Redox processes: electron transfer and changes in oxidation number (oxidation state)

Oxidation number rules (in priority order):

Rule	Oxidation number
Uncombined element	0 (e.g. Na, O₂, Cl₂)
Simple monatomic ion	its charge (e.g. Na⁺ = +1, O²⁻ = −2)
Fluorine (always)	−1
Oxygen	−2 (but −1 in peroxides, e.g. H₂O₂; +2 in OF₂)
Hydrogen	+1 (but −1 in metal hydrides, e.g. NaH)
Group 1 / Group 2 metals	+1 / +2

Then apply the balancing rule:

• Sum of oxidation numbers = 0 in a neutral compound.
• Sum = the overall charge in an ion.

Worked. Mn in $\text{MnO}_4^-$: O is $-2$ (×4 = −8); the ion charge is $-1$, so Mn $+ (-8) = -1$ → Mn = +7.

Worked. S in $\text{H}_2\text{SO}_4$: H ×2 = +2, O ×4 = −8; total must be 0, so S $= +6$.

Worked. N in $\text{NO}_3^-$: O ×3 = −6; ion charge −1, so N $= +5$.

Redox reactions involve the transfer of electrons. Use OIL RIG:

• Oxidation Is Loss of electrons → oxidation number increases.
• Reduction Is Gain of electrons → oxidation number decreases.

Both happen together (you can't have one without the other) — hence redox.

Worked. $\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}$:

• Zn: $0 \rightarrow +2$ (increase → oxidised, lost 2 e⁻).
• Cu: $+2 \rightarrow 0$ (decrease → reduced, gained 2 e⁻).

Half-equations show the electrons explicitly: $\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \quad\text{(oxidation)}$ $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \quad\text{(reduction)}$

Oxidising agent — a species that oxidises something else by accepting electrons, so it is itself reduced (its oxidation number decreases). Examples: $\text{MnO}_4^-$, $\text{Cr}_2\text{O}_7^{2-}$, $\text{Cl}_2$.

Reducing agent — a species that reduces something else by donating electrons, so it is itself oxidised (its oxidation number increases). Examples: reactive metals, $\text{I}^-$, $\text{Fe}^{2+}$.

Disproportionation — a reaction in which the same element is simultaneously oxidised and reduced (its atoms go to both a higher and a lower oxidation number).

Worked — chlorine with cold NaOH (links to 11.4): $\text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaOCl} + \text{H}_2\text{O}$ Chlorine goes from $0$ to −1 (in NaCl) and to +1 (in NaOCl) at the same time → disproportionation.

Balancing with oxidation numbers. In a redox equation, the total increase in oxidation number (oxidation) must equal the total decrease (reduction) — because electrons lost = electrons gained.

Method: find the oxidation-number change per atom, then choose coefficients so the increases and decreases balance.

Worked. Combining the half-equations $\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$ (gain 5 e⁻) and $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$ (lose 1 e⁻): multiply the iron half by 5 so electrons cancel: $\text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}$

Roman numerals indicate the magnitude of an oxidation number: iron(II) = Fe with +2, iron(III) = +3; manganese in $\text{MnO}_4^-$ is manganese(VII); chlorine in $\text{NaOCl}$ is chlorine(I).