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
