Q.Why do the following reactions proceed differently ? Pb3O4 + 8HCl → 3PbCl2 + Cl2 + 4H2O and Pb3O4 + 4HNO3 → 2Pb(NO3)2 + PbO2 + 2H2O
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Start your 14-day free trial to unlock the full solution →The difference arises because Pb₃O₄ is a mixed oxide (2PbO·PbO₂). With HCl (a reducing acid), Pb(IV) is reduced to Pb(II) and Cl⁻ is oxidised to Cl₂. With HNO₃ (an oxidising acid), Pb(IV) remains unchanged and precipitates as PbO₂, while Pb(II) simply forms the nitrate.
The Core Idea: Pb₃O₄ is Not a Simple Oxide
Pb₃O₄ (red lead or minium) is a mixed oxide — it behaves as if it contains two different oxidation states of lead: two Pb(II) atoms and one Pb(IV) atom. You can write its formula as .
This is the key to everything. The Pb(IV) centre is a strong oxidising agent. The Pb(II) centres are normal, stable lead ions. What happens in a reaction depends entirely on whether the acid you add can reduce the Pb(IV) or not.
Pb₃O₄ = 2PbO · PbO₂
Oxidation states: two Pb atoms at +2, one Pb atom at +4.
Step-by-Step Reasoning
1. Identify the two acids: one reducing, one oxidising
- HCl is a reducing acid. The chloride ion Cl⁻ is easily oxidised to Cl₂ gas.
- HNO₃ is a strong oxidising acid. It does not get oxidised further — it tends to oxidise other substances.
This difference in the acid's own redox behaviour decides the fate of the Pb(IV) in Pb₃O₄.
2. Reaction with HCl: Pb(IV) gets reduced
When Pb₃O₄ meets HCl, the Pb(IV) centre (+4) wants to drop to its more stable +2 state. To do that, it needs to accept electrons. Where do those electrons come from? From the chloride ions.
The half-reactions are:
- Reduction:
- Oxidation:
So the Pb(IV) oxidises Cl⁻ to Cl₂ gas, and itself becomes Pb(II). All three lead atoms end up as PbCl₂ — the overall reaction splits cleanly into an acid–base part and a redox part:
Adding the two gives the printed equation with 3PbCl₂ and Cl₂.
A common mistake is to think Pb₃O₄ is a single oxide with all Pb at the same oxidation state. If you treat it as Pb in +8/3, you lose the insight. Always remember: it is a mixed oxide — two distinct oxidation states coexist.
3. Reaction with HNO₃: Pb(IV) stays put
Now consider HNO₃. It is a powerful oxidising agent — it does not get oxidised by Pb(IV). In fact, it would rather oxidise something else. So the Pb(IV) centre has no reason to change its oxidation state. It remains as Pb(IV).
What happens to the Pb(II) centres? Being part of the basic oxide PbO, they react with the acid in a plain acid–base reaction:
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