Balancing using oxidation number method
For sulfur we can start by identifying the number of electrons lost in the oxidation half reaction:

$$\ce{PbS^{2-} -&gt;PbS^{6+}O4 +8e-}$$

Now we can balance the oxygens using water, which also leaves us with an excess of protons:

$$\ce{PbS + 4H2O -&gt; PbSO4 + 8e- +8H+}\tag1\label{Oxidation}$$

Now consider reduction half reaction:

$$\ce{H2O^{1-}2 +2e- -&gt; 2H2O^{2-} }$$

It will obviously require a proton source, so we balance with protons:

$$\ce{H2O2 + 2 H+ +2e- -&gt; 2H2O }$$

Multiply by a factor of 4 to match the number of electrons needed by the oxidation reaction:

$$\ce{4H2O2 + 8 H+ +8e- -&gt; 8H2O }\tag2\label{Reduction}$$

Combine the half reactions $(1)$ and $(2)$:

$$\ce{PbS + 4H2O2 -&gt; PbSO4 + 4H2O}$$