Redox Reactions: Oxidation, Reduction, and Everything You Need to Know
Why Your Phone Battery, Rusty Iron, and a Cut Apple All Share One Idea
Think about three everyday things: a freshly cut apple slowly browning on your kitchen counter, an iron railing turning orange-red after weeks of rain, and a smartphone battery quietly powering your screen without any fire or explosion.
These look completely unrelated. But they are all driven by exactly the same chemical process — the movement of electrons from one substance to another.
That process is called a redox reaction. And once you understand it, you will recognise it everywhere — in batteries, medicines, metabolism, metal extraction, water treatment, and combustion.
This article explains redox reactions from the ground up: what they are, how to identify them, how to find the oxidising and reducing agents, and what disproportionation means. By the end, you will be able to analyse any redox reaction with confidence.
Quick-Check Answers
What is redox reaction?
A redox reaction is a chemical reaction in which electrons are transferred from one substance to another.
Oxidation = loss of electrons → oxidation number increases
Reduction = gain of electrons → oxidation number decreases
Oxidation and reduction always happen at the same time — you cannot have one without the other.
The oxidising agent accepts electrons and is itself reduced.
The reducing agent donates electrons and is itself oxidised.
Memory Aid
OIL: Oxidation Is Loss,
RIG: Reduction Is Gain.
What Is an Oxidation Number?
Before you can identify oxidation and reduction in a reaction, you need a way to track what happens to electrons. That tracking tool is the oxidation number.
The oxidation number (also called the oxidation state) of an atom is the charge that atom would carry if all the bonds in its compound were completely ionic — that is, if the more electronegative atom in every bond were awarded full ownership of the shared electrons.
This is a bookkeeping device, not a physical measurement. In covalent compounds such as water (H₂O) or carbon dioxide (CO₂), electrons are shared rather than transferred, but we still assign oxidation numbers to keep track of electron ownership.
Remember: Oxidation numbers are written as signed integers with the sign before the number: +2, −1, 0, and so on. Do not confuse this with ionic charge, where the sign comes after the number , e.g. Ca2+ (2+ for a magnesium ion, not +2).
The Seven Rules for Assigning Oxidation Numbers
Apply these rules in order. Earlier rules take priority over later ones.
| Rule | Rule Name | Statement | Example |
| 1 | Pure elements | Every atom in a pure element has oxidation number = 0 | Zn(s), Cl₂(g), S₈(s): each atom = 0 |
| 2 | Monatomic ions | Oxidation number = ionic charge | Na⁺ = +1; Fe³⁺ = +3; Cl⁻ = −1 |
| 3 | Fluorine | Always −1 in all compounds | HF: F = −1; OF₂: F = −1 |
| 4 | Oxygen | −2 in almost all compounds. Exceptions: peroxides (−1); compounds with F (+2) | H₂O: O = −2; H₂O₂: O = −1; OF₂: O = +2 |
| 5 | Hydrogen | + 1 bonded to non-metals; −1 bonded to metals (metal hydrides) | HCl: H = +1; NaH: H = −1 |
| 6 | Neutral molecules | Sum of all oxidation numbers = 0 | H₂O: 2(+1) + (−2) = 0 ✓ |
| 7 | Polyatomic ions | Sum of all oxidation numbers = ion charge | SO₄²⁻: oxidation numbers sum to −2 |
Worked Example 1 — Assigning Oxidation Numbers to Sulfur
Worked Example 1 — Sulfur in Three Different Species
Problem:
Find the oxidation number of sulfur in (a) H2SO4 (b) SO32- (c) S8
Strategy: Apply the rules in order. Use Rule 6 for neutral molecules, Rule 7 for polyatomic ions, and Rule 1 for pure elements.
Part (a) — H2SO4:
Given values: H = +1 (Rule 5); O = −2 (Rule 4). Let x = oxidation number of S. Applying Rule 6: 2(+1) + x + 4(−2) = 0
Solving: 2 + x − 8 = 0 → x = +6
Part (b) — SO32-:
Given values: O = −2 (Rule 4). Let x = oxidation number of S. Applying Rule 7: x + 3(−2) = −2
Solving: x − 6 = −2 → x = +4
Part (c) — S8:
Sulfur is in its pure elemental form. Apply Rule 1: oxidation number = 0.
Final Answers:
(a) S in H2SO4 = +6 (b) S in SO32- = +4 (c) S in S8 = 0
Exam Tip
In examinations, always write the algebraic equation first (e.g. x + 3(−2) = −2), then solve for x. Stating only the final answer without the algebraic working will lose method marks.
Worked Example 2 — Oxidation Numbers in Complex Ions
Worked Example 2 — Cr, N, and Mn in Polyatomic Ions
Problem: Determine the oxidation number of (a) Cr in Cr₂O₇²⁻ (b) N in NO₃⁻ (c) Mn in MnO₄⁻
Part (a) — Cr in Cr₂O₇²⁻:
O = −2; two Cr atoms present. Applying Rule 7: 2x + 7(−2) = −2
2x − 14 = −2 → 2x = 12 → x = +6
Part (b) — N in NO₃⁻:
O = −2. Applying Rule 7: x + 3(−2) = −1
x − 6 = −1 → x = +5
Part (c) — Mn in MnO₄⁻: O = −2. Applying Rule 7: x + 4(−2) = −1
x − 8 = −1 → x = +7
Final Answers:
(a) Cr = +6 (b) N = +5 (c) Mn = +7
Did You Know?
Manganese can exist in oxidation states from −3 to +7 — one of the widest ranges of any element. That is why permanganate (MnO₄⁻, Mn = +7) is such a powerful oxidising agent: it can drop all the way down to Mn²⁺ (+2) or even MnO₂ (+4), accepting electrons at each step.
Oxidation, Reduction, and Redox Reactions
With oxidation numbers in hand, you can now define oxidation and reduction precisely. There are two ways to think about them: in terms of electron transfer, and in terms of changes in oxidation number. Both describe the same event.
Oxidation: Losing Electrons
A substance is oxidised when it loses electrons. When electrons leave, the oxidation number of that atom increases. Consider the conversion of iron(II) ions to iron(III) ions:
Fe2+ (aq) → Fe3+ (aq) + e⁻
Iron has lost one electron. Its oxidation number has increased from +2 to +3. Iron has been oxidised.
Notice something important: oxygen is not involved here. Historically, chemists named this process ‘oxidation’ because many early reactions involved oxygen. But the modern definition is purely about electron loss — oxygen is not required.
Reduction: Gaining Electrons
A substance is reduced when it gains electrons. When electrons arrive, the oxidation number of that atom decreases. Consider the deposition of copper metal from copper(II) ions:
Cu 2+(aq) + 2e⁻ → Cu(s)
Copper has gained two electrons. Its oxidation number has decreased from +2 to 0. Copper(II) has been reduced to copper metal.
Redox Reactions: Oxidation and Reduction Together
Oxidation and reduction never happen in isolation. Whenever one substance loses electrons, those electrons must go somewhere — another substance must gain them. A reaction in which both oxidation and reduction occur simultaneously is called a redox reaction (short for reduction–oxidation reaction).
A classic example is zinc metal reacting with copper(II) sulfate solution:
Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s)
In this reaction, the oxidation number of zinc (Zn) goes from 0 to +2 because zinc loses 2 electrons. This is why zinc is oxidised. Copper(II) (Cu²⁺) gains two electrons, so its oxidation number goes from +2 to 0. In this reaction, copper is reduced.
Since one species is oxidised while another is simultaneously reduced, this is a redox reaction. The number of electrons lost by zinc equals the number of electrons gained by copper — electrons are conserved.
Real-Life Application
This exact reaction — zinc and copper(II) sulfate — forms the basis of the Daniell cell, one of the earliest practical batteries. In a Daniell cell, the electron transfer happens through an external wire, producing an electric current you can use. Your smartphone battery uses a more sophisticated version of the same principle.