Consider the following reaction
xMnO−4+yC2O2−4+zH+⟶Mn+2+2yCO2+z2H2O
the values of x,y and z in the reaction are, respectively:
2,5 and 16
Ion electron method: First write the given equation in ionic form having ions with central atom (which has undergone a change in oxidation state).
xMnO−4+yC2O2−4+zH+⟶Mn+2+2yCO2+z2H2O
Note that O and H atoms attached to the central atom (shown in underlined) have to be retained.
Now, write the oxidation and reduction half reaction and balance them as shown:
Reduction:
MnO−4+C2O−24⟶Mn+2
a. First , make sure that the element undergone the change in oxidation state is balanced.
b. Balance O atoms by adding 4H2 on RHS.
MnO−4⟶Mn+2+4H2O
c.Now, RHS has excess of 8H atoms. Add 8H+ on LHS . Note, the medium is acidic due to the presence of H2SO4 .
2MnO−4+8H+⟶2Mn+2+4H2O
d. Now O and H are balanced . Balance the charge on both sides.
On LHS: Charge is 1× + 8×(+1) = +7
On RHS: Charge is 1×2 + 4×(0) = +2
Add 5e− in LHS (Note : each e− is equivqlent to acharge of -1)
MnO−4+8H++5e−⟶Mn+2+4H2O ...(i)
Oxidation:
C2O−24⟶CO2
Following the same procedure as above, we have:
a. Balance C atoms :C2O−24⟶CO2
b.Balance O atoms : Already balanced
c.Balance H atoms: No H atoms
d.C2O−24⟶2CO2+2e− ...(ii)
Multiply Eq(i)by 2 and Eq.(ii) by5 to balance the electrons transfer and add to get
2MnO−4+5C2O−24+16H+⟶2Mn+2+10CO2+8H2O