Consider the following redox reaction: 2Cu₂O (s) + Cu₂S (s) → 6Cu (s) + SO₂ (g) Identify the species among the following acting as oxidant and reductant, respectively:

Correct — A, Cu(I) and S of Cu₂S. Put oxidation numbers on both sides and the roles fall out. Copper is +1 in Cu₂O and +1 in Cu₂S, and 0 in the metal produced, so copper is reduced — it is the oxidant. Sulphur is −2 in Cu₂S and +4 in SO₂, a loss of six electrons, so the sulphide sulphur is oxidised and is the reductant. Oxygen stays at −2 on both sides and takes no part. The electron count balances exactly: six copper atoms gain one electron each, matching the six electrons the single sulphur atom gives up, which is why the equation produces exactly 6Cu. Chemistry names the species that is reduced the oxidising agent and the species that is oxidised the reducing agent, so the pair asked for is Cu(I) and the sulphur of Cu₂S.

• (b)Cu and S of SO₂ — Names the products. An oxidant and a reductant are reactants — they are what gets reduced and what gets oxidised. Copper metal and the sulphur in SO₂ are what those two reactants have become.
• (c)Cu and O of SO₂ — Wrong on both counts. Copper metal is a product, and oxygen holds −2 in Cu₂O and in SO₂ alike, so it is neither oxidised nor reduced anywhere in this equation.
• (d)Cu(I) and O of SO₂ — Gets the oxidant right and the reductant wrong. Oxygen's oxidation number never changes here, so it cannot be the species that gives up electrons; that is the sulphur, which climbs from −2 to +4.

A redox equation is read by assigning oxidation numbers to every element on both sides and looking for the two that move. The element whose number falls has been reduced, and the species carrying it is the oxidising agent; the element whose number rises has been oxidised, and its species is the reducing agent. The two changes must balance in electrons, which is a free check on the reading.

This particular equation is the self-reduction step of copper metallurgy. In the converter the copper(I) sulphide from the roasted ore meets copper(I) oxide made from part of it, and the sulphide's sulphur reduces the oxide's copper to the metal while itself leaving as sulphur dioxide — no coke or other reducing agent is needed, which is unusual and is why the reaction is set as a question. The trap the paper sets is the everyday habit of naming an agent by what it becomes: candidates see copper metal in the products and answer Cu. Fix the rule the other way round — the agent is the reactant, and it is named for what it does to the other species, not for its own fate.

• The oxidising agent is the species reduced; the reducing agent is the species oxidised.
• Copper is +1 in both Cu₂O and Cu₂S and 0 in the metal, a fall of one unit per copper atom.
• Sulphur rises from −2 in Cu₂S to +4 in SO₂, a loss of six electrons.
• Six copper atoms gaining one electron each balance the six lost by one sulphur, which fixes the coefficient 6Cu.
• Oxygen holds −2 throughout, so it is a spectator in this equation.

• Naming the oxidant or reductant from the products instead of the reactants.
• Assuming oxygen must be the oxidising agent because the word oxidation carries its name.
• Reading copper as +2 in Cu₂O or Cu₂S; in both compounds it is +1.

Asked with a single balanced equation and no other data, so the examiner is testing whether you assign oxidation numbers before answering.

• practice — not a real PYQ

  In the reaction MnO₂ + 4HCl → MnCl₂ + Cl₂ + 2H₂O, which species acts as the reducing agent?

  • (a)MnO₂
  • (b)Chlorine of HCl
  • (c)Hydrogen of HCl
  • (d)Water

  Answer(b) Chlorine of HCl — chloride rises from −1 to 0 in Cl₂, so it is oxidised and is the reducing agent, while manganese falls from +4 to +2.

• practice — not a real PYQ

  What is the oxidation number of sulphur in sodium thiosulphate, Na₂S₂O₃, on the conventional average reckoning?

  • (a)+2
  • (b)+4
  • (c)+6
  • (d)−2

  Answer(a) +2 — two sodium atoms give +2 and three oxygens give −6, so the two sulphur atoms must total +4, an average of +2 each.