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Redox Reactions

Redox reactions are electron-transfer reactions: one species loses electrons while another gains them. They power batteries, corrosion, and respiration — and once you can track the electrons, you can balance any of them.

Oxidation Numbers

An oxidation number is the charge an atom would have if all its bonds were ionic. The rules:

An element on its own is 0 (Na, O₂, S₈).
A monatomic ion equals its charge (Na⁺ is +1, Cl⁻ is −1).
Fluorine is always −1; oxygen is usually −2 (except in peroxides, −1, and OF₂, +2); hydrogen is usually +1 (except in metal hydrides, −1).
Group 1 metals are +1, Group 2 metals are +2, aluminum is +3.
The oxidation numbers in a compound sum to zero; in a polyatomic ion they sum to the ion's charge.

Example: Assign oxidation numbers in K₂Cr₂O₇. K is +1 (Group 1) and O is −2. Let Cr be $x$: $$2(+1) + 2x + 7(-2) = 0 \;\Rightarrow\; 2x = +12 \;\Rightarrow\; x = +6$$ So K is +1, Cr is +6, O is −2.

Oxidation, Reduction, and Agents

Oxidation is loss of electrons (oxidation number increases); reduction is gain of electrons (oxidation number decreases). Remember it as OIL RIG — Oxidation Is Loss, Reduction Is Gain. The species that is oxidized is the reducing agent; the species that is reduced is the oxidizing agent. In Zn + Cu²⁺ → Zn²⁺ + Cu, zinc is oxidized (0 → +2) so it is the reducing agent, and Cu²⁺ is reduced (+2 → 0) so it is the oxidizing agent.

Balancing in Acidic Solution — The Half-Reaction Method

Split the reaction into oxidation and reduction half-reactions, then for each: (1) balance atoms other than O and H; (2) balance O with H₂O; (3) balance H with H⁺; (4) balance charge with electrons; (5) multiply so the electrons cancel, then add the halves back together.

Example: Balance MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺ in acid. Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. Oxidation: Fe²⁺ → Fe³⁺ + e⁻ (× 5). Adding, the 5 electrons cancel: $$\text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} + 5\text{Fe}^{3+}$$ Check: Mn 1 = 1, O 4 = 4, H 8 = 8, Fe 5 = 5. Charge: left −1 + 8 + 10 = +17; right 2 + 15 = +17. Balanced.

Balancing in Basic Solution

Balance it as if it were acidic first, then neutralize every H⁺ by adding the same number of OH⁻ to both sides — each H⁺ + OH⁻ becomes H₂O — and cancel any H₂O that appears on both sides. (Equivalently, balance O with OH⁻ and H₂O directly in each half-reaction.)

Example: Balance ClO⁻ + CrO₂⁻ → Cl⁻ + CrO₄²⁻ in base. Chlorine goes +1 → −1 (reduction, gains 2 e⁻); chromium goes +3 → +6 (oxidation, loses 3 e⁻). Half-reactions in base: $$\text{ClO}^- + \text{H}_2\text{O} + 2\text{e}^- \rightarrow \text{Cl}^- + 2\text{OH}^- \qquad (\times 3)$$ $$\text{CrO}_2^- + 4\text{OH}^- \rightarrow \text{CrO}_4^{2-} + 2\text{H}_2\text{O} + 3\text{e}^- \qquad (\times 2)$$ Adding and cancelling 6 e⁻, 6 OH⁻, and 3 H₂O: $$3\text{ClO}^- + 2\text{CrO}_2^- + 2\text{OH}^- \rightarrow 3\text{Cl}^- + 2\text{CrO}_4^{2-} + \text{H}_2\text{O}$$ Check: Cl 3 = 3, Cr 2 = 2, O 3 + 4 + 2 = 9 = 8 + 1, H 2 = 2. Charge: left −3 − 2 − 2 = −7; right −3 − 4 = −7. Balanced.

Practice Problems

Assign oxidation numbers to each element in H₂SO₄.
In Zn + Cu²⁺ → Zn²⁺ + Cu, identify what is oxidized, what is reduced, the oxidizing agent, and the reducing agent.
Balance in acidic solution: Cr₂O₇²⁻ + Fe²⁺ → Cr³⁺ + Fe³⁺.
Balance in basic solution: MnO₄⁻ + SO₃²⁻ → MnO₂ + SO₄²⁻.