Lead sulfate undergoes an oxidation-reduction (redox) reaction, where lead(II) ions are reduced and sulfur ions are oxidized. This can be represented by the following half-reactions:

2 Pb2+ + 4 e- → 2 Pb0 **(reduction)**

S2- - 6 e- → S4+ **(oxidation)**

S6+ + 2 e- → S4+ **(reduction)**

## Word equation:

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

## Input interpretation:

PbS04 + PbS → Pb + SO2

## Balanced equation:

The overall balanced chemical equation for the reaction is:

**PbSO4 + PbS → 2Pb + SO2**

This can be balanced algebraically by adding stoichiometric coefficients, c, to the reactants and products, and setting the number of atoms in the reactants equal to the number of atoms in the products for oxygen, lead, and sulfur. The resulting system of equations can be solved for the coefficients:

c1 = 1 (arbitrary choice)

c2 = 1

c3 = 2

c4 = 2

Substituting the coefficients back into the chemical equation, we get the final balanced equation:

PbSO4 + PbS → 2Pb + SO2