This is an **oxidation-reduction** (redox) reaction:

* 2 PbII + 4 e- → 2 Pb0 (reduction)*

* S-II - 6 e- → SIV (oxidation)*

* SVI + 2 e- → SIV (reduction)*

### Word equation

* lead(II) sulfate + lead sulfide → lead + sulfur dioxide*

### Input interpretation

* PbS04 + PbS → Pb + SO2*

* lead(II) sulfate lead sulfide lead sulfur dioxide*

### Balanced equation

Balance the chemical equation algebraically:

* PbS04 + PbS → Pb + SO2*

Add stoichiometric coefficients, c, to the reactants and products:

* c1PbS04 + c2PbS → c3Pb + c4SO2*

Set the number of atoms in the reactants equal to the number of atoms in the products for ** O, Pb **and

*S:**O: 4c1 = 2c4*

*Pb: c1 + c2 = c3*

*S: c1+ c2 = c4*

Since the coefficients are relative quantities and underdetermined, choose a coefficient to set arbitrarily. To keep the coefficients small, the arbitrary value is ordinarily one. For instance, set ** c1 = 1** and solve the system of equations for the remaining coefficients:

*c1 = 1*

*c2 = 1*

*c3 = 2 *

*c4 = 2*

Substitute the coefficients into the chemical reaction to obtain the balanced

equation:

**Answer**:

* PbS04 + PbS → Pb + SO2*