Lesson 1 · 25 min

Oxidation states: bookkeeping for electrons

Burning methane involves no ions, yet electrons still change hands. A set of priority rules gives every atom a formal charge, and who is oxidised and who is reduced becomes plain.

In this lesson

State what an oxidation state means and assign it to every element of a formula or ion by the priority rules
Define oxidation and reduction by the change in oxidation state and identify the oxidising and reducing agents
Count the electrons transferred in a balanced equation and recognise disproportionation and comproportionation
Know that oxidation states are bookkeeping and that a fraction is an average

Dip a zinc strip into copper sulfate and within minutes it is coated with red-brown copper while the blue fades; light a gas stove and methane and oxygen become carbon dioxide and water. In the first reaction $\ce{Cu^2+}$ plainly gains electrons; in the second there is not a single ion — do electrons change hands at all?

Chemists keep the books with oxidation states: pretend every bond is ionic, hand the bonding electrons to the more electronegative partner, and the charge each atom is left with is its oxidation state. With this ledger, burning and displacement are described in the same language.

Predict first

What is the oxidation state of oxygen in hydrogen peroxide, $\ce{H2O2}$ ?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

−2 −1 0 +1

Rules that queue by priority

The “valence” of school chemistry is, in redox, more precisely the oxidation state. No structure is needed: apply the rules below in order, earlier rules taking priority, and let the last remaining element be fixed by “the sum equals the charge”:

An element on its own is 0 ( $\ce{O2}$ , $\ce{S8}$ , $\ce{Fe}$ ); a monatomic ion equals its charge ( $\ce{Fe^3+}$ is +3).
Fluorine is always −1 in compounds.
Group 1 metals are +1, group 2 metals +2, aluminium +3.
Hydrogen is +1 with non-metals and −1 when bonded only to metals (or B, Si): $\ce{NaH}$ , $\ce{LiAlH4}$ .
Oxygen is usually −2 (−1 in peroxides, +2 in $\ce{OF2}$ ).
Cl, Br and I are usually −1, except when bonded to oxygen or a more electronegative halogen.
The oxidation states of all atoms add up to the overall charge of the molecule or ion.

\sum_i n_i\,x_i = q

Rule 7 is the last step of every problem: once the other atoms are known, it solves for the one that is left.

$n_i$: number of atoms of element i in the formula
$x_i$: oxidation state of element i
$q$: overall charge (0 for a neutral molecule)

Priority rules, one element at a time

Oxidation-state finder

Formula or ion

Write multiple charges with ^: Cr2O7^2-, Fe^3+; complexes may use brackets: K4[Fe(CN)6].

Try

+1 K₂+6 Cr₂−2 O₇

Element	Atoms	State	Why
K	2	+1	Group 1 metals are +1 in compounds.
Cr	2	+6	From the balance: all the oxidation states add up to the overall charge.
O	7	−2	Oxygen is usually −2.

Check: Σ atoms × state = overall charge 2(+1) + 2(+6) + 7(−2) = 0

When an element appears more than once, the value is their average (the two C atoms of ethanol are really −3 and −1).

The number above each element is its oxidation state; the table says which rule set it. Try Fe₃O₄ (a fraction), NH₄NO₃ (one element in two states), NaH, OF₂ and K₄[Fe(CN)₆], or type a formula of your own.

Oxidation and reduction come in pairs

With oxidation states the definitions are simple:

Oxidation: the oxidation state rises — electrons are lost;
Reduction: the oxidation state falls — electrons are gained.

Electrons do not vanish, so whatever one species loses another gains. The substance that makes something else oxidised is the oxidising agent, and it is itself reduced; the one that makes something else reduced is the reducing agent, and it is itself oxidised. The mnemonic “OIL RIG” (Oxidation Is Loss, Reduction Is Gain) says the same thing.

One element can be both oxidised and reduced: bubble chlorine into cold sodium hydroxide and half the chlorine becomes $\ce{Cl-}$ (−1) while half becomes $\ce{ClO-}$ (+1). That is disproportionation; the reverse, a high and a low state meeting in the middle, is comproportionation.

Oxidation-number bookkeeping

Who is oxidised? Who is reduced?

Reaction Show oxidation numbers

A zinc strip in copper sulfate grows red-brown copper.

Zn+ Cu²⁺ → Zn²⁺+ Cu

Guess first: which reactant is the oxidising agent?

Guess the oxidising agent first, then read the change of every atom. Then switch to bleach making (disproportionation), photosynthesis and MnO₂ with concentrated HCl — where only 2 of the 4 Cl atoms are oxidised — and see why a neutralisation is not redox.

Worked example

The electron ledger of a breathalyser

In the old breathalyser tube, orange dichromate oxidises ethanol in the breath to ethanal and becomes green $\ce{Cr^3+}$ : $\ce{Cr2O7^2-} + \ce{3C2H5OH} + \ce{8H+} \mathrel{\ce{->}} \ce{2Cr^3+} + \ce{3CH3CHO} + \ce{7H2O}$ . Identify the oxidising and reducing agents and check the electron count.

Concept check

In the thermite reaction $\ce{2Al} + \ce{Fe2O3} \mathrel{\ce{->}} \ce{Al2O3} + \ce{2Fe}$ , what is the oxidising agent?

\ce{Al}

\ce{Fe2O3}

\ce{Al2O3}

The oxygen atoms

Try the calculation

What is the average oxidation state of sulfur in tetrathionate, $\ce{S4O6^2-}$ , formed in iodometry?

Key takeaways

An oxidation state is the charge an atom would carry if every bond were ionic; the priority rules fix most atoms and the charge balance fixes the last.
A rise is oxidation (electrons lost), a fall is reduction (electrons gained); the oxidising agent is reduced and the reducing agent oxidised.
In a balanced equation electrons lost equal electrons gained; one element going both up and down is disproportionation, meeting in the middle is comproportionation.
Oxidation states are bookkeeping, not real charges; a fraction means an average.

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Data and references

P. Karen, P. McArdle, J. Takats, “Comprehensive definition of oxidation state (IUPAC Recommendations 2016)”, Pure Appl. Chem. 88, 831–839 (2016).
IUPAC, Compendium of Chemical Terminology (the “Gold Book”), entries “oxidation state” and “oxidation number”.
C. E. Housecroft, A. G. Sharpe, Inorganic Chemistry, 5th ed., Pearson (2018), ch. 8 (reduction and oxidation).
人民教育出版社《普通高中教科书化学必修第一册》第一章第三节“氧化还原反应”（2019）。