Distinguish redox reactions using half-reactions and oxidation numbers

Alberta Grade 12 Chemistry

Electrochemical Changes

Alberta Chemistry 30 — Unit B1: Oxidation–Reduction

Drop a clean iron nail into a blue copper(II) sulfate solution and watch what happens over a few minutes: the blue colour fades, a reddish-brown solid coats the nail, and the nail itself slowly dissolves. No heat is applied, yet matter is visibly rearranging. Something is being transferred between the iron and the copper ions. That something is electrons. Reactions driven by electron transfer are called oxidation–reduction reactions, or redox reactions for short. In this lesson you will learn two connected tools — oxidation numbers and half-reactions — that let you identify, describe, and distinguish redox reactions from reactions where no electron transfer occurs. Both tools are essential throughout Chemistry 30 and appear regularly on the Alberta Chemistry 30 diploma examination.

What you will learn

Assign oxidation numbers to atoms in elements, compounds, and ions using established rules.
Identify oxidation and reduction by tracking changes in oxidation number.
Write a net ionic redox equation as two separate half-reactions, one for oxidation and one for reduction.
Distinguish redox reactions from non-redox reactions by determining whether electron transfer occurs.

Prerequisite Review: Ions, Bonding, and Net Ionic Equations

Before studying redox, recall three ideas from Chemistry 20. First, atoms can lose or gain electrons to become ions. A sodium atom loses one electron to form

\text{Na}^+

; a chlorine atom gains one electron to form

\text{Cl}^-

. Second, in a covalent bond, electrons are shared between atoms, but not always equally — the more electronegative atom draws the shared electrons closer to itself. Third, a net ionic equation shows only the species that actually change during a reaction; spectator ions that appear unchanged on both sides are left out.

These three ideas underpin oxidation numbers and half-reactions. You do not need to calculate formal bond energies or use any equation beyond basic atom counting and charge balancing.

Atoms can lose electrons (forming positive ions) or gain electrons (forming negative ions).
Covalent bonds involve shared electrons; the more electronegative atom in a bond is treated as though it owns those electrons.
A net ionic equation omits spectator ions and shows only the species that change.

Oxidation Numbers: A Bookkeeping System for Electrons

An oxidation number (also called oxidation state) is a whole-number value assigned to each atom in a substance. It represents how many electrons that atom has effectively gained or lost compared with its neutral elemental form. Think of oxidation numbers as an electron-accounting system: they do not always describe a real charge, but they let you track where electrons move during a reaction.

Apply the following rules in order when assigning oxidation numbers. Rule 1: A pure element has an oxidation number of 0. For example,

\text{Fe}\,\text{(s)}

,

\text{O}_{2(g)}

, and

\text{Cl}_{2(g)}

are all 0. Rule 2: For a monatomic ion, the oxidation number equals the ion charge.

\text{Fe}^{2+}

is

+2

;

\text{Cl}^-

is

-1

. Rule 3: Oxygen is almost always

-2

(except in peroxides, where it is

-1

, or when bonded to fluorine). Rule 4: Hydrogen is

+1

when bonded to a non-metal, and

-1

when bonded to a metal (metal hydrides). Rule 5: The sum of all oxidation numbers in a neutral compound equals 0. The sum in a polyatomic ion equals the ion's charge.

Use Rule 5 to find an unknown oxidation number by algebra. For example, in

\text{MnO}_4^-

: let the oxidation number of Mn be

x

. Oxygen is

-2

and there are four oxygen atoms, so

x + 4(-2) = -1

. Solving gives

x = +7

. Manganese is

+7

in permanganate ion.

\sum (oxidation numbers) = charge of species

Pure elements are always assigned oxidation number 0.
Monatomic ion oxidation number equals the ion charge.
Oxygen is $-2$ in most compounds; hydrogen is $+1$ bonded to non-metals.
Oxidation numbers in a neutral compound sum to 0; in a polyatomic ion they sum to the ion charge.
Use algebra with Rule 5 to find an unknown oxidation number.

Identifying Redox: Oxidation and Reduction

Once you assign oxidation numbers to every atom, look for changes as you go from reactants to products. If an atom's oxidation number increases, that atom has lost electrons — it has been oxidized. A helpful memory aid: OIL — Oxidation Is Loss (of electrons). If an atom's oxidation number decreases, that atom has gained electrons — it has been reduced. The paired memory aid: RIG — Reduction Is Gain (of electrons). OIL RIG together reminds you that these two processes always occur together.

Consider the iron–copper reaction from the introduction. Iron starts as

\text{Fe}\,\text{(s)}

(oxidation number 0) and ends as

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

(oxidation number

+2

). The oxidation number increased by 2, so iron was oxidized and lost 2 electrons. Copper starts as

\text{Cu}^{2+}\,\text{(aq)}

(oxidation number

+2

) and ends as

\text{Cu}\,\text{(s)}

(oxidation number 0). The oxidation number decreased by 2, so copper was reduced and gained 2 electrons. Electron transfer occurred, so this is a redox reaction.

A non-redox reaction shows no change in any oxidation number. For example, when sodium hydroxide neutralizes hydrochloric acid —

\text{NaOH}\,\text{(aq)} + \text{HCl}\,\text{(aq)} \rightarrow \text{NaCl}\,\text{(aq)} + \text{H}_2\text{O}\,\text{(l)}

— every atom keeps the same oxidation number throughout. No electrons are transferred, so this is not a redox reaction.

The substance that is oxidized is called the reducing agent because it provides the electrons that allow another substance to be reduced. The substance that is reduced is called the oxidizing agent because it accepts those electrons. In the iron–copper example, iron is the reducing agent and copper(II) ion is the oxidizing agent.

\Delta \text{Ox. No.} > 0 \Rightarrow \text{oxidized}; \quad \Delta \text{Ox. No.} < 0 \Rightarrow \text{reduced}

An increase in oxidation number means the atom is oxidized (loses electrons).
A decrease in oxidation number means the atom is reduced (gains electrons).
OIL RIG: Oxidation Is Loss, Reduction Is Gain.
The reducing agent is oxidized; the oxidizing agent is reduced.
If no oxidation numbers change, the reaction is not a redox reaction.

Half-Reactions: Separating the Electron Transfer

A half-reaction is an equation that shows either the oxidation step or the reduction step on its own, with electrons written explicitly as a reactant or product. Writing half-reactions lets you see exactly how many electrons are gained or lost and verify that the overall electron transfer balances.

To write a half-reaction, follow these steps. Step 1: Write the species being oxidized (or reduced) on the left and its product on the right. Step 2: Balance atoms other than oxygen and hydrogen first. Step 3: In acidic or aqueous solution, balance oxygen atoms by adding

\text{H}_2\text{O}

molecules, then balance hydrogen atoms by adding

\text{H}^+

ions. Step 4: Balance the charge by adding electrons (

e^-

) to whichever side needs them. In an oxidation half-reaction, electrons appear as products (released). In a reduction half-reaction, electrons appear as reactants (consumed).

For the iron–copper system, the two half-reactions are straightforward because no oxygen or hydrogen is involved. Oxidation:

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

. Reduction:

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

. Both atoms and charges are balanced in each half-reaction. Adding the two half-reactions and cancelling the electrons gives the net ionic equation:

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

.

When the number of electrons in each half-reaction is not equal, multiply one or both half-reactions by whole-number coefficients before adding. The electrons must cancel completely. For instance, if one half-reaction releases 2 electrons and the other consumes 6, multiply the first by 3 so both sides show 6 electrons. Never add the electrons from both sides — always cancel them.

\text{Oxidation: species} \rightarrow \text{product} + ne^-; \quad \text{Reduction: species} + ne^- \rightarrow \text{product}

An oxidation half-reaction shows electrons as products; a reduction half-reaction shows electrons as reactants.
Balance atoms, then charges, adding $e^-$ last.
Multiply half-reactions by integers so the number of electrons released equals the number consumed.
Add the balanced half-reactions and cancel electrons to get the net ionic equation.
The electrons must disappear completely when the half-reactions are combined.

Putting It Together: Distinguishing Redox from Non-Redox

To decide whether a reaction is a redox reaction, assign oxidation numbers to every atom in the equation, then check whether any oxidation number changes. If at least one atom is oxidized and at least one atom is reduced, the reaction is a redox reaction. You can then write the two half-reactions to describe it fully. If no oxidation numbers change, the reaction involves no electron transfer and is not a redox reaction — it belongs to a different category such as acid–base, precipitation, or simple dissociation.

Consider the reaction between zinc and dilute sulfuric acid:

\text{Zn}\,\text{(s)} + \text{H}_2\text{SO}_{4(aq)} \rightarrow \text{ZnSO}_{4(aq)} + \text{H}_{2(g)}

. Assign oxidation numbers: Zn goes from 0 to

+2

(oxidized); H goes from

+1

to 0 (reduced); S and O do not change. Electrons are transferred, so this is a redox reaction. The reducing agent is Zn and the oxidizing agent is

\text{H}^+

.

Now compare with a precipitation reaction:

\text{AgNO}_{3(aq)} + \text{NaCl}\,\text{(aq)} \rightarrow \text{AgCl}\,\text{(s)} + \text{NaNO}_{3(aq)}

. Assign oxidation numbers: Ag is

+1

in

\text{AgNO}_3

and

+1

in

\text{AgCl}

; Cl is

-1

in

\text{NaCl}

and

-1

in

\text{AgCl}

; Na, N, and O are unchanged. No oxidation number changes at all. This is not a redox reaction — it is a double displacement (precipitation) reaction with no electron transfer.

Assign oxidation numbers to every atom before deciding whether a reaction is redox.
A reaction is a redox reaction if and only if at least one atom is oxidized and at least one is reduced.
Precipitation, neutralization, and dissociation reactions are often non-redox.
Half-reactions can only be written for redox reactions where electron transfer occurs.

Oxidation Number Rules: Quick Reference

Rule	Assignment	Example
Pure element	0	$\text{Fe}\,\text{(s)}: 0$ ; $\text{Cl}_{2(g)}: 0$
Monatomic ion	= ion charge	$\text{Cu}^{2+}: +2$ ; $\text{S}^{2-}: -2$
Oxygen (most compounds)	$-2$	$\text{H}_2\text{O}: \text{O} = -2$
Hydrogen (bonded to non-metal)	$+1$	$\text{HCl}: \text{H} = +1$
Sum rule (neutral compound)	$\sum = 0$	$\text{H}_2\text{SO}_4$ : $2(+1)+x+4(-2)=0 \Rightarrow x=+6$
Sum rule (polyatomic ion)	$\sum =$ ion charge	$\text{MnO}_4^-$ : $x+4(-2)=-1 \Rightarrow x=+7$

Worked example

Identifying Redox and Writing Half-Reactions for a Permanganate Reaction

In acidic solution, permanganate ion oxidizes iron(II) ion to iron(III) ion, while permanganate is itself reduced to manganese(II) ion. The unbalanced net ionic equation is:

\text{MnO}_4^-\,\text{(aq)} + \text{Fe}^{2+}\,\text{(aq)} \rightarrow \text{Mn}^{2+}\,\text{(aq)} + \text{Fe}^{3+}\,\text{(aq)}

(oxygen and hydrogen not yet balanced). Confirm this is a redox reaction by assigning oxidation numbers, identify the oxidizing and reducing agents, and write balanced half-reactions. Then combine them into a balanced net ionic equation.

Step 1 — Assign oxidation numbers to Mn and Fe
In $\text{MnO}_4^-$ , oxygen is $-2$ and the ion charge is $-1$ . Let Mn be $x$ : $x + 4(-2) = -1$ , so $x = +7$ . In $\text{Mn}^{2+}$ , the oxidation number is $+2$ . Iron starts as $\text{Fe}^{2+}$ ( $+2$ ) and ends as $\text{Fe}^{3+}$ ( $+3$ ).
$x + 4(-2) = -1 \Rightarrow x = +7$
Step 2 — Identify oxidation and reduction
Mn decreases from $+7$ to $+2$ : a decrease of 5, so Mn is reduced (gains 5 electrons). Fe increases from $+2$ to $+3$ : an increase of 1, so Fe is oxidized (loses 1 electron). This confirms a redox reaction. $\text{MnO}_4^-$ is the oxidizing agent; $\text{Fe}^{2+}$ is the reducing agent.
\Delta Ox. No._{Mn} = +7 \to +2 = -5; \Delta Ox. No._{Fe} = +2 \to +3 = +1
Step 3 — Write the oxidation half-reaction
Iron(II) loses one electron to become iron(III). No oxygen or hydrogen is involved, so atoms are already balanced. Balance charge by placing one electron on the right (products side).
$\text{Fe}^{2+}\,\text{(aq)} \rightarrow \text{Fe}^{3+}\,\text{(aq)} + e^-$
Step 4 — Write the reduction half-reaction
Permanganate is reduced to $\text{Mn}^{2+}$ . First balance Mn (already 1 on each side). Next, balance the 4 oxygen atoms by adding 4 water molecules to the right. Then balance the 8 hydrogen atoms that appeared by adding 8 $\text{H}^+$ to the left. Finally, check the charge: left side is $(-1 + 8) = +7$ ; right side is $+2$ . Add 5 electrons to the left to make both sides $+2$ .
$\text{MnO}_4^-\,\text{(aq)} + 8\text{H}^+\,\text{(aq)} + 5e^- \rightarrow \text{Mn}^{2+}\,\text{(aq)} + 4\text{H}_2\text{O}\,\text{(l)}$
Step 5 — Equalise electrons transferred
The oxidation half-reaction releases 1 electron; the reduction half-reaction consumes 5 electrons. Multiply the oxidation half-reaction by 5 so both half-reactions involve exactly 5 electrons.
$5 \times [\text{Fe}^{2+}\,\text{(aq)} \rightarrow \text{Fe}^{3+}\,\text{(aq)} + e^-] \Rightarrow 5\text{Fe}^{2+}\,\text{(aq)} \rightarrow 5\text{Fe}^{3+}\,\text{(aq)} + 5e^-$
Step 6 — Add half-reactions and cancel electrons
Add the two half-reactions. The 5 electrons appear on both sides and cancel completely, giving the balanced net ionic equation.
$\text{MnO}_4^-\,\text{(aq)} + 8\text{H}^+\,\text{(aq)} + 5\text{Fe}^{2+}\,\text{(aq)} \rightarrow \text{Mn}^{2+}\,\text{(aq)} + 5\text{Fe}^{3+}\,\text{(aq)} + 4\text{H}_2\text{O}\,\text{(l)}$

Answer: The reaction is confirmed as a redox reaction. The oxidizing agent is

\text{MnO}_4^-

and the reducing agent is

\text{Fe}^{2+}

. The balanced net ionic equation is

\text{MnO}_4^-\,\text{(aq)} + 8\text{H}^+\,\text{(aq)} + 5\text{Fe}^{2+}\,\text{(aq)} \rightarrow \text{Mn}^{2+}\,\text{(aq)} + 5\text{Fe}^{3+}\,\text{(aq)} + 4\text{H}_2\text{O}\,\text{(l)}

.

Check: Atom check — Mn: 1 = 1 ✓; O: 4 = 4 ✓; H: 8 = 8 ✓; Fe: 5 = 5 ✓. Charge check — Left:

(-1) + (+8) + 5(+2) = +17

; Right:

(+2) + 5(+3) + 0 = +17

✓. Electrons cancelled completely ✓.

Common mistakes and how to avoid them

Forgetting to multiply half-reactions before adding them, leaving electrons that do not cancel completely.

Correction: Always check that the number of electrons in the oxidation half-reaction equals the number in the reduction half-reaction before adding. Multiply by the smallest whole-number factor needed.

Placing electrons on the wrong side of a half-reaction — for example, writing electrons as reactants in the oxidation step.

Correction: In oxidation, electrons are lost and appear as products (right side). In reduction, electrons are gained and appear as reactants (left side). Use OIL RIG as your check.

Forgetting to balance oxygen and hydrogen atoms in half-reactions that involve oxygen-containing species, leaving unbalanced equations.

Correction: In acidic aqueous solution, balance O atoms by adding

\text{H}_2\text{O}

, then balance the H atoms introduced by adding

\text{H}^+

. Only then balance charge with electrons.

Calling a reaction redox because it looks like it involves a metal, without actually checking oxidation number changes.

Correction: Always assign oxidation numbers to every atom and compare reactant values to product values. If no oxidation number changes, the reaction is not redox, regardless of what type of substances are present.

Confusing the oxidizing agent with the species that is oxidized.

Correction: The oxidizing agent is the substance that causes oxidation in another species — it does so by accepting electrons, which means the oxidizing agent itself is reduced. The reducing agent is oxidized.

Lesson summary

Oxidation numbers are assigned using a fixed set of rules; the sum rule (neutral compound sums to 0; ion sums to ion charge) is used to find unknown values by algebra.
A reaction is a redox reaction when at least one atom's oxidation number increases (oxidation) and at least one atom's oxidation number decreases (reduction); if no oxidation numbers change, no electron transfer occurs and the reaction is not redox.
OIL RIG: Oxidation Is Loss of electrons; Reduction Is Gain of electrons. These two processes always occur together.
The oxidizing agent accepts electrons and is itself reduced; the reducing agent donates electrons and is itself oxidized.
Half-reactions isolate either the oxidation or the reduction step and show electrons explicitly; they must be multiplied so that the electrons cancel completely when the two half-reactions are added.
A balanced net ionic equation is obtained by adding the two balanced half-reactions after ensuring the electron count is equal on both sides.

Check your understanding

Question 1

In the reaction

2\text{Mg}\,\text{(s)} + \text{O}_{2(g)} \rightarrow 2\text{MgO}\,\text{(s)}

, what is the oxidation number of magnesium in the product

\text{MgO}

?

0
+1
+2
+4

Show answer and explanation

+2

In

\text{MgO}

, oxygen is

-2

. The compound is neutral, so

x + (-2) = 0

, giving

x = +2

. Magnesium is

+2

in

\text{MgO}

. Since Mg started at 0 (pure element) and is now

+2

, it was oxidized.

Question 2

Which statement correctly describes the reducing agent in a redox reaction?

It gains electrons and is oxidized.
It gains electrons and is reduced.
It loses electrons and is oxidized.
It loses electrons and is reduced.

Show answer and explanation

It loses electrons and is oxidized.

The reducing agent donates (loses) electrons to another species, enabling that species to be reduced. Because it loses electrons, the reducing agent itself is oxidized — its oxidation number increases.

Question 3

A student writes the following half-reaction for the reduction of dichromate ion to chromium(III) in acidic solution:

\text{Cr}_2\text{O}_7^{2-}\,\text{(aq)} + 14\text{H}^+\,\text{(aq)} + 6e^- \rightarrow 2\text{Cr}^{3+}\,\text{(aq)} + 7\text{H}_2\text{O}\,\text{(l)}

. What is the oxidation number of chromium in

\text{Cr}_2\text{O}_7^{2-}

?

+3
+4
+6
+7

Show answer and explanation

+6

Let the oxidation number of Cr be

x

. Oxygen is

-2

and there are 7 oxygen atoms. The ion charge is

-2

:

2x + 7(-2) = -2 \Rightarrow 2x = +12 \Rightarrow x = +6

. Chromium is

+6

in dichromate, and it is reduced to

+3

in

\text{Cr}^{3+}

.

Question 4

Which of the following reactions is NOT a redox reaction?

$\text{Zn}\,\text{(s)} + 2\text{HCl}\,\text{(aq)} \rightarrow \text{ZnCl}_{2(aq)} + \text{H}_{2(g)}$
$\text{Fe}\,\text{(s)} + \text{CuSO}_{4(aq)} \rightarrow \text{FeSO}_{4(aq)} + \text{Cu}\,\text{(s)}$
$\text{NaOH}\,\text{(aq)} + \text{HCl}\,\text{(aq)} \rightarrow \text{NaCl}\,\text{(aq)} + \text{H}_2\text{O}\,\text{(l)}$
$2\text{H}_2\text{O}_{2(aq)} \rightarrow 2\text{H}_2\text{O}\,\text{(l)} + \text{O}_{2(g)}$

Show answer and explanation

\text{NaOH}\,\text{(aq)} + \text{HCl}\,\text{(aq)} \rightarrow \text{NaCl}\,\text{(aq)} + \text{H}_2\text{O}\,\text{(l)}

In the neutralization reaction

\text{NaOH}\,\text{(aq)} + \text{HCl}\,\text{(aq)} \rightarrow \text{NaCl}\,\text{(aq)} + \text{H}_2\text{O}\,\text{(l)}

, every atom keeps the same oxidation number throughout (Na:

+1

; O:

-2

; H:

+1

; Cl:

-1

). No electrons are transferred, so this is not a redox reaction. All other options involve at least one change in oxidation number.

Key terms

Oxidation number (oxidation state): A whole-number value assigned to an atom that represents the effective number of electrons it has lost or gained compared with its neutral elemental form. Used to track electron transfer in reactions.
Oxidation: A process in which an atom's oxidation number increases, indicating a loss of electrons.
Reduction: A process in which an atom's oxidation number decreases, indicating a gain of electrons.
Redox reaction: A reaction in which electrons are transferred from one species to another; both oxidation and reduction occur simultaneously.
Half-reaction: An equation that shows either the oxidation step or the reduction step of a redox reaction, with electrons written explicitly as a reactant or product.
Oxidizing agent: The substance that accepts electrons from another species during a redox reaction. The oxidizing agent is itself reduced.
Reducing agent: The substance that donates electrons to another species during a redox reaction. The reducing agent is itself oxidized.
Net ionic equation: A balanced chemical equation that includes only the species that actually change during a reaction; spectator ions are omitted.

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30-B1.3k · Distinguish redox reactions using half-reactions and oxidation numbers

Alberta Grade 12 Chemistry

Alberta Chemistry 30 — Unit B1: Oxidation–Reduction

What you will learn

Prerequisite Review: Ions, Bonding, and Net Ionic Equations

Oxidation Numbers: A Bookkeeping System for Electrons

Identifying Redox: Oxidation and Reduction

Half-Reactions: Separating the Electron Transfer

Putting It Together: Distinguishing Redox from Non-Redox

Oxidation Number Rules: Quick Reference

Identifying Redox and Writing Half-Reactions for a Permanganate Reaction

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