Redox

In a nutshell

Redox is any reaction where electrons move from one species to another: one species is oxidised (loses electrons) while another is reduced (gains them).

This subtopic gives you the tool for tracking those electrons, the oxidation number. You use it to name compounds, write formulae from names, and decide what is oxidised and what is reduced in any reaction, including ones you have never seen.

Assumed knowledge: Compounds, formulae and equations: ion charges and balanced equations; Acids: what a salt is.

Core content

Oxidation number: assigning and calculating

An oxidation number tells you how many electrons an atom has lost (a positive number) or gained (a negative number) compared with the uncombined element.

It is written with the sign first: +2, -1, +7. (An ion's charge is written the other way round: 2+, 1-.)

Every oxidation number comes from four rules:

1. An atom in an uncombined element has oxidation number 0: Zn, Cl₂, O₂, S₈.
2. In a compound, the oxidation numbers add up to 0.
3. In a simple (monatomic) ion, the oxidation number equals the charge: Fe³⁺ is +3, Cl^- is -1.
4. In a polyatomic ion, the oxidation numbers add up to the charge on the ion: in SO₄^2-, S + 4(-2) = -2, so S = +6.

Some elements almost always have the same oxidation number in compounds. When two of them clash, the one higher in this list wins:

The priority order explains both exceptions OCR expects you to know:

• In NaH, Na (+1) outranks H, so H must be -1.
• In H₂O₂, H (+1) outranks O, so 2(+1) + 2(O) = 0 gives O = -1.

Any element not in the table (C, N, S, P, most d-block metals) takes whatever value makes the total come out right.

Always work out the oxidation number of one atom. In CaCl₂, chlorine is -1, not -2: the two Cl atoms together contribute -2.

Elements such as nitrogen and sulfur have many oxidation numbers, depending on what they are bonded to:

Still don't get it? · what an oxidation number actually is

Imagine every atom starts a game holding its own set of tokens, and its score starts at 0. Each token is worth minus one point. Hand a token away and your score goes up by one; take a token from someone else and your score goes down by one.

Now rebuild it in chemistry, one step at a time:

1. The tokens are electrons, and electrons are negative. That is why losing one makes the score more positive.
2. An atom in an uncombined element is holding exactly its own electrons, so its score is 0.
3. A magnesium atom that gives away 2 electrons to become Mg²⁺ now scores +2. A chlorine atom that takes 1 electron to become Cl^- scores -1. For a simple ion, the score is just the charge.
4. In a covalent bond the electrons are shared, not handed over. So we pretend: each shared pair is counted as belonging to the atom that pulls harder on it. The rules in the table (F is -1, O is -2, H is +1) are that pretending already done for you.
5. Electrons are never created or destroyed, so in a neutral compound the scores must add up to 0, and in an ion they add up to its charge.

Back to the exam wording: oxidation is loss of electrons, so the oxidation number goes up. Reduction is gain of electrons, so the oxidation number goes down.

Oxidation number: writing formulae

If you know the oxidation number of each element, you can write the formula: choose the smallest whole numbers of atoms that make the oxidation numbers add up to zero (or to the ion's charge).

This works for covalent compounds too (sulfur(VI) oxide is covalent), which is why oxidation numbers are more useful than ion charges alone.

Oxidation number: Roman numerals in names

When an element can have more than one oxidation number in its compounds, its name carries a Roman numeral showing the oxidation number.

• iron(II) chloride is FeCl₂, with Fe at +2.
• iron(III) chloride is FeCl₃, with Fe at +3.

The same idea names oxyanions: negative ions containing oxygen. The Roman numeral gives the oxidation number of the element that is not oxygen. Within each family the charge on the ion stays the same; only the number of oxygen atoms changes.

Two things to remember:

• The Roman numeral is the oxidation number, never the number of oxygen atoms. Chlorate(V) is ClO₃^-, not ClO₅^-.
• If a question just says "nitrate" or "sulfate" with no Roman numeral, assume NO₃^- and SO₄^2-.

Redox reactions: electron transfer and oxidation number

OCR describes oxidation and reduction in two equivalent ways:

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

A redox reaction is one in which oxidation and reduction happen at the same time. Electrons lost by one species are gained by another, so the total electrons lost = total electrons gained.

Redox happens right across the periodic table:

• s-block: 2Na(s) + Cl₂(g) → 2NaCl(s). Each Na loses 1 electron (0 → +1, oxidised); each Cl gains 1 electron (0 → -1, reduced).
• p-block: in the same reaction, chlorine (a p-block element) is the one reduced. Aluminium is a p-block metal that is oxidised (0 → +3) when it reacts with acids (see below).
• d-block: 2Fe²⁺(aq) + Cl₂(aq) → 2Fe³⁺(aq) + 2Cl^-(aq). Each Fe²⁺ loses 1 electron (+2 → +3, oxidised); each Cl gains 1 electron (0 → -1, reduced).

You can show the electron loss or gain on its own, for example Na → Na⁺ + e^- and Cl₂ + 2e^- → 2Cl^-.

Not every reaction is redox. In a neutralisation (acid with a base, carbonate or alkali, from 2.1.4), no oxidation numbers change, so it is not redox.

You will also meet two terms for the species that cause the change (they are used later in the course, e.g. 3.1.3 and 4.2.1):

• An oxidising agent takes electrons from another species, so the oxidising agent is itself reduced.
• A reducing agent gives electrons to another species, so the reducing agent is itself oxidised.

Still don't get it? · oxidised vs oxidising agent

Think of a pickpocket and a victim. The victim loses a wallet. The pickpocket gains it. And the pickpocket is the one who caused the victim's loss.

Now map it onto electrons, one step at a time:

1. The wallet is the electrons.
2. The victim loses electrons, so the victim is oxidised (OIL).
3. The pickpocket gains electrons, so the pickpocket is reduced (RIG).
4. The pickpocket made the oxidation happen, so the pickpocket is the oxidising agent. The oxidising agent is always the species that gets reduced.
5. Turn it round: the victim "supplied" the electrons that reduced the pickpocket, so the victim is the reducing agent, and it is always the species that gets oxidised.

Back to the exam wording: in Mg + 2HCl, magnesium is oxidised (0 to +2, loses electrons). Never write "magnesium is oxidising": that phrase means magnesium is the oxidising agent, which is the opposite. Say "oxidised" or "reduced" for what happens to a species.

Redox reactions: metals with acids

A reactive metal reacts with a dilute acid such as hydrochloric acid or dilute sulfuric acid to form a salt and hydrogen:

metal + acid → salt + hydrogen

This is a redox reaction:

• The metal is oxidised: its oxidation number goes from 0 to a positive value, as it loses electrons to form a positive ion in the salt.
• The hydrogen is reduced: its oxidation number goes from +1 in the acid to 0 in H₂, as it gains electrons.
• The other part of the acid (Cl or SO₄) does not change oxidation number.

The full equations you need, for metals from the s-, p- and d-blocks:

Points that carry marks:

• Hydrogen gas is H₂, never H.
• Iron with these dilute acids forms the iron(II) salt (FeCl₂, FeSO₄), not the iron(III) salt.
• Aluminium forms Al³⁺, so its equations need the larger numbers: 2Al, 3H₂.
• OCR wants full equations here (ionic equations are not required for these reactions).

What you would see: effervescence (fizzing, as hydrogen is given off) and the metal dissolving.

Redox reactions: unfamiliar reactions

You can be given a redox reaction you have never met (including ones with nitric acid or concentrated sulfuric acid) and asked to interpret it. The method is always the same:

1. Write the oxidation number of every element on both sides.
2. Find the elements whose oxidation number changes. Ignore the rest.
3. Increase = oxidised (lost electrons). Decrease = reduced (gained electrons).
4. The change in oxidation number per atom = the number of electrons lost or gained per atom.
5. Check: total increase = total decrease, because every electron lost is gained by something else. This also lets you predict the ratio in which species react (Worked examples 4 and 5).

Watch for an element that appears in two products with different oxidation numbers: only some of its atoms change, as with nitrogen in Worked example 4.

Worked examples

These are oxidation-number calculations, so the answers are signed whole numbers with no units. Every step is on its own line.

1. Assigning oxidation numbers

(a) Find the oxidation number of Mn in KMnO₄.

$\text{K} = +1, \quad \text{O} = -2$ $(+1) + \text{Mn} + 4(-2) = 0$ $\text{Mn} = 0 - 1 + 8 = +7$

(b) Find the oxidation number of Cr in the ion Cr₂O₇^2-.

$2\,\text{Cr} + 7(-2) = -2 \quad \text{(sum equals the charge on the ion, not 0)}$ $2\,\text{Cr} = -2 + 14 = +12$ $\text{Cr} = +6$

Dividing by 2 in the last line is where marks go. +12 is the total for both Cr atoms; the answer is per atom.

(c) Find the oxidation number of O in Na₂O₂, and of H in CaH₂.

$\text{Na}_2\text{O}_2: \quad 2(+1) + 2\,\text{O} = 0 \;\Rightarrow\; \text{O} = -1 \quad \text{(a peroxide)}$ $\text{CaH}_2: \quad (+2) + 2\,\text{H} = 0 \;\Rightarrow\; \text{H} = -1 \quad \text{(a metal hydride)}$

Writing O = -2 or H = +1 here out of habit gives a total that does not add up to zero. That is your signal that an exception applies.

2. Names and formulae with Roman numerals

(a) Write the formula of sodium chlorate(III).

$\text{Cl} = +3, \quad \text{O} = -2, \quad \text{chlorate ion charge} = 1-$ $(+3) + n(-2) = -1 \;\Rightarrow\; n = 2$ $\text{ion} = \text{ClO}_2^{-} \;\Rightarrow\; \text{formula} = \text{NaClO}_2$

(b) Name KClO₃.

$(+1) + \text{Cl} + 3(-2) = 0 \;\Rightarrow\; \text{Cl} = +5$

Name: potassium chlorate(V).

(c) Name Cu₂O.

$2\,\text{Cu} + (-2) = 0 \;\Rightarrow\; \text{Cu} = +1$

Name: copper(I) oxide. The numeral is the oxidation number (+1), not the 2 in the formula.

3. A metal with an acid

Write the full equation, with state symbols, for aluminium reacting with dilute sulfuric acid, and use oxidation numbers to show it is redox.

Aluminium forms Al³⁺ and sulfate is SO₄^2-, so the salt is aluminium sulfate, Al₂(SO₄)₃.

2Al(s) + 3H₂SO₄(aq) → Al₂(SO₄)₃(aq) + 3H₂(g)

Check the electrons balance:

$\text{Al}: 0 \rightarrow +3 \quad \text{(oxidised; 2 atoms lose } 2 \times 3 = 6 \text{ electrons)}$ $\text{H}: +1 \rightarrow 0 \quad \text{(reduced; 6 atoms gain } 6 \times 1 = 6 \text{ electrons)}$ $\text{S stays } +6, \quad \text{O stays } -2$

Electrons lost = electrons gained = 6, so the equation is consistent. Balance SO₄ as one unit: 3 on each side.

4. Interpreting an unfamiliar reaction

Copper reacts with concentrated nitric acid:

Cu(s) + 4HNO₃(aq) → Cu(NO₃)₂(aq) + 2NO₂(g) + 2H₂O(l)

Identify what is oxidised and what is reduced, and explain the 1 : 2 ratio of Cu to NO₂.

$\text{Cu}: 0 \rightarrow +2 \quad \text{(oxidised, loses 2 electrons)}$ $\text{N in HNO}_3: +5$ $\text{N in NO}_2: +4 \quad \text{(reduced, each gains 1 electron)}$ $\text{N in Cu(NO}_3)_2: +5 \quad \text{(no change)}$ $\text{H stays } +1, \quad \text{O stays } -2$

Copper is oxidised. Nitrogen is reduced, but only the 2 N atoms that end up in NO₂; the other 2 stay as nitrate.

Each Cu loses 2 electrons and each N that is reduced gains only 1, so 2 NO₂ must form for every Cu:

$\text{electrons lost} = 1 \times 2 = 2 \qquad \text{electrons gained} = 2 \times 1 = 2$

Saying "nitric acid is oxidised" or "oxygen is reduced" loses the mark: name the element and give its oxidation numbers.

5. Using oxidation numbers to balance an equation

Balance: MnO₂ + HCl → MnCl₂ + Cl₂ + H₂O

Step 1, find the changes:

$\text{Mn}: +4 \rightarrow +2 \quad \text{(decrease of 2: gains 2 electrons)}$ $\text{Cl}: -1 \rightarrow 0 \text{ in Cl}_2 \quad \text{(increase of 1 per Cl atom)}$

Step 2, make total increase = total decrease. One Cl₂ contains 2 Cl atoms, an increase of 2 in total, which matches the decrease of 2 for one Mn:

$1\,\text{MnO}_2 : 1\,\text{Cl}_2$

Step 3, balance the atoms that do not change. MnCl₂ needs 2 more Cl, so 4 HCl in total; the 4 H make 2 H₂O:

MnO₂(s) + 4HCl(aq) → MnCl₂(aq) + Cl₂(g) + 2H₂O(l)

Final check: Mn 1, O 2, H 4 and Cl 4 on each side. Only 2 of the 4 Cl atoms are oxidised; the other 2 stay at -1 in MnCl₂.

Common exam mistakes

• Giving the total, not the per-atom value. Students write Cl = -2 in CaCl₂, or Cr = +12 in Cr₂O₇^2-. An oxidation number is always for one atom: Cl is -1, Cr is +6.
• Missing the two exceptions. O is -1 in peroxides (H₂O₂, Na₂O₂) and H is -1 in metal hydrides (NaH, CaH₂). Using -2 and +1 by habit gives a total that does not add up.
• Summing to zero in an ion. In SO₄^2- or Cr₂O₇^2- the oxidation numbers add up to the charge on the ion, not to 0.
• Writing the sign in the wrong place. The oxidation number of iron in Fe³⁺ is +3; "3+" is the charge on the ion. Use the sign-first form for oxidation numbers.
• Reading the Roman numeral as a count of atoms. Fe₂O₃ is iron(III) oxide, not iron(II) oxide because of the 2. Chlorate(V) is ClO₃^-, not ClO₅^-.
• Leaving the Roman numeral off. "Sodium chlorate" or "iron chloride" is ambiguous when the element has more than one oxidation number; write sodium chlorate(I), iron(II) chloride.
• Defining oxidation as "gaining oxygen". At this level the answer needs loss of electrons or increase in oxidation number.
• Writing "oxidising" for "oxidised". "Magnesium is oxidising" means it is an oxidising agent, the opposite of what happens. Write "magnesium is oxidised".
• Naming the compound instead of the element. "The acid is reduced" or "HNO₃ is reduced" is vague; say "nitrogen is reduced from +5 to +4".
• Wrong salt or wrong hydrogen in metal + acid equations. Iron gives FeCl₂ / FeSO₄ with dilute acids, hydrogen is H₂, and aluminium sulfate is Al₂(SO₄)₃.
• Calling neutralisation redox. Acid + base, carbonate or alkali involves no change in oxidation number, so it is not redox. Only acid + metal (giving hydrogen) is redox here.

Key definitions

• Oxidation: loss of electrons, or an increase in oxidation number.
• Reduction: gain of electrons, or a decrease in oxidation number.
• Redox reaction: a reaction in which oxidation and reduction take place at the same time.
• Oxidation number: the number of electrons an atom has lost (positive) or gained (negative) compared with the uncombined element (counting shared electrons as belonging to the atom that attracts them more strongly).
• Oxidising agent: a species that takes electrons from another species, and is itself reduced.
• Reducing agent: a species that gives electrons to another species, and is itself oxidised.
• Roman numeral in a name: shows the oxidation number of the element it follows, e.g. iron(III) = iron with oxidation number +3; chlorate(I) = ClO^-.

Specification

• I can assign and calculate the oxidation number of an atom in an element, a compound or an ion, including O in peroxides (-1) and H in metal hydrides (-1). (2.1.5 a)
• I can write formulae using oxidation numbers. (2.1.5 b)
• I can use a Roman numeral to show the oxidation number of an element that has more than one, e.g. iron(II) and iron(III), and write formulae from names such as chlorate(I) and chlorate(III), and names from formulae. (2.1.5 c)
• I can describe oxidation and reduction in terms of electron transfer, with examples from the s-, p- and d-blocks. (2.1.5 d i)
• I can describe oxidation and reduction in terms of changes in oxidation number. (2.1.5 d ii)
• I can write full equations for the redox reactions of metals such as Mg, Al, Fe and Zn with acids to form a salt and hydrogen. (2.1.5 e)
• I can interpret these equations, and unfamiliar redox reactions (including ones involving nitric acid or concentrated sulfuric acid), to predict what is oxidised and reduced in terms of oxidation numbers and electron loss or gain. (2.1.5 f)

Related notes

• Compounds, formulae and equations: ion charges, formulae and balancing
• Amount of substance
• Acids: salts and neutralisation (not redox)
• Group 2: redox reactions of the Group 2 metals
• The halogens: halogens as oxidising agents, and disproportionation
• Alcohols: oxidation of alcohols by an oxidising agent