Chemistry · 6 Chemical reactions

6.4 Redox

Before you start

Ions and their charges, e.g. $Fe^{2 +}$ , $Cl^{-}$ , $SO_{4}^{2 -}$ .
Balancing simple equations and reading state symbols.
That a positive ion has lost electrons and a negative ion has gained them.

Syllabus objectives covered (13)

Use a Roman numeral to indicate the oxidation number of an element in a compound
Define redox reactions as involving simultaneous oxidation and reduction
Define oxidation as gain of oxygen and reduction as loss of oxygen
Identify redox reactions as reactions involving gain and loss of oxygen
Identify oxidation and reduction in redox reactions
Define oxidation in terms of: (a) loss of electrons (b) an increase in oxidation number
Define reduction in terms of: (a) gain of electrons (b) a decrease in oxidation number
Identify redox reactions as reactions involving gain and loss of electrons
Identify redox reactions by changes in oxidation number using: (a) the oxidation number of elements in their uncombined state is zero (b) the oxidation number of a monatomic ion is the same as the charge on the ion (c) the sum of the oxidation numbers in a compound is zero (d) the sum of the oxidation numbers in an ion is equal to the charge on the ion
Identify redox reactions by the colour changes involved when using acidified aqueous potassium manganate(VII) or aqueous potassium iodide
Define an oxidising agent as a substance that oxidises another substance and is itself reduced
Define a reducing agent as a substance that reduces another substance and is itself oxidised
Identify oxidising agents and reducing agents in redox reactions

Part 1 of 6

Three definitions of the same thing

Redox is one process seen three ways. As you move through the syllabus the definition widens, but it never changes what is actually happening.
In terms of oxygen. Oxidation is gain of oxygen; reduction is loss of oxygen. This is the oldest definition and the easiest to spot.
In terms of electrons. Oxidation is loss of electrons; reduction is gain of electrons. This is the definition that covers reactions with no oxygen anywhere.
The mnemonic is worth having: OIL RIG — Oxidation Is Loss, Reduction Is Gain. Of electrons.
In terms of oxidation number. Oxidation is an increase in oxidation number; reduction is a decrease.
The word 'redox' exists because the two always happen together: if something loses electrons, something else must gain them. Electrons do not go nowhere.
That is the definition of a redox reaction — one involving simultaneous oxidation and reduction. The word simultaneous is a marking point; do not leave it out.

Part 2 of 6

Roman numerals and what they mean

A Roman numeral in a name gives the oxidation number of that element in the compound. Iron(II) chloride contains $Fe^{2 +}$ ; iron(III) chloride contains $Fe^{3 +}$ .
It is used where an element forms more than one ion, which is mostly the transition elements. Sodium is always $+ 1$ , so nobody writes sodium(I) chloride.
The numeral belongs to the named element only, not to the whole compound. In copper(II) sulfate the II refers to the copper, not the sulfate.
Two on the syllabus catch people out: potassium manganate(VII), where manganese is $+ 7$ , and vanadium(V) oxide, where vanadium is $+ 5$ .
Written into a formula, the numeral tells you the charge, which tells you the formula. Iron(III) oxide has $Fe^{3 +}$ and $O^{2 -}$ , so it must be $Fe_{2} O_{3}$ .

Part 3 of 6

Working out oxidation numbers

Four rules do all the work, and every question is one of them applied carefully.
Rule 1. An element in its uncombined state has oxidation number zero. $Fe$ , $O_{2}$ , $Cl_{2}$ , $S_{8}$ — all zero, however many atoms are joined together.
Rule 2. A monatomic ion has an oxidation number equal to its charge. $Na^{+}$ is $+ 1$ , $Cl^{-}$ is $- 1$ , $Mg^{2 +}$ is $+ 2$ .
Rule 3. The oxidation numbers in a compound sum to zero. A compound is electrically neutral overall.
Rule 4. The oxidation numbers in an ion sum to the charge on the ion. For $SO_{4}^{2 -}$ they sum to $- 2$ .
In practice you use two familiar values and solve for the unknown: oxygen is usually $- 2$ and hydrogen usually $+ 1$ .
Find the oxidation number of sulfur in $H_{2} SO_{4}$ : two hydrogens give $+ 2$ , four oxygens give $- 8$ , and the total must be zero. So sulfur is $+ 6$ .
$2 (+ 1) + x + 4 (- 2) = 0 \Rightarrow x = + 6$
Write the sign every time. $+ 6$ and $6$ are not the same answer, and a bare number often loses the mark.

Part 4 of 6

Spotting redox in an equation

By oxygen. Compare each substance before and after. In $2 CuO + C \overset{}{\to} 2 Cu + CO_{2}$ , the copper oxide has lost oxygen, so copper(II) oxide is reduced; the carbon has gained oxygen, so carbon is oxidised.
Both happen in the same equation, which is what makes it redox rather than just 'a reaction'.
By electrons. $Mg + Cl_{2} \overset{}{\to} MgCl_{2}$ contains no oxygen at all, yet it is redox. Magnesium atoms become $Mg^{2 +}$ , losing two electrons each; chlorine atoms become $Cl^{-}$ , gaining one each.
Half-equations make this explicit, and are often worth writing even when not demanded:
$Mg \overset{}{\to} Mg^{2 +} + 2 e^{-} Cl_{2} + 2 e^{-} \overset{}{\to} 2 Cl^{-}$
By oxidation number. The most general method: work out the oxidation number of each element on both sides, and look for one that goes up and one that goes down.
If nothing changes, it is not a redox reaction. Neutralisation and precipitation reactions are the usual non-redox examples — the ions swap partners but no oxidation number moves.

Part 5 of 6

Oxidising and reducing agents

An oxidising agent oxidises another substance and is itself reduced. A reducing agent reduces another substance and is itself oxidised.
Both halves of each definition are needed. Writing only 'an oxidising agent oxidises something' gets half a definition and, usually, half the marks.
The wording feels back to front and that is exactly why it is examined. An oxidising agent gets reduced — because it is the thing taking the electrons.
So the shortcut when identifying: find the substance that gains electrons (or goes down in oxidation number). That one is the oxidising agent.
In $2 CuO + C \overset{}{\to} 2 Cu + CO_{2}$ , copper goes from $+ 2$ to $0$ — a decrease, so it is reduced, so copper(II) oxide is the oxidising agent. Carbon goes from $0$ to $+ 4$ , so carbon is the reducing agent.
Name the whole substance, not the element, when the question gives you a compound: 'copper(II) oxide is the oxidising agent', not 'copper is'.

Part 6 of 6

Two colour changes that reveal redox

Some redox reactions announce themselves with a colour change, and two are on the syllabus.
Acidified aqueous potassium manganate(VII) is purple. When it acts as an oxidising agent it is reduced and the purple disappears — the solution is decolourised.
So decolourising acidified potassium manganate(VII) is evidence that the other substance is a reducing agent. You met this with sulfite ions and sulfur dioxide in 12.5.
Aqueous potassium iodide is colourless. When it acts as a reducing agent, the iodide ions are oxidised to iodine, and the solution turns brown.
$2 I^{-} \overset{}{\to} I_{2} + 2 e^{-}$
So a colourless solution turning brown on adding potassium iodide is evidence that the added substance is an oxidising agent.
The two tests are mirror images: manganate(VII) losing its colour detects a reducing agent, potassium iodide gaining colour detects an oxidising agent. Getting these the right way round is the whole of the question.

Worked examples

Example 1. Determine the oxidation number of manganese in the manganate(VII) ion, $MnO_{4}^{-}$ .

Show the working

Identify which rule applies. This is an ion, so the oxidation numbers sum to the charge on the ion, which is $- 1$ .
Use the value you know: each oxygen is $- 2$ , and there are four of them, giving $- 8$ .
Let the oxidation number of manganese be $x$ and write the equation.
$x + 4 (- 2) = - 1$
Solve: $x - 8 = - 1$ , so $x = + 7$ .
Check it against the name. Potassium manganate(VII) — the Roman numeral says $+ 7$ , and it agrees.

Answer: $+ 7$ .

Example 2. For the reaction $Fe_{2} O_{3} + 3 CO \overset{}{\to} 2 Fe + 3 CO_{2}$ , identify what is oxidised, what is reduced, and name the oxidising and reducing agents. Justify your answer.

Show the working

Use the oxygen definition first, since oxygen is present throughout. Iron(III) oxide loses oxygen to become iron, so it is reduced.
Carbon monoxide gains oxygen to become carbon dioxide, so it is oxidised.
Confirm with oxidation numbers. Iron goes from $+ 3$ in $Fe_{2} O_{3}$ to $0$ in $Fe$ — a decrease, so reduction, which agrees.
Carbon goes from $+ 2$ in $CO$ to $+ 4$ in $CO_{2}$ — an increase, so oxidation. Both methods agree, which is a good sign.
Now apply the definitions. The substance that is reduced is the oxidising agent, so iron(III) oxide is the oxidising agent; carbon monoxide is the reducing agent.

Answer: Iron(III) oxide is reduced (loses oxygen; Fe goes from $+ 3$ to $0$ ) and is the oxidising agent. Carbon monoxide is oxidised (gains oxygen; C goes from $+ 2$ to $+ 4$ ) and is the reducing agent.

Example 3. Chlorine is bubbled through colourless aqueous potassium iodide. The solution turns brown. Deduce which substance is the oxidising agent, and explain your reasoning.

Show the working

Interpret the observation first. A brown colour in this context means iodine has been formed from the colourless iodide ions.
So the iodide ions have lost electrons: $2 I^{-} \overset{}{\to} I_{2} + 2 e^{-}$ . Loss of electrons is oxidation, so the iodide has been oxidised.
Those electrons must have gone somewhere. The chlorine has taken them: $Cl_{2} + 2 e^{-} \overset{}{\to} 2 Cl^{-}$ , so chlorine has been reduced.
Apply the definition. The substance that is reduced is the one doing the oxidising, so chlorine is the oxidising agent.
Sanity-check the direction: chlorine is above iodine in Group VII and is the more reactive halogen, so it displacing iodine is what you would expect.

Answer: Chlorine is the oxidising agent. It gains electrons and is reduced to chloride ions, while iodide ions lose electrons and are oxidised to iodine, which gives the brown colour.

Where people go wrong

Mistake: Defining an oxidising agent as 'a substance that is oxidised'.

Instead: An oxidising agent oxidises something else and is itself reduced.

The name describes what it does to others, not what happens to it. Reversing this is the most common error in the whole subtopic.

Mistake: Giving OIL RIG in terms of oxygen rather than electrons.

Instead: Oxidation Is Loss, Reduction Is Gain — of electrons. In terms of oxygen it is the opposite way round: oxidation is gain of oxygen.

Two definitions with opposite directions are easy to conflate. Always say which one you are using.

Mistake: Writing an oxidation number without its sign.

Instead: Always write $+ 2$ or $- 2$ , never a bare $2$ .

The sign is the information. An unsigned number is ambiguous and is usually not credited.

Mistake: Assuming a reaction cannot be redox if there is no oxygen in it.

Instead: Check for electron transfer or a change in oxidation number instead.

$Mg + Cl_{2} \overset{}{\to} MgCl_{2}$ has no oxygen and is unambiguously redox.

Mistake: Treating the Roman numeral as the charge on the whole compound.

Instead: It gives the oxidation number of the element it follows.

Compounds are neutral overall. In iron(III) chloride, the III is the iron's oxidation number, not a charge on $FeCl_{3}$ .

Mistake: Mixing up which reagent gains colour and which loses it.

Instead: Manganate(VII) is purple and is decolourised by a reducing agent; potassium iodide is colourless and turns brown with an oxidising agent.

The tests detect opposite things, so swapping them inverts the conclusion entirely.

Check yourself

Define oxidation in terms of oxygen, electrons and oxidation number.
Answer
Gain of oxygen; loss of electrons; an increase in oxidation number.
Why must oxidation and reduction always occur together?
Answer
Electrons lost by one substance must be gained by another, so the two processes are simultaneous.
What is the oxidation number of vanadium in vanadium(V) oxide, and what is the formula of the compound?
Answer
$+ 5$ ; the formula is $V_{2} O_{5}$ .
Determine the oxidation number of sulfur in $SO_{4}^{2 -}$ .
Answer
$x + 4 (- 2) = - 2$ , so $x = + 6$ .
State the oxidation number of chlorine in $Cl_{2}$ and in $NaCl$ .
Answer
$0$ in $Cl_{2}$ , because it is an uncombined element; $- 1$ in $NaCl$ .
Define a reducing agent.
Answer
A substance that reduces another substance and is itself oxidised.
In $ZnO + C \overset{}{\to} Zn + CO$ , state what is oxidised and what is reduced.
Answer
Carbon gains oxygen, so carbon is oxidised. Zinc oxide loses oxygen, so zinc oxide is reduced.
A solution decolourises acidified aqueous potassium manganate(VII). What does this show about the solution?
Answer
It contains a reducing agent, which reduces the manganate(VII) and removes the purple colour.

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. Iron is extracted in a blast furnace by the reaction $Fe_{2} O_{3} + 3 CO \overset{}{\to} 2 Fe + 3 CO_{2}$

(a) Identify the substance that is oxidised in this reaction, and give a reason. 2 marks No calculator

Mark scheme

B1 carbon monoxide / CO
B1 it gains oxygen, or the oxidation number of carbon increases from +2 to +4

(b) Determine the oxidation number of iron in $Fe_{2} O_{3}$ and in $Fe$ . 2 marks No calculator

Mark scheme

B1 +3 in Fe2O3
B1 0 in Fe

(c) Define an oxidising agent. 2 marks No calculator

Mark scheme

B1 a substance that oxidises another substance
B1 and is itself reduced

(d) State which substance in the equation is the oxidising agent. 1 mark No calculator

Mark scheme

B1 iron(III) oxide / Fe2O3; ecf from (a)

Question 2. Magnesium burns in chlorine gas. $Mg + Cl_{2} \overset{}{\to} MgCl_{2}$

(a) Explain why this is a redox reaction, even though no oxygen is involved. 3 marks No calculator

Mark scheme

B1 magnesium loses electrons, so it is oxidised
B1 chlorine gains electrons, so it is reduced
B1 oxidation and reduction occur simultaneously, which defines a redox reaction

(b) Give the half-equation for the change undergone by magnesium. 1 mark No calculator

Mark scheme

B1 Mg → Mg²⁺ + 2e⁻

Question 3. A student adds a colourless solution of substance X to acidified aqueous potassium manganate(VII). The purple colour disappears. A second sample of X is added to aqueous potassium iodide and no colour change is seen.

(a) Deduce whether X is an oxidising agent or a reducing agent, using both observations. 3 marks No calculator

Mark scheme

B1 X is a reducing agent
B1 it reduces the manganate(VII), which is decolourised
B1 the potassium iodide is unchanged, so X does not oxidise iodide ions to iodine, ORA

(b) Predict the colour change if chlorine were added to the aqueous potassium iodide instead, and explain it. 2 marks No calculator

Mark scheme

B1 colourless to brown
B1 chlorine oxidises iodide ions to iodine / 2I⁻ → I₂ + 2e⁻

Words to know

Redox reaction: A reaction involving simultaneous oxidation and reduction.
Oxidation: Gain of oxygen, loss of electrons, or an increase in oxidation number.
Reduction: Loss of oxygen, gain of electrons, or a decrease in oxidation number.
Oxidation number: A signed number describing the combining state of an element in a compound or ion.
Oxidising agent: A substance that oxidises another substance and is itself reduced.
Reducing agent: A substance that reduces another substance and is itself oxidised.

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