnO2+zO2 x, y, and z respectively are:
2, 4, 3
Mn2O7 is splitted into two parts ( +7×2Mn2 and −2×7O7 )
Let the following equation be numbered ...(i)
(i). 6e−+Mn2→2Mn
2x=14 2x=8
so change in oxidation state require =14−8=6e−
Let the following equation be numbered ...(ii)
(ii) 2O7 → 7O2+28e−
14x=−28 14x=0
so change in oxidation state require 0−(−28)=28e−
Now, multiply equation (i) by 28 and equation (ii) by 6 and add them as
28×6e−+28Mn7+2→56Mn4+
12O2−7→42O2+28×6e−
overall reaction: 28Mn7+2+12O2−7→56Mn4++42O2
So,
28Mn2O7+12Mn2O7⟶56MnO2+42O2 ...(iii)
In the above equation (iii), balance the Mn atoms disregarding O atoms.
40Mn2O7⟶56MnO2+24MnO2+42O2 ...(iv)
In the above equation, now balance the O atoms to get
40Mn2O7⟶80MnO2+42O2+18O2
The final, balanced equation is:
2Mn2O7⟶4MnO2+3O2