# Reaction between lead dioxide and nitric acid

A wikipedia article suggests that lead dioxide is produced by this reaction: $$\ce{Pb3O4 + 4 HNO3 → PbO2 + 2 Pb(NO3)2 + 2 H2O}$$

The same article also suggests that lead dioxide reacts with nitric acid in the following way: $$\ce{2 PbO2 + 4 HNO3 → 2 Pb(NO3)2 + 2 H2O + O2}$$

I am unable to understand why then, in the first reaction, $\ce{PbO2}$ doesn't further react with the residual nitric acid. Is it because the quantity of the acid is regulated? Or is there some error in the reactions?

## 2 Answers

Exact statements are:

Because of the instability of its $\ce {Pb^4+}$ cation, lead dioxide reacts with warm acids, converting to the more stable $\ce {Pb^2+}$ state and liberating oxygen $$\ce{2 PbO2 + 4 HNO3 → 2 Pb(NO3)2 + 2 H2O + O2}$$

and

Lead dioxide is produced commercially by several methods, which include ... or reacting $\ce {Pb3O4}$ with dilute nitric acid: $$\ce{Pb3O4 + 4 HNO3 → PbO2 + 2 Pb(NO3)2 + 2 H2O}$$

Since nitric acid in warmer in first reaction, the following reaction occurs easily, which is actually a decomposition reaction:

When heated, lead(II) nitrate crystals decompose to lead(II) oxide, oxygen and nitrogen dioxide. $$\ce{Pb(NO3)2->PbO +NO2 +O2}$$

Hence the oxygen might be liberated with decomposition-cum-oxidation reaction.

You need to consider the reaction conditions.

The Wikipedia Lead Dioxide article is saying that in dilute nitric acid the first reaction occurs, while in warm nitric acid the second reaction occurs.

So the first reaction would only proceed to the second reaction under harsher conditions, heat and more concentrated acid.