Self-Reduction (Autoreduction) Reactions
A handful of metal sulfides reduce their own oxide directly during roasting, needing no external reducing agent at all — but this self-reduction trick does not work for every sulfide ore, and recognising exactly which reaction qualifies is the recurring test here.
Chemistry questions first require the material, particles or reaction condition to be identified before a formula or rule is selected.
Self-reduction works for Cu, Pb and Hg sulfides, but not Zn
If the remaining unreacted sulfide can then react DIRECTLY with that newly formed oxide (2MO + MS → 3M + SO₂), no external reducing agent like carbon is needed at all — this is self-reduction (autoreduction). This works cleanly for copper (Cu₂S + 2Cu₂O → 6Cu + SO₂) and lead (PbS + 2PbO → 3Pb + SO₂), and mercury sulfide is reduced even more directly by roasting alone (HgS + O₂ → Hg + SO₂, since HgO itself is thermally unstable and readily decomposes back to the metal). Zinc is the exception: ZnS roasts to ZnO, but ZnO is thermodynamically too stable relative to ZnS for the analogous self-reduction step (2ZnO + ZnS → 3Zn + SO₂) to proceed — zinc extraction instead requires a genuinely separate carbon-reduction step after roasting.
Recognising the pattern that makes a reaction a true autoreduction
A true autoreduction reaction has one signature form: an oxide and a sulfide of the SAME metal reacting together, with sulfur ending up oxidized to SO₂ while the metal ends up reduced to its elemental form, with no separate external reducing agent (like carbon or another metal) involved at all
This is genuinely different from a standard carbon-monoxide blast-furnace reduction (Fe₃O₄ + 4CO → 3Fe + 4CO₂, which uses an EXTERNAL reducing agent, CO), a standard carbon reduction (Cu₂O + C → 2Cu + CO, again an external reducing agent), or a simple metal-displacement reaction (Cu₂+(aq) + Fe(s) → Cu(s) + Fe²⁺, a completely different kind of redox reaction with no oxide-sulfide pairing at all). Spot autoreduction specifically by the oxide-plus-sulfide-of-the-same-metal pattern.
Identifying SO₂ from a set of linked clues
A gas that (i) is released when an amorphous solid burns in air/O₂ and turns limewater milky, (ii) is released when a metal sulfide is roasted, and (iii) decolourises acidified KMnO₄ and reduces Fe³⁺ to Fe²⁺, is consistently identifiable as SO₂ across all three clues, not CO₂. CO₂ also turns limewater milky, but CO₂ does NOT decolourise KMnO₄ or reduce Fe³⁺ (it has no reducing power in this sense) — SO₂, by contrast, is a well-known mild reducing agent that does both. Reading this kind of multi-clue identification question means checking EVERY clue against the candidate gas, not stopping at the first clue that seems to fit (limewater going milky, which both CO₂ and SO₂ can do) — the amorphous solid burning in air to give this gas is then identified as sulfur itself (S₈, an amorphous/rhombic solid), not a metal oxide or sulfide.
OlympIQ exam lens
- ✓Self-reduction (oxide + sulfide of the same metal to metal + SO₂) works for Cu, Pb and Hg sulfides — not for Zn, whose oxide is too stable relative to its sulfide
- ✓A true autoreduction reaction pairs an oxide and a sulfide of the SAME metal, with no external reducing agent involved
- ✓External-reducing-agent reactions (carbon, CO, or another metal) are NOT autoreduction, even if they also reduce a metal compound
- ✓SO₂ (not CO₂) turns limewater milky AND decolourises acidified KMnO₄ AND reduces Fe³⁺ to Fe²⁺ — check every clue, not just the limewater test
- ✓An amorphous solid that burns in air/O₂ to give SO₂ is elemental sulfur, not a metal oxide or sulfide
Do not identify a gas as CO₂ purely because it turns limewater milky. SO₂ does this too — check the other diagnostic clues (decolourising acidified KMnO₄, reducing Fe³⁺ to Fe₂⁺) that CO₂ cannot produce, since only SO₂ shows this reducing behaviour.
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