In the reaction, H₂O₂ + PbS → H₂O + PbSO₄, the increase in the oxidation number of sulphur is

${\stackrel{+1}{\mathrm{H}}}_2{\stackrel{-1}{\mathrm{O}}}_2+\stackrel{+2}{\mathrm{Pb}}\stackrel{-2}{\mathrm{S}}\to \stackrel{+2}{\mathrm{Pb}}\stackrel{-6}{\mathrm{S}}{\mathrm{O}}_4+{\stackrel{+1}{\mathrm{H}}}_2\stackrel{-2}{\mathrm{O}}$

Increase in oxidation state of 'S' (PbS to PbSO₄) = -2 to +6 = 8 units.