Balance the following equation using oxidation number method: $\ce{PbS + H2O2 -> PbSO4 + H2O}$


I have written down the oxidation states of all atoms and we can observe reduction of peroxide and oxidation of sulphur.

$\ce{Pb^{II}S^{II-} + H^{I}_2O^{-I}2 ->Pb^{II}S^{VI}O^{-II}_4 + H^{I}2O^{-II} }$

The specific problem I am facing with this question is that oxygen is going from $-1$ to $-2$ in both the reaction products. How do I handle that?


1 Answer 1


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

$$\ce{PbS^{2-} ->PbS^{6+}O4 +8e-}$$

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

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

Now consider reduction half reaction:

$$\ce{H2O^{1-}2 +2e- -> 2H2O^{2-} }$$

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

$$\ce{H2O2 + 2 H+ +2e- -> 2H2O }$$

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

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

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

$$\ce{PbS + 4H2O2 -> PbSO4 + 4H2O}$$


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