PbS+H2O2→PbSO4+H2O In the balance equation of the above reaction, number of electrons transferred from each molecule of H2O2  to lead sulphide is

PbS+4H2O2→PbSO4+4H2O From each H2O2  molecule, 2 electrons are transferred such that a total of 8 electrons to PbS. The change in the oxidation state of the sulphur is from – 2 to +6.