The Concept of Oxidation and Reduction, Oxidation States

Every chemical equation balanced so far in this course has been read as a rearrangement of atoms. This chapter opens a new lens on the same reactions: tracking not where atoms go, but where electrons go. A huge fraction of chemistry — corrosion, combustion, batteries, electroplating, respiration and photosynthesis among them — is driven by electrons transferring from one species to another, and the formalism built here, the oxidation state, is the single bookkeeping tool that makes that transfer visible even in reactions, such as those of covalent compounds, where no discrete ions are actually exchanged.

This chapter builds only the concepts and the bookkeeping: what oxidation and reduction mean, how to assign an oxidation state to any atom in any species, and how to use the resulting numbers to identify which species is oxidised, which is reduced, and which reagent is responsible for each. Writing and balancing the resulting redox equations — including the trickier half-equation method needed in acidic or basic solution — is deliberately left to the next chapter, which takes the oxidation-state toolkit built here as its starting point.

1. Oxidation and Reduction as Electron Transfer

The oldest definition of oxidation is the gain of oxygen by a substance, and of reduction the loss of oxygen (literally, a reduction in mass as oxygen is removed) — language that survives from the early chemistry of combustion and metal smelting. $2\text{Mg}(s) + \text{O}_2(g) \longrightarrow 2\text{MgO}(s)$ is still correctly described as magnesium being oxidised. This definition is too narrow, however: magnesium reacts with chlorine, $\text{Mg}(s) + \text{Cl}_2(g) \longrightarrow \text{MgCl}_2(s)$, in a way chemically identical in spirit to its reaction with oxygen, yet no oxygen is involved at all.

The modern, general definition replaces "oxygen" with electrons:

Definition. Oxidation is the loss of electrons by a species. Reduction is the gain of electrons by a species. The mnemonic OIL RIG — Oxidation Is Loss, Reduction Is Gain — fixes the direction permanently.

In $\text{Mg}(s) + \text{Cl}_2(g) \longrightarrow \text{MgCl}_2(s)$, each magnesium atom loses two electrons to become $\text{Mg}^{2+}$ (oxidation), and each chlorine atom gains one electron to become $\text{Cl}^{-}$ (reduction):

These two equations are half-equations: each shows only the electron transfer undergone by one species, and neither can occur in isolation, since every electron lost by one species must be gained by another. A reaction in which both an oxidation and a reduction occur together is called a redox reaction (reduction + oxidation), and the two half-equations added together, with electrons cancelling exactly, must reproduce the overall equation.

Key distinction. Oxidation and reduction are not independent events; they are the two halves of a single electron transfer, and one cannot happen without the other. "An oxidation reaction" in isolation, with no corresponding reduction, is not physically possible.

The species that causes another species to be oxidised — by accepting the electrons it loses — is called the oxidising agent (or oxidant); it is itself reduced in the process. Conversely, the species that causes another species to be reduced is the reducing agent (or reductant); it is itself oxidised. In the magnesium–chlorine reaction, $\text{Cl}_2$ is the oxidising agent (it oxidises magnesium, and is itself reduced to $\text{Cl}^{-}$), and $\text{Mg}$ is the reducing agent.

2. The Limitation of Electron Counting in Covalent Compounds

The electron-transfer definition works cleanly for reactions that form clearly ionic products, where whole electrons visibly move from one atom to become part of another ion. It becomes ambiguous for covalent compounds, where electrons are shared, not transferred outright. Consider the combustion of methane:

No ions are formed anywhere in this equation, yet chemically this is unmistakably a redox reaction: carbon is being "oxidised" in the combustion sense (gaining oxygen), just as surely as magnesium was. To extend the electron-transfer idea to covalent bonding, chemists assign each shared pair of electrons, on paper, entirely to the more electronegative atom of the pair — as if every bond were ionic, even when it is not. This bookkeeping device is the oxidation state (sometimes called the oxidation number), and it is what the rest of this chapter develops.

3. Rules for Assigning Oxidation States

The oxidation state of an atom in a species is the charge that atom would carry if every bond to a more electronegative atom were treated as fully ionic (the electron pair assigned entirely to the more electronegative partner) and every bond between identical atoms were split evenly. In practice, this definition is applied through a short, ordered list of rules, each rule overriding those below it only when they would otherwise conflict:

1. A free (uncombined) element has an oxidation state of $0$. This includes elements that exist as diatomic or polyatomic molecules, such as $\text{O}_2$, $\text{Cl}_2$, $\text{P}_4$ and $\text{S}_8$ — the atoms are identical, so no electron pair is assigned preferentially to either.
2. A monoatomic ion has an oxidation state equal to its actual ionic charge: $\text{Na}^{+}$ is $+1$, $\text{Cl}^{-}$ is $-1$, $\text{Fe}^{3+}$ is $+3$.
3. Fluorine is always $-1$ in every compound, with no exception — it is the most electronegative element, and so in any bond it forms, the shared pair is assigned entirely to it.
4. Oxygen is usually $-2$. The recognised exceptions are: in peroxides (containing the $\text{O}\!-\!\text{O}$ linkage, such as $\text{H}_2\text{O}_2$ or $\text{Na}_2\text{O}_2$), each oxygen is $-1$; in superoxides (such as $\text{KO}_2$), each oxygen is $-\tfrac{1}{2}$; and in compounds with fluorine (such as $\text{OF}_2$), oxygen is $+2$, since fluorine (rule 3) outranks it.
5. Hydrogen is usually $+1$ when bonded to a non-metal. In metal hydrides (such as $\text{NaH}$ or $\text{CaH}_2$), where hydrogen is bonded to a less electronegative metal, hydrogen is $-1$ instead.
6. The sum of the oxidation states of every atom in a neutral species is $0$; the sum for a polyatomic ion equals the ion's overall charge. This is the rule that is actually solved, algebraically, for the one atom (almost always a transition metal, or an atom whose oxidation state is not fixed by rules 3–5) whose value is not already known.

Warning. Oxidation state is a bookkeeping convention, not a measured physical charge. $\text{Mn}$ in $\text{MnO}_4^{-}$ does not carry a literal $+7$ charge — the permanganate ion is covalently bonded throughout — but assigning it the oxidation state $+7$ correctly tracks how electron density is distributed relative to the far more electronegative oxygen atoms, and that is all the convention is designed to do.

4. Worked Examples of Assigning Oxidation States

Worked example — potassium permanganate, $\text{KMnO}_4$. Potassium, an alkali metal always found as $\text{K}^{+}$, is $+1$ (rule 2, since it is effectively a monoatomic ion within the lattice). Oxygen is $-2$ (rule 4; no peroxide or fluorine exception applies). Let $x$ be the oxidation state of manganese. The compound is neutral, so by rule 6:

Manganese in $\text{KMnO}_4$ has oxidation state $+7$ — the maximum possible for manganese, corresponding to it having "given up," on paper, all seven of its valence electrons to the more electronegative oxygens surrounding it.

Worked example — dichromate ion, $\text{Cr}_2\text{O}_7^{2-}$. Oxygen is $-2$ throughout (rule 4). Let $x$ be the oxidation state of each chromium atom (both chromiums are chemically equivalent in this ion, so the same value applies to each). The ion carries an overall charge of $-2$, so by rule 6:

Each chromium in $\text{Cr}_2\text{O}_7^{2-}$ has oxidation state $+6$.

Worked example — hydrogen peroxide, $\text{H}_2\text{O}_2$. Hydrogen bonded to a non-metal is $+1$ (rule 5). Because this molecule contains the $\text{O}\!-\!\text{O}$ peroxide linkage, oxygen takes its exception value of $-1$ (rule 4), not the usual $-2$. Checking rule 6 for the neutral molecule: $2(+1) + 2(-1) = 0$. ✓ — oxygen genuinely is $-1$ here, a value worth memorising as the single most common exam trap in this topic.

Worked example — iron in $\text{Fe}_3\text{O}_4$ (magnetite). Oxygen is $-2$ (rule 4). If every iron atom carried the same oxidation state $x$, rule 6 would require $3x + 4(-2) = 0$, giving the non-integer $x = +\tfrac{8}{3}$. A fractional oxidation state is a signal, not an answer: $\text{Fe}_3\text{O}_4$ is chemically a mixed-valence oxide, $\text{FeO}\cdot\text{Fe}_2\text{O}_3$, containing one $\text{Fe}^{2+}$ and two $\text{Fe}^{3+}$ per formula unit. The average oxidation state of iron is indeed $+\tfrac{8}{3}$, and that average is what rule 6 correctly returns, but it is not the oxidation state of any single iron atom in the structure.

Key distinction. A non-integer result from rule 6 nearly always indicates a mixture of two integer oxidation states for chemically distinct atoms of the same element, averaged by the formula — not a genuinely fractional charge on any individual atom.

5. Oxidation States and Organic-Style Covalent Molecules

The same rules apply, in principle, to any covalently bonded atom, including carbon — useful preparation for the organic chemistry later in this block, where tracking the oxidation state of carbon identifies oxidation and reduction steps that involve no ions whatsoever.

Worked example — carbon in methane, carbon dioxide, and ethanol. In $\text{CH}_4$, each hydrogen is $+1$ (rule 5), so by rule 6, $x + 4(+1) = 0 \implies x = -4$: carbon is $-4$, its most reduced possible state, since it is bonded only to atoms (hydrogen) less electronegative than itself. In $\text{CO}_2$, oxygen is $-2$ (rule 4), so $x + 2(-2) = 0 \implies x = +4$: carbon is $+4$, its most oxidised possible state. The combustion $\text{CH}_4 \longrightarrow \text{CO}_2$ is therefore a genuine oxidation of carbon, from $-4$ to $+4$, even though it is written with no ions at all. In ethanol, $\text{C}_2\text{H}_5\text{OH}$, the two carbons are not chemically equivalent: the $\text{CH}_3$ carbon is bonded only to hydrogen and the other carbon, while the $\text{CH}_2\text{OH}$ carbon is additionally bonded to oxygen; a full assignment (treating each $\text{C}\!-\!\text{C}$ bond as contributing $0$ to either carbon) gives $-3$ for the methyl carbon and $-1$ for the carbon bonded to oxygen, consistent with the overall neutral-molecule check once every hydrogen ($+1$ each) and the single oxygen ($-2$) are included.

6. Identifying a Redox Reaction from Oxidation State Changes

With oxidation states assigned to every atom on both sides of an equation, a reaction is identified as a redox reaction precisely when at least one element's oxidation state changes between reactants and products. If every oxidation state is unchanged, the reaction is not a redox reaction (it may instead be, for example, an acid–base or precipitation reaction, where ions merely recombine without any change in their oxidation states).

Worked example. Classify $\text{Zn}(s) + \text{CuSO}_4(aq) \longrightarrow \text{ZnSO}_4(aq) + \text{Cu}(s)$.

Zinc goes from $0$ (free element, rule 1) to $+2$ (as $\text{Zn}^{2+}$ in $\text{ZnSO}_4$): zinc is oxidised. Copper goes from $+2$ (as $\text{Cu}^{2+}$ in $\text{CuSO}_4$) to $0$ (free element): copper is reduced. Sulfur and oxygen, present as the sulfate ion $\text{SO}_4^{2-}$ throughout, keep the same oxidation states on both sides ($+6$ for sulfur, $-2$ for each oxygen) — sulfate is a spectator ion in the redox sense as well as the precipitation sense of the earlier chapter on ionic equations. This is a redox reaction: $\text{Zn}$ is the reducing agent, and $\text{CuSO}_4$ (specifically its $\text{Cu}^{2+}$ ion) is the oxidising agent.

Counter-example. $\text{NaOH}(aq) + \text{HCl}(aq) \longrightarrow \text{NaCl}(aq) + \text{H}_2\text{O}(l)$ is a neutralisation, not a redox reaction: sodium stays $+1$, chlorine stays $-1$, oxygen stays $-2$, and hydrogen stays $+1$ throughout. No oxidation state changes, so no electron transfer has occurred — only $\text{H}^{+}$ and $\text{OH}^{-}$ ions have recombined to form water.

Key distinction. Checking oxidation states is the general method for telling whether any reaction — ionic, covalent, or organic — is a redox reaction. A reaction that forms a precipitate or neutralises an acid is not automatically a redox reaction; the oxidation-state check must still be performed explicitly.

7. Disproportionation Reactions

Occasionally, a single element in a single reactant is simultaneously oxidised and reduced, ending up as two different products with two different oxidation states. This special case is called disproportionation.

Worked example. Chlorine gas reacts with cold, dilute sodium hydroxide:

Chlorine enters as the free element, oxidation state $0$ (rule 1). In the product $\text{NaCl}$, chlorine is the chloride ion, oxidation state $-1$ — it has been reduced. In the product $\text{NaOCl}$ (sodium hypochlorite), oxygen is $-2$ and sodium is $+1$, so chlorine must be $+1$ to satisfy rule 6 — it has been oxidised. The same element, starting from the same oxidation state in the same reactant molecule, has split into one reduced product and one oxidised product: a disproportionation.

Worked example. The decomposition of hydrogen peroxide, $2\text{H}_2\text{O}_2(aq) \longrightarrow 2\text{H}_2\text{O}(l) + \text{O}_2(g)$, is a second classic disproportionation. Oxygen enters at $-1$ (the peroxide exception, Section 3). In water, oxygen returns to its usual $-2$ — a reduction. In $\text{O}_2$, oxygen is the free element, $0$ — an oxidation. Hydrogen peroxide is simultaneously its own oxidising and reducing agent, which is exactly why it decomposes spontaneously rather than needing an external partner.

8. Consistency Between Electrons Transferred and Oxidation State Change

The oxidation-state and electron-transfer pictures are not two competing descriptions — they must agree, and checking that they do is a useful consistency test, as well as the method the next chapter relies on to balance redox equations by the half-equation method. The magnitude of the change in oxidation state of one atom equals the number of electrons it loses or gains, and the total electrons lost by all oxidised atoms in a balanced equation must equal the total electrons gained by all reduced atoms.

Worked example. In $\text{Zn}(s) + \text{Cu}^{2+}(aq) \longrightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s)$, zinc's oxidation state rises by $2$ (from $0$ to $+2$), matching the $2$ electrons explicitly lost in its half-equation $\text{Zn} \longrightarrow \text{Zn}^{2+} + 2e^{-}$. Copper's oxidation state falls by $2$ (from $+2$ to $0$), matching the $2$ electrons gained in $\text{Cu}^{2+} + 2e^{-} \longrightarrow \text{Cu}$. The electrons lost ($2$) exactly equal the electrons gained ($2$), as they must for any correctly balanced redox equation — a one-to-one match that will not always hold so simply once the oxidising and reducing agents involve different numbers of electrons, which is precisely the balancing problem taken up next.

9. Common Exam Pitfalls

• Forgetting the peroxide exception for oxygen. $\text{O}$ is $-1$, not $-2$, whenever an $\text{O}\!-\!\text{O}$ bond is present (as in $\text{H}_2\text{O}_2$ or any peroxide salt); this is the single most frequently tested exception in the whole topic.
• Treating oxidation state as a real, measurable charge. It is a formal bookkeeping convention based on electronegativity, useful for tracking electron transfer, not a literal ionic charge on a covalently bonded atom such as manganese in $\text{MnO}_4^{-}$.
• Assuming every reaction that forms a precipitate or neutralises an acid is a redox reaction. The oxidation-state check must be performed explicitly; many such reactions involve no change in any oxidation state at all.
• Missing a disproportionation by assuming an element must either be wholly oxidised or wholly reduced across a reaction. When the same element appears in two different products with two different oxidation states, both oxidation and reduction are occurring to the same starting species.
• Mis-assigning hydrogen's oxidation state in a metal hydride. Hydrogen is usually $+1$, but becomes $-1$ in compounds such as $\text{NaH}$ or $\text{CaH}_2$, where it is bonded to a less electronegative metal rather than a non-metal.
• Dividing a fractional oxidation state (such as $+\tfrac{8}{3}$ for iron in $\text{Fe}_3\text{O}_4$) evenly among identical-looking atoms, rather than recognising it as the average of two genuinely different integer oxidation states carried by chemically distinct atoms of the same element.

Summary

• Oxidation is the loss of electrons; reduction is the gain of electrons (OIL RIG); the two always occur together in a single redox reaction, linked by a pair of half-equations whose electrons cancel when combined.
• The oxidising agent is reduced and causes another species to be oxidised; the reducing agent is oxidised and causes another species to be reduced.
• The oxidation state extends electron counting to covalent bonds by assigning each shared electron pair, on paper, to the more electronegative atom.
• Assignment follows an ordered rule set: free elements are $0$; monoatomic ions equal their charge; $\text{F}$ is always $-1$; $\text{O}$ is usually $-2$ (exceptions: $-1$ in peroxides, $-\tfrac{1}{2}$ in superoxides, $+2$ with fluorine); $\text{H}$ is usually $+1$ (exception: $-1$ in metal hydrides); and the oxidation states of a neutral species sum to $0$, of an ion to its charge.
• A reaction is a redox reaction exactly when at least one element's oxidation state changes between reactants and products; unchanged oxidation states throughout signal a non-redox process such as acid–base neutralisation.
• Disproportionation occurs when the same element, starting from one oxidation state, ends up simultaneously oxidised in one product and reduced in another.
• The magnitude of an oxidation-state change equals the number of electrons transferred by that atom, and total electrons lost by all oxidised species must equal total electrons gained by all reduced species — the consistency check the next chapter uses to balance redox equations by the half-equation method, in both acidic and basic solution.