Define redox agents, oxidation numbers, half-reactions, and disproportionation

Alberta Grade 12 Chemistry

Electrochemical Changes

Alberta Chemistry 30 — Outcome 30-B1.2k

Drop a clean iron nail into a solution of copper(II) sulfate and watch: the blue colour fades, and a reddish coating of copper metal appears on the nail. No heat, no light — just two substances exchanging electrons at room temperature. This is a redox reaction. Redox reactions involve the transfer of electrons from one substance to another. They explain corrosion, batteries, bleaching, and metabolism. This lesson builds the four key definitions you need for Alberta Chemistry 30: oxidizing agents, reducing agents, oxidation numbers, half-reactions, and disproportionation. Each definition is a tool. By the end of this lesson, you will use all four tools together to analyse any redox equation you encounter.

What you will learn

Define oxidizing agent and reducing agent and identify each in a redox reaction.
Assign oxidation numbers to atoms in compounds and ions using the Alberta Chemistry 30 rules.
Write and balance oxidation and reduction half-reactions, including electrons and charges.
Define disproportionation and identify a disproportionation reaction from a given equation.

Prerequisite Review: Ions, Charges, and Electron Transfer

Before defining redox terms, recall two ideas from Chemistry 20. First, atoms can gain or lose electrons to form ions. When a neutral sodium atom loses one electron, it becomes a sodium ion with a charge of 1+. When a neutral chlorine atom gains one electron, it becomes a chloride ion with a charge of 1−. Second, in a chemical bond, electrons are shared or transferred between atoms. These two ideas are the entire foundation of redox chemistry.

In a redox reaction, electrons move from one substance to another. The substance that loses electrons is oxidized. The substance that gains electrons is reduced. A helpful memory phrase is OIL RIG: Oxidation Is Loss, Reduction Is Gain — of electrons. Keep this phrase available; you will use it for every redox problem.

Oxidation = loss of electrons by a substance.
Reduction = gain of electrons by a substance.
OIL RIG: Oxidation Is Loss, Reduction Is Gain.
Both processes always occur together in a redox reaction.

Oxidizing Agents and Reducing Agents

An oxidizing agent is the substance that accepts electrons from another substance during a redox reaction. Because it accepts electrons, the oxidizing agent itself becomes reduced. Think of it as the electron thief — it pulls electrons away from something else, causing that something else to be oxidized.

A reducing agent is the substance that donates electrons to another substance during a redox reaction. Because it donates electrons, the reducing agent itself becomes oxidized. It is the electron donor.

In the iron-nail demonstration, iron metal donates electrons to copper(II) ions. Iron is the reducing agent (it gets oxidized). Copper(II) ions accept those electrons and become copper metal. The copper(II) ion is the oxidizing agent (it gets reduced). Notice the language carefully: the oxidizing agent is reduced, and the reducing agent is oxidized. This is a deliberate and important flip that students often find confusing at first.

The overall ionic equation for the iron and copper(II) reaction is:

\text{Fe}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Fe}^{2+}(aq) + \text{Cu}(s)

. Iron loses 2 electrons; copper(II) gains 2 electrons. Atom counts and charges balance on both sides.

\text{Fe}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Fe}^{2+}(aq) + \text{Cu}(s)

Oxidizing agent: gains electrons; is reduced in the reaction.
Reducing agent: loses electrons; is oxidized in the reaction.
The oxidizing agent and reducing agent are always different substances (except in disproportionation — covered later).
Identifying the agent means identifying the substance, not just the element.

Oxidation Numbers

Many redox reactions involve covalent compounds, not simple ions. You cannot directly 'see' an electron transfer in a molecule the way you can with ions. Chemists use oxidation numbers (also called oxidation states) as a bookkeeping tool to track electron ownership in any substance, ionic or covalent. An oxidation number is an integer assigned to an atom that represents how many electrons it has gained or lost compared to its neutral, elemental form. It is a useful accounting device, not a real charge.

The Alberta Chemistry 30 Data Booklet provides the rules for assigning oxidation numbers, summarized here. Rule 1: Any pure element has an oxidation number of zero. For example,

\text{O}_2

,

\text{Fe}

, and

\text{Cl}_2

all have oxidation numbers of 0. Rule 2: For a monatomic ion, the oxidation number equals the ion's charge.

\text{Na}^+

has an oxidation number of

+1

;

\text{S}^{2-}

has

-2

. Rule 3: In a neutral compound, all oxidation numbers must sum to zero. Rule 4: In a polyatomic ion, the oxidation numbers must sum to the ion's charge. Rule 5: Specific elements have fixed oxidation numbers in almost all compounds — fluorine is always

-1

; oxygen is almost always

-2

(except in peroxides where it is

-1

); hydrogen bonded to nonmetals is almost always

+1

(except in metal hydrides where it is

-1

); Group 1 metals are always

+1

; Group 2 metals are always

+2

.

To find an unknown oxidation number, set up an algebraic equation using Rule 3 or 4. For example, in sulfuric acid

\text{H}_2\text{SO}_4

: hydrogen is

+1

(two H atoms contribute

+2

total), oxygen is

-2

(four O atoms contribute

-8

total). The oxidation number of sulfur is found by solving

2(+1) + x + 4(-2) = 0

, giving

x = +6

. So sulfur has an oxidation number of

+6

in

\text{H}_2\text{SO}_4

.

A change in oxidation number tells you that a redox reaction has occurred. If an atom's oxidation number increases, that atom has been oxidized. If it decreases, that atom has been reduced. An increase in oxidation number means electrons were lost; a decrease means electrons were gained.

2(+1) + x + 4(-2) = 0 \implies x = +6

Oxidation number is a bookkeeping integer showing electron ownership relative to the neutral element.
Pure elements: oxidation number = 0.
Monatomic ions: oxidation number = ion charge.
Sum of oxidation numbers = 0 for neutral compounds; = ion charge for polyatomic ions.
Oxidation number increases → oxidized (lost electrons); decreases → reduced (gained electrons).

Half-Reactions

A half-reaction is an equation that shows only the oxidation step or only the reduction step of a redox reaction, written separately. Half-reactions are useful because they make electron transfer visible and they allow you to balance complex redox equations systematically.

Every half-reaction must be balanced for both atoms and charge. Electrons are written explicitly as products (oxidation half-reaction) or reactants (reduction half-reaction). Consider the iron–copper reaction again. The oxidation half-reaction shows iron losing electrons:

\text{Fe}(s) \rightarrow \text{Fe}^{2+}(aq) + 2e^-

. The reduction half-reaction shows copper(II) gaining electrons:

\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)

. In each half-reaction, count atoms on both sides (balanced) and count charge on both sides (also balanced). On the left of the oxidation half-reaction, charge = 0. On the right, charge =

(+2) + 2(-1) = 0

. Balanced.

To obtain the overall ionic equation, add both half-reactions together and cancel species that appear on both sides. The electrons must cancel exactly: the number of electrons lost in the oxidation half-reaction must equal the number of electrons gained in the reduction half-reaction. If they do not match, multiply one or both half-reactions by a whole number until the electron counts are equal before adding.

Half-reactions are also the format used in the Alberta Chemistry 30 Data Booklet Standard Reduction Potentials table, so recognising and writing them fluently is essential for the rest of the electrochemistry unit.

\text{Fe}(s) \rightarrow \text{Fe}^{2+}(aq) + 2e^-

A half-reaction shows only oxidation or only reduction, with electrons written explicitly.
Balance atoms first, then balance charge by adding electrons to the more positive side.
Electrons appear as products in oxidation half-reactions and as reactants in reduction half-reactions.
When combining half-reactions, electrons must cancel completely — multiply if needed.
Check: atoms balanced AND charge balanced on both sides of every half-reaction.

Disproportionation

Most redox reactions involve two different substances — one is the oxidizing agent and the other is the reducing agent. Disproportionation is a special case where a single substance acts as both the oxidizing agent and the reducing agent at the same time. One portion of the substance is oxidized while another portion of the same substance is reduced.

A classic example is the decomposition of hydrogen peroxide,

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

. In pure water, hydrogen peroxide slowly breaks down:

2\text{H}_2\text{O}_2(aq) \rightarrow 2\text{H}_2\text{O}(l) + \text{O}_2(g)

. Check the oxidation numbers: in

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

, oxygen has an oxidation number of

-1

(hydrogen is

+1

, and two H plus two O must sum to 0, so each O =

-1

). In water, oxygen is

-2

. In

\text{O}_2

, oxygen is

0

. So oxygen goes from

-1

to

-2

in water (reduced) and from

-1

to

0

in oxygen gas (oxidized). The same element, in the same starting compound, simultaneously increases and decreases its oxidation number. That is disproportionation.

Disproportionation reactions can be split into two half-reactions just like any other redox reaction. The oxidation half-reaction for hydrogen peroxide disproportionation is

\text{H}_2\text{O}_2(aq) \rightarrow \text{O}_2(g) + 2\text{H}^+(aq) + 2e^-

, and the reduction half-reaction is

\text{H}_2\text{O}_2(aq) + 2\text{H}^+(aq) + 2e^- \rightarrow 2\text{H}_2\text{O}(l)

. Adding these and cancelling

2\text{H}^+

and

2e^-

from both sides reproduces the balanced overall equation above.

The key diagnostic question for disproportionation is: does one element start at a single oxidation number and end at two different oxidation numbers in the products? If yes, you have a disproportionation reaction.

2\text{H}_2\text{O}_2(aq) \rightarrow 2\text{H}_2\text{O}(l) + \text{O}_2(g)

Disproportionation: one substance is simultaneously the oxidizing agent and the reducing agent.
The same element in the same starting substance is both oxidized (oxidation number increases) and reduced (oxidation number decreases).
Disproportionation reactions can still be split into two half-reactions.
Diagnostic check: one element moves from one oxidation number to two different oxidation numbers in the products.

Summary of Oxidation Number Assignment Rules (Alberta Chemistry 30)

Rule	Condition	Oxidation Number	Example
1	Pure element (any form)	$0$	$\text{Fe}(s)$ , $\text{Cl}_2(g)$ , $\text{O}_2(g)$
2	Monatomic ion	= ion charge	$\text{Ca}^{2+}$ is $+2$ ; $\text{Cl}^-$ is $-1$
3	Fluorine in compounds	$-1$ always	$\text{HF}$ , $\text{CF}_4$
4	Oxygen in most compounds	$-2$ (except peroxides: $-1$ )	$\text{H}_2\text{O}$ is $-2$ ; $\text{H}_2\text{O}_2$ is $-1$
5	Hydrogen bonded to nonmetal	$+1$ (metal hydrides: $-1$ )	$\text{HCl}$ is $+1$ ; $\text{NaH}$ is $-1$
6	Neutral compound (sum)	$= 0$	$\text{H}_2\text{SO}_4$ : $2(+1)+x+4(-2)=0$
7	Polyatomic ion (sum)	= ion charge	$\text{SO}_4^{2-}$ : sum $= -2$

Worked example

Analysing a Redox Reaction: Permanganate Oxidising Bromide

In acidic solution, permanganate ion reacts with bromide ion according to the following skeleton equation:

\text{MnO}_4^-(aq) + \text{Br}^-(aq) \rightarrow \text{Mn}^{2+}(aq) + \text{Br}_2(l)

(unbalanced, acidic solution). (a) Assign oxidation numbers to Mn and Br on each side. (b) Identify the oxidizing agent and the reducing agent. (c) Write a balanced half-reaction for each process. (d) State whether this is a disproportionation reaction and justify your answer.

Step 1 — Assign oxidation numbers to Mn
In MnO₄⁻, oxygen is −2 and the ion has a charge of −1. Set up the equation: oxidation number of Mn plus four times (−2) equals −1. Solve for Mn.
$x + 4(-2) = -1 \implies x = +7$
Step 2 — Assign oxidation number to Mn in product
Mn²⁺ is a monatomic ion. Its oxidation number equals its charge.
Mn in Mn^{2+} = +2
Step 3 — Assign oxidation numbers to Br
Br⁻ is a monatomic ion; its oxidation number equals its charge. In Br₂, bromine is a pure element.
Br in Br^- = -1 ; Br in Br_2 = 0
Step 4 — Identify oxidation and reduction
Mn decreases from +7 to +2: oxidation number decreases, so Mn is reduced. Br increases from −1 to 0: oxidation number increases, so Br is oxidized. Apply OIL RIG to confirm.
$\text{Mn: } +7 \rightarrow +2 \text{ (decrease, reduction)} \quad ; \quad \text{Br: } -1 \rightarrow 0 \text{ (increase, oxidation)}$
Step 5 — Identify the agents
The oxidizing agent is the substance that gets reduced: MnO₄⁻. The reducing agent is the substance that gets oxidized: Br⁻.
$\text{Oxidizing agent: } \text{MnO}_4^- \quad ; \quad \text{Reducing agent: } \text{Br}^-$
Step 6 — Write and balance the reduction half-reaction
Mn goes from +7 to +2, a gain of 5 electrons. Balance Mn atoms (already 1 each side), then balance O by adding 4 H₂O to the right, then balance H by adding 8 H⁺ to the left, then add 5 electrons to the left to balance charge. Left charge: (−1) + 8(+1) + 5(−1) = +2. Right charge: +2. Balanced.
$\text{MnO}_4^-(aq) + 8\text{H}^+(aq) + 5e^- \rightarrow \text{Mn}^{2+}(aq) + 4\text{H}_2\text{O}(l)$
Step 7 — Write and balance the oxidation half-reaction
Each Br goes from −1 to 0, losing 1 electron. Two Br⁻ produce one Br₂, releasing 2 electrons. Balance Br atoms (2 each side), then check charge. Left: 2(−1) = −2. Right: 0 + 2(−1) = −2. Balanced.
$2\text{Br}^-(aq) \rightarrow \text{Br}_2(l) + 2e^-$
Step 8 — Check for disproportionation
In this reaction, Mn is reduced and Br is oxidized. These are two different elements in two different substances. Therefore, neither substance acts as both oxidizing and reducing agent simultaneously.
$\text{Two different substances} \Rightarrow \text{not a disproportionation reaction}$

Answer: (a) Mn: +7 (in MnO₄⁻) → +2 (in Mn²⁺); Br: −1 (in Br⁻) → 0 (in Br₂). (b) Oxidizing agent: MnO₄⁻; Reducing agent: Br⁻. (c) Reduction:

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

; Oxidation:

2\text{Br}^-(aq) \rightarrow \text{Br}_2(l) + 2e^-

. (d) Not a disproportionation reaction — two different substances act as the two agents.

Check: Charge check on reduction half-reaction: left =

(−1) + 8(+1) + 5(−1) = −1 + 8 − 5 = +2

; right =

+2 + 0 = +2

. ✓ Charge check on oxidation half-reaction: left =

2(−1) = −2

; right =

0 + 2(−1) = −2

. ✓ Atom counts balance in both half-reactions. ✓

Common mistakes and how to avoid them

Confusing which agent is oxidized and which is reduced. Students write 'the oxidizing agent is oxidized' — this is backwards.

Correction: The oxidizing agent is reduced (it gains electrons). The reducing agent is oxidized (it loses electrons). Connect with OIL RIG and practise saying it aloud: 'oxidizing agent gets reduced.'

Forgetting to balance charge when writing half-reactions, only balancing atoms.

Correction: Always count total charge on each side after balancing atoms. Add electrons to the more positive side until both sides have equal total charge.

Applying the oxygen oxidation number rule of −2 to peroxides, giving an incorrect oxidation number.

Correction: In peroxides (compounds containing the O−O bond, such as H₂O₂), oxygen is −1, not −2. Check whether the compound is a peroxide before applying the −2 rule.

Misidentifying disproportionation: labelling any reaction where one substance reacts with itself as disproportionation.

Correction: Disproportionation requires one element to change to two different oxidation numbers in the products. Check oxidation numbers in reactants and both products — both an increase and a decrease must occur for the same element.

Adding half-reactions without first equalising the number of electrons, so electrons do not fully cancel.

Correction: Before adding, multiply one or both half-reactions by whole numbers so that the electron count is identical in both. Only then add and cancel.

Lesson summary

An oxidizing agent accepts electrons from another substance and is itself reduced during the reaction. A reducing agent donates electrons to another substance and is itself oxidized during the reaction.
Oxidation numbers are integer bookkeeping values that track electron ownership. They are assigned using a fixed set of rules from the Alberta Chemistry 30 Data Booklet. An increase in oxidation number signals oxidation; a decrease signals reduction.
A half-reaction shows only the oxidation step or only the reduction step of a redox reaction. Electrons are written explicitly. Both atoms and charge must be balanced in every half-reaction.
Disproportionation is a special redox process where a single substance acts as both oxidizing and reducing agent at the same time. One portion of the substance is oxidized while another portion is reduced, so one element starts at one oxidation number and ends at two different oxidation numbers.
OIL RIG (Oxidation Is Loss, Reduction Is Gain) is the core memory tool. Apply it first, then use oxidation numbers to confirm which atoms changed and by how much.
Always verify half-reactions by checking that both atom counts and total charges are equal on the left and right sides before combining them into an overall equation.

Check your understanding

Question 1

In the reaction

\text{Zn}(s) + 2\text{HCl}(aq) \rightarrow \text{ZnCl}_2(aq) + \text{H}_2(g)

, which species is the oxidizing agent?

$\text{Zn}(s)$ , because it loses electrons
$\text{H}^+(aq)$ , because it gains electrons and is reduced
$\text{Cl}^-(aq)$ , because it forms a compound with zinc
$\text{ZnCl}_2(aq)$ , because it is a product of the reaction

Show answer and explanation

\text{H}^+(aq)

, because it gains electrons and is reduced

Zinc goes from an oxidation number of 0 to +2 (oxidized; it is the reducing agent). Hydrogen goes from +1 in HCl to 0 in H₂ (reduced; it gains electrons). The species that is reduced is the oxidizing agent, so H⁺ is the oxidizing agent. Chloride does not change oxidation number and is not involved in electron transfer.

Question 2

What is the oxidation number of nitrogen in the nitrate ion,

\text{NO}_3^-

?

$-3$
$0$
$+3$
$+5$

Show answer and explanation

+5

Oxygen is −2 in most compounds. The ion has a charge of −1. Setting up the equation:

x + 3(-2) = -1

, so

x - 6 = -1

, giving

x = +5

. Nitrogen is +5 in NO₃⁻.

Question 3

Which of the following is the correctly balanced oxidation half-reaction for the conversion of

\text{Fe}^{2+}(aq)

to

\text{Fe}^{3+}(aq)

?

$\text{Fe}^{2+}(aq) + e^- \rightarrow \text{Fe}^{3+}(aq)$
$\text{Fe}^{2+}(aq) \rightarrow \text{Fe}^{3+}(aq) + e^-$
$\text{Fe}^{2+}(aq) \rightarrow \text{Fe}^{3+}(aq) - e^-$
$\text{Fe}^{3+}(aq) + e^- \rightarrow \text{Fe}^{2+}(aq)$

Show answer and explanation

\text{Fe}^{2+}(aq) \rightarrow \text{Fe}^{3+}(aq) + e^-

Fe²⁺ loses one electron to become Fe³⁺ — this is oxidation. In an oxidation half-reaction, electrons are written as products (on the right). Option B is correct: left charge = +2, right charge = (+3) + (−1) = +2. Balanced. Option A adds an electron as a reactant, which would be reduction. Option D is the reduction half-reaction (going from +3 back to +2).

Question 4

Chlorine gas reacts with water:

\text{Cl}_2(g) + \text{H}_2\text{O}(l) \rightarrow \text{HCl}(aq) + \text{HOCl}(aq)

. The oxidation number of Cl in Cl₂ is 0. In HCl it is −1 and in HOCl it is +1. What type of reaction is this?

A precipitation reaction, because two products form from one reactant
A disproportionation reaction, because Cl goes from one oxidation number to two different oxidation numbers
A synthesis reaction, because Cl₂ and H₂O combine to form new substances
A simple oxidation reaction, because Cl₂ loses electrons overall

Show answer and explanation

A disproportionation reaction, because Cl goes from one oxidation number to two different oxidation numbers

Chlorine starts at oxidation number 0 in Cl₂. It ends at −1 in HCl (reduced) and +1 in HOCl (oxidized). One element starts at one oxidation number and finishes at two different oxidation numbers in the products — the defining feature of disproportionation. Cl₂ is both the oxidizing agent and the reducing agent.

Key terms

Oxidation: The loss of electrons by a substance in a chemical reaction, shown by an increase in oxidation number.
Reduction: The gain of electrons by a substance in a chemical reaction, shown by a decrease in oxidation number.
Oxidizing agent: The substance in a redox reaction that accepts electrons from another substance. The oxidizing agent is itself reduced during the reaction.
Reducing agent: The substance in a redox reaction that donates electrons to another substance. The reducing agent is itself oxidized during the reaction.
Oxidation number: An integer assigned to an atom in a compound or ion that represents the number of electrons the atom has effectively gained or lost compared with its neutral elemental form. It is a bookkeeping tool, not a real measured charge.
Half-reaction: An equation that shows only the oxidation step or only the reduction step of a redox reaction, with electrons written explicitly as either a product (oxidation) or a reactant (reduction).
Disproportionation: A type of redox reaction in which a single substance acts as both the oxidizing agent and the reducing agent. One portion of the substance is oxidized while another portion is reduced, so one element ends up at two different oxidation numbers in the products.
OIL RIG: A memory phrase for redox reactions: Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons).

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30-B1.2k · Define redox agents, oxidation numbers, half-reactions, and disproportionation

Alberta Grade 12 Chemistry

Alberta Chemistry 30 — Outcome 30-B1.2k

What you will learn

Prerequisite Review: Ions, Charges, and Electron Transfer

Oxidizing Agents and Reducing Agents

Oxidation Numbers

Half-Reactions

Disproportionation

Summary of Oxidation Number Assignment Rules (Alberta Chemistry 30)

Analysing a Redox Reaction: Permanganate Oxidising Bromide

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Question 4

Key terms

Continue through CHEM30

About this lesson and its review