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

1. Overview

A redox reaction moves electrons from one substance to another. Redox reactions run batteries, extract metals from their ores, rust iron and release energy in respiration. This section gives you two ways to follow the electrons: electron transfer and oxidation numbers. You will calculate oxidation numbers and use them to spot redox and disproportionation reactions, identify oxidising and reducing agents, name compounds with Roman numerals and balance equations.

Key Definitions

• Oxidation: the loss of electrons by a species, or an increase in the oxidation number of an element.
• Reduction: the gain of electrons by a species, or a decrease in the oxidation number of an element.
• Redox reaction: a reaction in which oxidation and reduction happen together, because the electrons lost by one species are gained by another.
• Oxidation number (oxidation state): the charge an atom would have if all of its bonds were fully ionic, with the electrons of each bond given to the more electronegative atom (or shared equally between identical atoms); it is 0 for an uncombined element and is written with the sign first, for example +3.
• Oxidising agent: a species that oxidises another species by taking electrons from it; the oxidising agent is itself reduced, so an element in it decreases in oxidation number.
• Reducing agent: a species that reduces another species by giving electrons to it; the reducing agent is itself oxidised, so an element in it increases in oxidation number.
• Disproportionation: a redox reaction in which atoms of one element, all starting in the same oxidation state, are both oxidised and reduced, so some of them end in a higher oxidation state and some in a lower one.
• Half-equation: an equation that shows only the oxidation or only the reduction part of a redox reaction, with the electrons lost or gained written as e⁻.

Content

Oxidation and reduction as electron transfer

Put magnesium ribbon into silver nitrate solution. Silver metal coats the ribbon and the magnesium dissolves:

Mg(s) + 2Ag⁺(aq) → Mg²⁺(aq) + 2Ag(s)

Split the reaction into two half-equations:

• Mg(s) → Mg²⁺(aq) + 2e⁻. Magnesium loses electrons, so it is oxidised.
• Ag⁺(aq) + e⁻ → Ag(s). Each silver ion gains an electron, so silver ions are reduced.

A memory aid is OIL RIG: Oxidation Is Loss, Reduction Is Gain (of electrons).

Electrons cannot be lost unless something takes them, so oxidation and reduction always happen together: the reaction is redox. Electrons lost must equal electrons gained: each Mg atom loses two electrons and each Ag⁺ ion takes one, so two Ag⁺ react per Mg. The nitrate ions take no part. No oxygen is involved: "gain of oxygen" is only a special case of oxidation.

Oxidation numbers

In 2H₂(g) + O₂(g) → 2H₂O(l) no ions form: the O-H bonds are covalent, with the shared electrons closer to oxygen, the more electronegative atom. Oxidation numbers turn this into bookkeeping. Pretend every bond is ionic, and give both electrons of each bond to the more electronegative atom.

• In H₂O, oxygen is given two extra electrons, so it is −2. Each hydrogen atom is +1.
• In a bond between identical atoms, such as H-H or O=O, the electrons are shared equally, so every atom of an uncombined element is 0.

This defines oxidation and reduction for any reaction:

• Oxidation is an increase in oxidation number. Losing negative electrons makes the number more positive.
• Reduction is a decrease in oxidation number.

When water forms, hydrogen goes from 0 to +1 (oxidised) and oxygen from 0 to −2 (reduced). It is redox even though no ions form.

Write an oxidation number with its sign first (+2, −1), and the charge on an ion with its sign last (Mg²⁺, Cl⁻). Manganese in MnO₄⁻ is +7, but there is no Mn⁷⁺ ion in it, so never write an oxidation number as "7+" or "Mn⁷⁺".

Rules for calculating oxidation numbers

Rule	Examples
1. An uncombined element is 0.	Na, Fe, H₂, O₂, Cl₂, S₈
2. A simple (one-atom) ion has an oxidation number equal to its charge.	Na⁺ +1, Fe³⁺ +3, Cl⁻ −1, O²⁻ −2
3. The oxidation numbers in a formula add up to its overall charge: 0 for a neutral compound, the ion's charge for an ion.	SO₄²⁻: (+6) + 4(−2) = −2
4. Fluorine is −1 in all its compounds. Group 1 metals are +1, Group 2 metals +2 and aluminium +3 in their compounds.	NaF, MgCl₂, Al₂O₃
5. Hydrogen is +1 in most compounds, but −1 in metal hydrides and in NaBH₄ (boron is less electronegative than hydrogen).	+1 in HCl and NH₃; −1 in NaH and CaH₂
6. Oxygen is −2 in most compounds. The exceptions to know are −1 in peroxides and +2 in OF₂.	−2 in CO₂; −1 in H₂O₂
7. Chlorine, bromine and iodine are −1 in most compounds, but positive when combined with oxygen or a more electronegative halogen.	−1 in HBr; +7 in Cl₂O₇; +3 in ClF₃

If two rules clash, the one higher in the table wins, which is where the exceptions come from: in NaH, rule 4 beats rule 5, so hydrogen is −1. Use rule 3 to find any element the rules do not cover. If no rule covers either element, give the negative number to the more electronegative one.

Worked example 1: oxidation numbers in five species

Question: Calculate the oxidation number of: (a) Fe in K₂FeO₄; (b) P in P₂O₇⁴⁻; (c) O in BaO₂; (d) H in NaH; (e) O in OF₂.

Call the unknown $x$ and apply rule 3.

(a) K is +1 and O is −2. $2(+1) + x + 4(-2) = 0$, so $x = +6$.

(b) O is −2 and the ion's charge is −4. $2x + 7(-2) = -4$, so $2x = +10$. Divide by the two P atoms: $x = +5$.

(c) Ba is +2 (rule 4 beats rule 6). $+2 + 2x = 0$, so $x = -1$. BaO₂ is a peroxide: it contains the O₂²⁻ ion.

(d) Na is +1. $+1 + x = 0$, so $x = -1$. NaH is a metal hydride: it contains the H⁻ ion.

(e) F is −1 (rule 4 beats rule 6). $x + 2(-1) = 0$, so $x = +2$.

Answer: (a) +6; (b) +5; (c) −1; (d) −1; (e) +2.

Two sense checks:

• Upper limit. For a Period 3 element, the oxidation number cannot be higher than its number of outer-shell electrons: phosphorus is at most +5 and sulfur at most +6 (see 9.2).
• Fractions. A fraction is an average. In Mn₃O₄, $3x + 4(-2) = 0$ gives $x = +\frac{8}{3}$, because the oxide contains manganese in the +2 and +3 states in the ratio 1 : 2.

Is a reaction redox?

Find the oxidation number of every element on both sides. If any element changes, the reaction is redox, and there must be at least one increase and one decrease. If nothing changes, it is not redox: neutralisation, precipitation and many decompositions are not.

Reaction	Oxidation numbers	Redox?
CaCO₃(s) → CaO(s) + CO₂(g)	Ca +2, C +4, O −2 on both sides	no
SO₃(g) + H₂O(l) → H₂SO₄(aq)	S +6, H +1, O −2 on both sides	no
2H₂(g) + O₂(g) → 2H₂O(l)	H: 0 to +1; O: 0 to −2	yes
2Ca(NO₃)₂(s) → 2CaO(s) + 4NO₂(g) + O₂(g)	N: +5 to +4; the O that forms O₂: −2 to 0	yes

Oxidising agents and reducing agents

• The oxidising agent contains the element whose oxidation number falls. It takes electrons, so it is reduced.
• The reducing agent contains the element whose oxidation number rises. It gives electrons, so it is oxidised.

In Mg + 2Ag⁺, Ag⁺ is the oxidising agent (+1 to 0) and Mg is the reducing agent (0 to +2). Name the agent as the whole reactant species, such as Ag⁺ or Cr₂O₇²⁻. A species whose oxidation numbers do not change, such as H⁺ in an acidified solution, is neither.

Oxidising agent	Change	Reducing agent	Change
O₂	O: 0 to −2	reactive metals, such as Mg	Mg: 0 to +2
Cl₂	Cl: 0 to −1	H₂	H: 0 to +1
acidified MnO₄⁻ (purple to colourless)	Mn: +7 to +2	C or CO, in extracting metals	C: 0 or +2 to +4
acidified Cr₂O₇²⁻ (orange to green)	Cr: +6 to +3	I⁻	I: −1 to 0
H₂O₂	O: −1 to −2	Fe²⁺	Fe: +2 to +3
concentrated H₂SO₄	S: +6 to +4 or lower	SO₂	S: +4 to +6

An element in an in-between oxidation state can go either way, depending on its partner: oxygen in H₂O₂ (−1) can fall to −2 or rise to 0. An element already at its highest oxidation number, such as manganese at +7 in MnO₄⁻, can only be reduced. Comparing the strengths of agents comes later (11.2, 11.3 and 24.2).

Disproportionation

In disproportionation, one element is both oxidised and reduced in the same reaction. Hydrogen peroxide slowly decomposes:

2H₂O₂(aq) → 2H₂O(l) + O₂(g)

Oxygen starts at −1 in H₂O₂. Some oxygen atoms fall to −2 in H₂O (reduced) and others rise to 0 in O₂ (oxidised). In electron terms, the H₂O₂ molecules that are reduced gain the electrons that the others lose. Chlorine with cold or hot aqueous sodium hydroxide is another example (see 11.4).

Check all three points:

1. The same element is both oxidised and reduced.
2. Its atoms start in one oxidation state.
3. Some end in a higher state and some in a lower state than the start.

These are redox but not disproportionation:

• 2Ca(NO₃)₂ → 2CaO + 4NO₂ + O₂: nitrogen is reduced but oxygen is oxidised, two different elements.
• 2H₂S(g) + SO₂(g) → 3S(s) + 2H₂O(l): sulfur is oxidised (−2 to 0) and reduced (+4 to 0), but it starts in two states and ends in one, the reverse of disproportionation.

Roman numerals in names

Many elements have more than one oxidation number, so a name must say which one. A Roman numeral in brackets, straight after the element's name with no space, gives the size of the oxidation number. In an anion whose name ends in "-ate", the numeral refers to the central atom: manganate(VII) means manganese is +7.

Formula	Oxidation number	Name
FeSO₄ and Fe₂O₃	Fe: +2 and +3	iron(II) sulfate and iron(III) oxide
PCl₃ and PCl₅	P: +3 and +5	phosphorus(III) chloride and phosphorus(V) chloride
KMnO₄	Mn: +7	potassium manganate(VII)
K₂Cr₂O₇	Cr: +6	potassium dichromate(VI)
NaClO and NaClO₃	Cl: +1 and +5	sodium chlorate(I) and sodium chlorate(V)
Na₂SO₄ and Na₂SO₃	S: +6 and +4	sodium sulfate(VI) and sodium sulfate(IV), also called sodium sulfite

• In names like these the oxidation number is positive, so the numeral has no sign.
• Leave the numeral out when an element has only one usual oxidation number (sodium chloride, zinc sulfate). Keep it whenever it matters: "sodium chlorate" alone could be NaClO or NaClO₃.
• The traditional names sulfite (SO₃²⁻) and nitrite (NO₂⁻) are also used, and plain "sulfate" and "nitrate" mean sulfate(VI) and nitrate(V).
• From name to formula, the numeral gives the ion's charge: chromium(III) oxide contains Cr³⁺ and O²⁻, so it is Cr₂O₃ (see 2.3).

Balancing equations with oxidation numbers

Every electron lost is gained by something else, so the total increase in oxidation number must equal the total decrease. Use this to find the ratio of the reacting species first:

1. Write the formulae of the species that change.
2. Find the oxidation number of each element that changes, before and after, and the change for each species: change per atom × number of atoms of that element in the formula.
3. Choose numbers in front so that the total increase equals the total decrease.
4. Balance the atoms that do not change. In acid, add H₂O for oxygen and H⁺ for hydrogen; in alkali, use OH⁻ and H₂O.
5. Check that both the atoms and the total charges balance.

Worked example 2: acidified dichromate(VI) and tin(II) ions

Question: Acidified dichromate(VI) ions oxidise tin(II) ions to tin(IV) ions and are reduced to chromium(III) ions. Use oxidation numbers to construct the ionic equation.

Step 1: the species that change.
Cr₂O₇²⁻ + Sn²⁺ → Cr³⁺ + Sn⁴⁺

Step 2: oxidation numbers. In Cr₂O₇²⁻, $2x + 7(-2) = -2$, so Cr is +6.

• Cr: +6 to +3, a decrease of 3 per atom. Each Cr₂O₇²⁻ contains two Cr atoms, so the decrease is 2 × 3 = 6 per ion.
• Sn: +2 to +4, an increase of 2 per ion.

Step 3: match the changes. One Cr₂O₇²⁻ (decrease 6) needs three Sn²⁺ (increase 3 × 2 = 6):
Cr₂O₇²⁻ + 3Sn²⁺ → 2Cr³⁺ + 3Sn⁴⁺

Step 4: balance oxygen and hydrogen. The 7 oxygen atoms become 7H₂O, which need 14H⁺ on the left.

Step 5: check the charges. Left: (−2) + 3(+2) + 14(+1) = +18. Right: 2(+3) + 3(+4) = +18. They balance.

Answer: Cr₂O₇²⁻(aq) + 14H⁺(aq) + 3Sn²⁺(aq) → 2Cr³⁺(aq) + 3Sn⁴⁺(aq) + 7H₂O(l)

The solution turns from orange to green. In a titration, 1 mol of Cr₂O₇²⁻ reacts with 3 mol of Sn²⁺.

Worked example 3: sulfur in hot alkali

Question: When sulfur is heated with concentrated aqueous sodium hydroxide, one reaction that takes place forms sulfide ions, S²⁻, and sulfite ions, SO₃²⁻. (a) Explain, using oxidation numbers, why this is disproportionation. (b) Construct the ionic equation.

(a) Sulfur starts at 0, as the element. In S²⁻ it is −2. In SO₃²⁻, $x + 3(-2) = -2$, so it is +4. From a single starting state, sulfur is reduced from 0 to −2 and oxidised from 0 to +4 in the same reaction, so this is disproportionation.

(b) Step 1: match the changes. Forming S²⁻ is a decrease of 2 per atom; forming SO₃²⁻ is an increase of 4. Two S²⁻ (decrease 2 × 2 = 4) balance one SO₃²⁻ (increase 4):
3S → 2S²⁻ + SO₃²⁻

Step 2: balance the charge with OH⁻. The right side has charge 2(−2) + (−2) = −6, so put 6OH⁻ on the left.

Step 3: balance hydrogen with water. The 6 hydrogen atoms form 3H₂O on the right. Oxygen check: 6 on the left; 3 + 3 = 6 on the right.

Answer: 3S(s) + 6OH⁻(aq) → 2S²⁻(aq) + SO₃²⁻(aq) + 3H₂O(l)

When one reactant does two jobs

Silver dissolves in dilute nitric acid, giving nitrogen monoxide, NO. Silver rises from 0 to +1 and nitrogen falls from +5 to +2, so three Ag are oxidised for each NO formed. The three AgNO₃ also need three nitrate groups, from nitric acid that is not reduced, so 4HNO₃ react in all, and their 4 hydrogen atoms form 2H₂O:

3Ag(s) + 4HNO₃(aq) → 3AgNO₃(aq) + NO(g) + 2H₂O(l)

Oxidation numbers fix the ratio of the species that change (Ag to NO, 3 : 1); count the unchanged atoms afterwards.

Half-equations

A half-equation shows one half of a redox reaction with its electrons. Electrons on the right are lost (oxidation); electrons on the left are gained (reduction). The number of electrons equals the total change in oxidation number, so you can balance a half-equation for an ion in acid:

1. Balance the element that changes.
2. Add electrons equal to its total change in oxidation number: on the left for a reduction, on the right for an oxidation.
3. Add H₂O to balance oxygen, then H⁺ to balance hydrogen.
4. Check that the charges balance.

For the nitrate reduced by silver above, nitrogen falls from +5 to +2, so 3e⁻ go on the left. The two spare oxygen atoms form 2H₂O, which need 4H⁺:

NO₃⁻(aq) + 4H⁺(aq) + 3e⁻ → NO(g) + 2H₂O(l)

Charges: (−1) + 4(+1) + 3(−1) = 0 on the left, 0 on the right. When a question gives you two half-equations, multiply them so the electrons lost equal the electrons gained, then add them and cancel the electrons. Three times Ag(s) → Ag⁺(aq) + e⁻ supplies the 3e⁻, giving 3Ag(s) + NO₃⁻(aq) + 4H⁺(aq) → 3Ag⁺(aq) + NO(g) + 2H₂O(l), the ionic form of the equation above. Choosing the relevant half-equations from electrode potential data comes in 24.2.

Worked example 4: a new oxidation number from reacting amounts

Question: A solution contains ions of a metal M, in which M has oxidation number +2. In acid, 10.0 cm³ of 0.0600 mol dm⁻³ M²⁺(aq) reacts exactly with 16.0 cm³ of 0.0150 mol dm⁻³ MnO₄⁻(aq). Each MnO₄⁻ ion is reduced to Mn²⁺. Deduce the oxidation number of M after the reaction.

Step 1: amounts. $$n(\text{M}^{2+}) = \frac{10.0}{1000} \times 0.0600 = 6.00 \times 10^{-4}\ \text{mol}$$ $$n(\text{MnO}_4^-) = \frac{16.0}{1000} \times 0.0150 = 2.40 \times 10^{-4}\ \text{mol}$$

Step 2: electrons gained by manganese. Mn falls from +7 to +2, so each MnO₄⁻ gains 5 electrons. $$n(\text{e}^-) = 5 \times 2.40 \times 10^{-4} = 1.20 \times 10^{-3}\ \text{mol}$$

Step 3: electrons lost by each M ion. M loses the electrons that manganese gains. $$\frac{1.20 \times 10^{-3}}{6.00 \times 10^{-4}} = 2.00$$

Step 4: new oxidation number. Losing 2 electrons raises the oxidation number by 2: +2 + 2 = +4.

Answer: M has oxidation number +4 after the reaction.

Check: MnO₄⁻ and M²⁺ react in the ratio 2.40 : 6.00 = 2 : 5, and two MnO₄⁻ gain 10 electrons while five M²⁺ lose 10.

Key Equations

Sum of oxidation numbers (each element's oxidation number times its number of atoms in the formula; the overall charge is 0 for a neutral compound): $$\sum (\text{oxidation number} \times \text{number of atoms}) = \text{overall charge}$$

Change for one species (for Cr₂O₇²⁻ going to Cr³⁺, 3 × 2 = 6): $$\text{change per species} = \text{change per atom} \times \text{atoms of that element in the formula}$$

Balancing rule (electrons lost = electrons gained): $$\text{total increase in oxidation number} = \text{total decrease in oxidation number}$$

Moles of electrons transferred ($n$ in mol): $$n(\text{e}^-) = n(\text{species}) \times \text{change in oxidation number per species}$$

Two changes worth knowing (in acid): MnO₄⁻ → Mn²⁺ gains 5 electrons per ion; Cr₂O₇²⁻ → 2Cr³⁺ gains 6 electrons per ion.

Common Mistakes to Avoid

• Mistake: setting the sum of oxidation numbers to zero for an ion. Fix: the sum equals the ion's charge. In Worked example 1(b), using 0 instead of −4 gives P = +7, impossible for an atom with five outer electrons.
• Mistake: writing an oxidation number like a charge, as "6+" or "Cr⁶⁺". Fix: write +6, or the Roman numeral VI. Chromium in Cr₂O₇²⁻ is +6, but no Cr⁶⁺ ion exists in it.
• Mistake: forgetting to multiply by the number of atoms. Fix: Cr₂O₇²⁻ → 2Cr³⁺ is a decrease of 3 for each of two Cr atoms, 6 in total. Using 3 in Worked example 2 gives 2Cr₂O₇²⁻ to 3Sn²⁺ instead of 1 to 3.
• Mistake: explaining why a reaction is redox with only one half. Fix: give both changes, naming the element and its species: "magnesium is oxidised from 0 to +2, and silver in Ag⁺ is reduced from +1 to 0".
• Mistake: naming a product, just the element, or the acid as the oxidising agent. Fix: in Worked example 2 the oxidising agent is Cr₂O₇²⁻, not Cr³⁺ or "chromium". The H⁺ ions are not an agent: hydrogen is +1 in both H⁺ and H₂O.
• Mistake: calling any redox decomposition disproportionation, or saying a compound is "oxidised and reduced". Fix: in the decomposition of Ca(NO₃)₂, two different elements change, so it is not disproportionation. For H₂O₂ decomposing, say "oxygen is oxidised from −1 to 0 and reduced from −1 to −2".
• Mistake: calling a reaction redox because it involves oxygen or an oxide. Fix: check every element. In the acid-base reaction CO₂ + Ca(OH)₂ → CaCO₃ + H₂O, carbon stays +4 and nothing else changes, so it is not redox.

Exam Tips

• Explain why a reaction is redox by giving, for each element that changes, its species, its oxidation number before and after, and whether it is oxidised or reduced. If asked for the type of reaction, write "redox", not just "oxidation".
• State or deduce an oxidation number with its sign in front, such as +4 (not 4+). When asked to calculate, show the sum that gives it.
• Define terms in the way the question asks: "in terms of electron transfer" needs electrons lost or gained. An oxidising agent takes electrons and is itself reduced.
• Identify an agent by giving the formula of the reactant species and a reason: "Cr₂O₇²⁻, because chromium is reduced from +6 to +3".
• Construct a redox equation by matching the oxidation number changes first, then adding H₂O and H⁺ (acid) or OH⁻ (alkali). Check atoms and charges, and leave out spectator ions.
• Name a compound with the Roman numeral in brackets straight after the element or the "-ate" ending, such as iron(III) chloride or sodium chlorate(V).