Question

The reaction for the combustion of acetylene is shown. Explain why this reaction is a redox reaction by identifying the oxidation numbers that change, what is oxidized, what is reduced, and what happens to the electrons during the reaction.

2C2H2 + 5O2 → 4CO2 + 2H2O

Answer 1

Assign oxidation numbers:
- In C2H2: H = +1, so 2C + 2(+1) = 0 → each C = −1.
- In O2: O = 0 (elemental).
- In CO2: O = −2, so C = +4.
- In H2O: H = +1, O = −2.

Which change:
- Carbon: −1 → +4 (increase by 5 per C) — carbon is oxidized (it loses electrons).
- Oxygen: 0 → −2 (decrease by 2 per O) — oxygen is reduced (it gains electrons).
- Hydrogen stays +1 (no change).

Electron bookkeeping:
- There are 4 C atoms in 2 C2H2; each C goes from −1 to +4, a change of +5, so total electrons lost = 4 × 5 = 20 e−.
- There are 10 O atoms from 5 O2; each O goes from 0 to −2, a change of −2, so total electrons gained = 10 × 2 = 20 e−.
- Electrons lost by carbon = electrons gained by oxygen (conservation of charge).

Agents:
- Acetylene (the carbon in it) is the reducing agent (it is oxidized).
- O2 is the oxidizing agent (it is reduced).

In short: this combustion is a redox reaction because oxidation numbers change (C is oxidized, O2 is reduced) and electrons are transferred from carbon to oxygen (20 electrons in the balanced reaction).