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Chemistry 30 · Redox

Oxidation and reduction

Redox is bookkeeping for electrons. Assign oxidation numbers, see which ones moved, split the reaction into half-reactions, and a table of reduction potentials will tell you whether the reaction happens at all.

Oxidation numbers

The idea: an oxidation number is a bookkeeping charge — what an atom's charge would be if every bond were fully ionic. Comparing before and after tells you which atoms lost or gained electrons.

RuleExample
An element by itself is 0Na, O₂, Cl₂, Fe all 0
A monatomic ion equals its chargeNa⁺ is +1, S²⁻ is −2
Oxygen is −2 (−1 in peroxides)−2 in H₂O; −1 in H₂O₂
Hydrogen is +1 (−1 with metals)+1 in HCl; −1 in NaH
The numbers add to the overall charge0 for a compound, the ion charge for an ion

Worked: sulfur in SO₄²⁻

Four oxygens at −2 total −8. The ion is −2 overall, so sulfur must make up the difference: x + (−8) = −2, giving x = +6.

The same method handles anything: in Cr₂O₇²⁻, seven oxygens give −14, the ion is −2, so two chromiums share +12 — each is +6.

Reading the change: oxidation number up means electrons were lost — oxidation. Down means electrons were gained — reduction. OIL RIG: Oxidation Is Loss, Reduction Is Gain.

Who does what to whom

The idea: the agents are named for what they do to their partner, not for what happens to them — which is why they always sound backwards at first.

SubstanceElectronsOxidation numberCalled
Is oxidizedLoses themIncreasesThe reducing agent
Is reducedGains themDecreasesThe oxidizing agent

In 2Mg(s) + O₂(g) → 2MgO(s), magnesium goes 0 → +2 (oxidized, so it is the reducing agent) and oxygen goes 0 → −2 (reduced, so it is the oxidizing agent). Every redox reaction has one of each: electrons have to come from somewhere and go somewhere.

Watch out: not every reaction is redox. In neutralization and precipitation, ions swap partners but no oxidation number changes. Check the numbers before reaching for half-reactions.

Half-reactions and balancing

The idea: split the reaction into the losing half and the gaining half, balance each one completely, then scale them so the electrons cancel.

  1. Balance the main atom in the half-reaction.
  2. Balance oxygen with H₂O, adding a water for each oxygen needed.
  3. Balance hydrogen with H⁺.
  4. Balance the charge with electrons, added to whichever side makes both sides match.
  5. Scale both halves so the electrons lost equal the electrons gained, then add and cancel.
  6. In basic solution, finish by adding OH⁻ to both sides to neutralize every H⁺, and simplify the water.

Worked: permanganate in acid

MnO₄⁻ → Mn²⁺

  1. Manganese is already balanced.
  2. Four oxygens on the left, so add four waters on the right: MnO₄⁻ → Mn²⁺ + 4H₂O.
  3. Eight hydrogens on the right, so add 8H⁺ on the left.
  4. Charge is +7 on the left and +2 on the right, so add 5 electrons to the left.

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. The 5 electrons match manganese dropping from +7 to +2, which is the quickest way to check the answer.

Watch out: electrons never appear in a final balanced equation. If any are left, the two halves were not scaled to match.

Predicting whether it happens

The idea: a table of standard reduction potentials ranks every half-reaction. The strongest oxidizing agent present takes electrons from the strongest reducing agent present — if the numbers allow it.

  1. List every species in the mixture, including water if it is an aqueous solution.
  2. Find the strongest oxidizing agent — the one highest on the left of the table, with the most positive reduction potential.
  3. Find the strongest reducing agent — the one lowest on the right.
  4. If the oxidizing agent sits above the reducing agent, the reaction is spontaneous.
  5. Check with the numbers: E°cell = E°(cathode) − E°(anode). A positive value means spontaneous.

Worked: zinc and copper

Cu²⁺ + 2e⁻ → Cu has E° = +0.34 V; Zn²⁺ + 2e⁻ → Zn has E° = −0.76 V.

Copper's half-reaction is higher, so Cu²⁺ is reduced and zinc metal is oxidized:

E°cell = 0.34 − (−0.76) = +1.10 V, so a strip of zinc dropped into copper sulfate really does plate with copper.

Reverse the pairing — copper metal in a zinc solution — and E°cell would be −1.10 V, which is the table's way of saying nothing happens.

Never multiply E° by anything. Doubling a half-reaction doubles the electrons but not the voltage: potential is energy per charge, so it does not scale with amount — unlike ΔH, which does.

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