X-method

 Advanced Methods for Balancing Equations by Inspection

The general inspection method to balance equations is, no doubt, very powerful. However, in case of some complicated chemical reactions, the inspection can not accomplish the job due to the fact that it is not able to locate an atom that can be uniquely balanced without some specific treatments. There are three more sophisticated ways to continue the balancing. They are:

1. The use of charge conditions

2. X-Method

3. The use of oxidation numbers

The following examples will illustrate how to use these methods.

### The Use of Charge Conditions

Pb + PbO2 + H+ -> Pb2+ + H2O

In the first step of normal inspection, the oxygen is balanced:

Pb + 1 PbO2 + H+ -> Pb2+ + 2 H2O

Next step is to balance hydrogen:

Pb + 1 PbO2 + 4 H+ -> Pb2+ + 2 H2O

The balance can't be continued because the coefficients of Pb and Pb2+ can't be determined. But if the charge conditions are considered, in which the charges of both sides of a balanced equation should be same, the equation becomes:

Pb + 1 PbO2 + 4 H+ -> 2 Pb2+ + 2 H2O

The coefficient 1 adding to Pb finished the balance:

Pb + PbO2 + 4 H+ = 2 Pb2+ + 2 H2O

(The coefficient 1 is drop out in the final equation according to the customs.)

### X-Method

For equation:

Cu + H2SO4 -->  CuSO4 + SO2 + H2O

The first step of inspection gives:

Cu + 1 H2SO4 -->  CuSO4 + SO2 + 1 H2O  (for H)

After this, there are no more atoms can be balanced according to the inspection rules. In this case, we will introduce an unknown coefficient x for the most complex compound, CuSO4 in this example.

Cu + 1 H2SO4 -->  x CuSO4 + SO2 + 1 H2O

After introducing x, the balance can be continued. In above example, Cu can be determined. Therefore, the equation becomes:

x Cu + 1 H2SO4 --> x CuSO4 + SO2 + 1 H2O

Then, by S

x Cu + 1 H2SO4 -->  x CuSO4 + (1-x)SO2 + 1 H2

Until to know, the all coefficients have been determined by numbers and the functions of x. The x has to be solved by one more unused relation of atoms from the equation. In the above equation, The H are balanced by step 1. The Cu and S are balanced by x and its function. Only O is not used. By considering O in above equation, an algebraic equation can be formed:

4 = 4x + 2(1-x) +1

Solving the equation:

x = 1/2

Substituting it into the latest form of above chemical equation:

1/2 Cu + H2SO4 = 1/2CuSO4 + 1/2 SO2 + H2O

Then , the final balanced equation can be written:

Cu + 2 H2SO4 = CuSO4 + SO2 + 2 H2O

### The Use of Oxidation Numbers

For balanced Redox equations, besides the atoms in the both sides of the equation have to be equal, the changes of oxidation numbers between the both sides have to be same, too. To balance some complex Redox equation with the inspection method, the oxidation number conditions have to be used. For example:

MnO4- + H2O2 + H+ -> Mn2+ + O2 + H2O

The first step in the inspection method is to balance Mn:

1 MnO4- + H2O2 + H+ -> 1 Mn2+ + O2 + H2O

The second step uses the charge condition to get coefficient to H+:

1 MnO4- + H2O2 + 3 H+ -> 1 Mn2+ + O2 + H2O

Although the charge conditions are used, the x way still is necessary to continue:

1 MnO4- + x H2O2 + 3 H+ -> 1 Mn2+ + O2 + H2O

Balance H:

1 MnO4- + x H2O2 + 3 H+ -> 1 Mn2+ + O2 + (2x+3)/2 H2O

Balance O:

1 MnO4- + x H2O2 + 3 H+ -> 1 Mn2+ + (2x+5)/4 O2 + (2x+3)/2 H2O

In the above equation, all the coefficients are gotten (by numbers or functions of x) and relations of atoms (Mn, H and O ) and the charge condition are used. To getting x, only condition can be used is the oxidation number relations. In the original equation, the oxidation numbers of each atom of all molecules involved in the reaction are:

MnO4- + H2O2 + H+ -> Mn2+ + O2 + H2O

+7 -2 +1 -1 +1 +2 0 +1 -2

The oxidation number of Mn atom from +7 changed into +2 and that of O in H2O2 from -1 to 0 in O2. To equal the change of oxidation number in the both sides, the algebraic equation can be written:

1 * 7 (from MnO4-) + 2 * x * (-1) (from H2O2) = 1 * (+2) (from Mn2+) + (2x+5)/4 * 2 * 0 (from O2)

Solving the equation gets:

x = 5/2

Substituting back to the latest equation to get:

MnO4- + 5/2 H2O2 + 3 H+ -> Mn2+ + 5/2 O2 + 4 H2O

The final balanced equation is:

2 MnO4- + 5 H2O2 + 6 H+ = 2 Mn2+ + 5 O2 + 8 H2O