Notes · ChemistryUK · A-Levels

Oxidation, reduction and redox equations

Redox reactions - those involving the transfer of electrons - underlie combustion, corrosion, extraction of metals, electrochemistry and much of inorganic chemistry. This chapter defines oxidation and reduction in terms of electron transfer and of oxidation state, sets out the rules for assigning oxidation states, and shows how to build balanced half-equations and combine them into overall redox equations.

4 sections~11 min reading time3 competenciesLevel Foundation 1 · Standard 2 · Advanced 1

T·0777 / 18

Exam profile

AO1 · Define oxidation and reduction in terms of electrons and oxidation state, and state the assignment rulesAO2 · Assign oxidation states and construct balanced half-equations and overall redox equationsAO3 · Use oxidation states to identify the oxidising and reducing agents in an unfamiliar reaction

Operators:statedefinededuceidentifyconstructbalance

basic level

AS-Level requires oxidation and reduction as electron transfer, the oxidation-state rules, identifying oxidising and reducing agents, and constructing half-equations and overall equations.

higher level

The full A-Level applies this constantly - in Group 7 chemistry, transition-metal redox titrations, electrode potentials and organic oxidation - so fluency with half-equations here pays off across the course.

Depth

Text

Contents · 4 sections

Oxidation, reduction and redox equations

Oxidation and reduction as electron transfer#

FoundationAQA 7405 3.1.7.1DfE GCE Chemistry - oxidation and reduction

At A-Level, oxidation and reduction are defined in terms of electrons: oxidation is the loss of electrons and reduction is the gain of electrons. The mnemonic OIL RIG captures it - Oxidation Is Loss, Reduction Is Gain. The two always happen together, because the electrons lost by one species must be gained by another, which is why these are called redox (reduction-oxidation) reactions. When magnesium burns, each Mg atom loses two electrons (oxidation) and each oxygen atom gains two (reduction).

This electron definition supersedes the earlier GCSE idea of oxidation as 'gain of oxygen' and reduction as 'loss of oxygen' (recap), which is only a special case. The electron definition covers reactions with no oxygen at all - the reaction of sodium with chlorine,

2\text{Na} + \text{Cl}_2 \rightarrow 2\text{NaCl}

, is redox because sodium loses electrons and chlorine gains them, even though no oxygen is involved.

It is essential to track which species loses and which gains electrons. The species that loses electrons is oxidised; the species that gains them is reduced. In a displacement reaction such as

\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}

, zinc atoms lose two electrons (oxidised to

\text{Zn}^{2+}

) and copper ions gain two electrons (reduced to

\text{Cu}

). Writing out where the electrons go makes any redox reaction transparent.

Because electrons cannot simply disappear, the number lost in the oxidation must equal the number gained in the reduction. This conservation of electrons is the principle that lets us balance redox equations by combining half-equations, as developed later in the chapter. Keeping the electron bookkeeping exact is the whole skill.

Worked example

Identifying oxidation and reduction

In the reaction Zn + Cu2+ -> Zn2+ + Cu, state what is oxidised and what is reduced, in terms of electrons.

Track the zinc
Zn goes to Zn2+, losing two electrons, so zinc is oxidised.
Track the copper
Cu2+ goes to Cu, gaining two electrons, so the copper ion is reduced.
Check the balance
The two electrons lost by zinc are exactly the two gained by copper - electrons are conserved.

Result: Zinc is oxidised (loses 2 electrons) and Cu2+ is reduced (gains 2 electrons).

Typical mistakes

Reversing the definitions - remember OIL RIG, oxidation is loss and reduction is gain.
Thinking a reaction must involve oxygen to be redox.

Active revision

For the reaction $\text{Mg} + \text{Cl}_2 \rightarrow \text{MgCl}_2$ , state which species is oxidised and which is reduced, and how many electrons each atom transfers.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for the sciences (Department for Education) · AQA A-level Chemistry 7405 specification (AQA)

Oxidation states and the assignment rules#

StandardAQA 7405 3.1.7.1DfE GCE Chemistry - oxidation states

Oxidation states of nitrogen

Fig. 1Nitrogen shows a wide range of oxidation states, from -3 in ammonia to +5 in nitric acid.

The oxidation state (oxidation number) of an atom is the charge it would have if all the bonds in the species were completely ionic - a bookkeeping device that lets us track electron transfer even in covalent compounds. It is written with the sign first (

+2

-1

) to distinguish it from an ionic charge. A change in oxidation state is the surest way to spot a redox reaction: an increase means oxidation (electrons lost), a decrease means reduction (electrons gained).

A short set of rules assigns oxidation states reliably. An uncombined element has an oxidation state of

0

(including diatomic elements such as

\text{O}_2

). A simple monatomic ion has an oxidation state equal to its charge (

\text{Na}^+

+1

\text{Cl}^-

-1

). In a compound, the sum of all the oxidation states equals the overall charge (zero for a neutral compound, the ion charge for an ion). Combined oxygen is usually

-2

, and combined hydrogen usually

+1

; the more electronegative element in a bond takes the negative value.

A few exceptions must be memorised. Oxygen is

-1

in peroxides (such as

\text{H}_2\text{O}_2

) and

+2

when bonded to fluorine (in

\text{OF}_2

). Hydrogen is

-1

in metal hydrides (such as

\text{NaH}

). Fluorine, being the most electronegative element, is always

-1

in compounds. Group 1 metals are always

+1

and Group 2 metals always

+2

in their compounds. These fixed values are your anchors when working out an unknown oxidation state.

To find an unknown oxidation state, set it as

x

, assign the known values to the other atoms, multiply each by how many atoms are present, and set the total equal to the overall charge. For the manganate(VII) ion

\text{MnO}_4^{-}

: the four oxygens contribute

4 \times (-2) = -8

, and the total must equal the

-1

charge, so

x + (-8) = -1

, giving manganese

= +7

. The Roman numeral in a name, as in manganate(VII) or iron(III), is exactly this oxidation state.

\sum (\text{oxidation states}) = \text{overall charge}

The summation rule

Set the unknown as x, add the known contributions, and equate the total to the charge on the species (0 for a neutral compound).

Worked example

Oxidation state of chromium in the dichromate ion

Deduce the oxidation state of chromium in the dichromate(VI) ion, Cr2O7 2-.

Assign the oxygens
Each of the seven oxygens is -2, giving $7 \times (-2) = -14$ .
Use the charge sum
The two chromiums (each x) plus the oxygens equal the -2 charge: $2x + (-14) = -2$ .
Solve
$2x = -2 + 14 = 12$ , so $x = +6$ .
$2x - 14 = -2 \;\Rightarrow\; x = +6$

Result: Chromium has an oxidation state of +6, consistent with the name dichromate(VI).

Typical mistakes

Assuming oxygen is always -2 or hydrogen always +1, forgetting peroxides and metal hydrides.
Confusing the oxidation state (written +2) with the ionic charge, or forgetting to multiply by the number of atoms.

Active revision

Deduce the oxidation state of sulfur in (a) $\text{H}_2\text{SO}_4$ , (b) the $\text{SO}_3^{2-}$ ion and (c) $\text{S}_8$ .

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Chemistry 7405 specification (AQA)

Oxidising and reducing agents#

StandardAQA 7405 3.1.7.1DfE GCE Chemistry - oxidising and reducing agents

Electron transfer in a displacement reaction

Fig. 2Zinc (reducing agent) donates electrons and is oxidised; the copper ion (oxidising agent) accepts them and is reduced.

An oxidising agent is a species that oxidises something else, and to do so it must accept electrons - so the oxidising agent is itself reduced. A reducing agent reduces something else by donating electrons, so it is itself oxidised. The terminology is deliberately counter-intuitive: the agent that causes oxidation is the one being reduced. Reading it as 'the oxidising agent takes electrons, so its own oxidation state goes down' keeps it straight.

You can identify the agents by tracking oxidation-state changes. The species whose oxidation state falls has been reduced, so it is the oxidising agent; the species whose oxidation state rises has been oxidised, so it is the reducing agent. In the reaction of chlorine with iron(II),

\text{Cl}_2

(chlorine goes from

0

-1

, reduced) is the oxidising agent and

\text{Fe}^{2+}

(iron goes from

+2

+3

, oxidised) is the reducing agent.

Common oxidising agents include the halogens, oxygen, concentrated sulfuric acid, potassium manganate(VII) and potassium dichromate(VI); common reducing agents include reactive metals, hydrogen, carbon and the halide ions (especially iodide). Recognising a reagent's usual role speeds up analysis of an unfamiliar reaction, though the safest method is always to work out the oxidation-state changes.

The strength of an oxidising or reducing agent is periodic. Down Group 7 the halogens become weaker oxidising agents (they hold their electrons less tightly, so accept new ones less readily), while the halide ions become stronger reducing agents. This trend, developed fully in the Group 7 chapter, is why chlorine can displace bromine and iodine from their salts but iodine cannot displace chlorine - the stronger oxidising agent displaces the weaker.

Worked example

Identifying the agents by oxidation state

In the reaction 2Fe3+ + 2I- -> 2Fe2+ + I2, identify the oxidising and reducing agents.

Track iron
Fe goes from +3 to +2: its oxidation state falls, so Fe3+ is reduced.
Track iodine
I goes from -1 to 0: its oxidation state rises, so I- is oxidised.
Assign the roles
The species reduced (Fe3+) is the oxidising agent; the species oxidised (I-) is the reducing agent.

Result: Fe3+ is the oxidising agent (reduced to Fe2+) and I- is the reducing agent (oxidised to I2).

Typical mistakes

Labelling the species that is oxidised as the oxidising agent - it is the reducing agent.
Guessing an agent's role from its name rather than checking the oxidation-state change.

Active revision

In the reaction $\text{Cl}_2 + 2\text{KBr} \rightarrow 2\text{KCl} + \text{Br}_2$ , identify the oxidising agent and the reducing agent, justifying your answer with oxidation states.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Chemistry 7405 specification (AQA)

Half-equations and balanced redox equations#

AdvancedAQA 7405 3.1.7.1DfE GCE Chemistry - half-equations and redox equations

Combining half-equations

Fig. 3Multiplying the iron half-equation by five equalises the electrons, which then cancel to give the overall equation.

A half-equation shows either the oxidation or the reduction on its own, including the electrons transferred. To build one, first balance the atoms being oxidised or reduced, then balance oxygen by adding water, balance hydrogen by adding

\text{H}^+

ions (in acidic conditions), and finally balance the charge by adding electrons to the more positive side. The reduction of manganate(VII) in acid, for instance, is

\text{MnO}_4^{-} + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}

The number of electrons in a half-equation must be correct, because it links to the oxidation-state change: manganese falls from

+7

+2

, a change of five, so five electrons appear. Checking that both the atoms and the total charge balance on each side is the surest test that a half-equation is right - the charges on the two sides of

\text{MnO}_4^{-} + 8\text{H}^+ + 5e^-

(

-1 + 8 - 5 = +2

) match the

\text{Mn}^{2+}

on the right.

To construct the overall equation, combine an oxidation half-equation with a reduction half-equation so that the electrons cancel. Multiply each half-equation by whatever factor makes the number of electrons equal, then add them and cancel the electrons (and any species appearing on both sides). Because the electrons lost must equal the electrons gained, this guarantees a balanced overall equation with no electrons left in it.

This method is the backbone of redox-titration calculations later in the course, where manganate(VII) or dichromate(VI) reacts with iron(II) or another reductant. A worked half-equation approach also handles disproportionation, where a single element is simultaneously oxidised and reduced (as when chlorine reacts with water), by writing separate half-equations for the two fates of the element. Always finish by checking that atoms and charge both balance.

\text{MnO}_4^{-} + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}

Reduction of manganate(VII) in acid

Oxygen balanced by water, hydrogen by H+, charge by five electrons (Mn falls from +7 to +2).

Worked example

Combining half-equations

Combine MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O with Fe2+ -> Fe3+ + e- to give the overall equation.

Equalise the electrons
The manganate half-equation needs 5 electrons; multiply the iron half-equation by 5 to give 5Fe2+ -> 5Fe3+ + 5e-.
Add and cancel electrons
Adding the two, the 5 electrons cancel.
Write the overall equation
MnO4- + 8H+ + 5Fe2+ -> Mn2+ + 4H2O + 5Fe3+.
$\text{MnO}_4^{-} + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} + 5\text{Fe}^{3+}$
Check
Charge: left $-1 + 8 + 10 = +17$ ; right $+2 + 15 = +17$ . Atoms balance too.

Result: MnO4- + 8H+ + 5Fe2+ -> Mn2+ + 4H2O + 5Fe3+, with charge (+17) and atoms balanced.

Typical mistakes

Leaving electrons in the final overall equation because the half-equations were not scaled to equal electrons.
Balancing atoms but forgetting to balance the charge (or vice versa).

Active revision

The half-equations are $\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$ and $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$ . Construct the overall redox equation.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Chemistry 7405 specification (AQA)

References & sources

Sources

Department for Education

GCE AS and A level subject content for the sciences

AQA

AQA A-level Chemistry 7405 specification

Oxidation, reduction and redox equations

4 sections~11 min reading time3 competenciesLevel Foundation 1 · Standard 2 · Advanced 1

Oxidation, reduction and redox equations

Key points

Identifying oxidation and reduction

Key points

Oxidation state of chromium in the dichromate ion

Key points

Identifying the agents by oxidation state

Key points

Combining half-equations

Sources

Oxidation, reduction and redox equations

Key points

Identifying oxidation and reduction

Key points

Oxidation state of chromium in the dichromate ion

Key points

Identifying the agents by oxidation state

Key points

Combining half-equations

Sources