Redox reactions and electrochemistry basics

IB Chemistry HL· R3: Mechanisms of chemical change > Unit 6: Redox processes· 20 min read

1. Core Definitions of Oxidation and Reduction★★☆☆☆⏱ 5 min

Originally, oxidation described reaction with oxygen, and reduction described removal of oxygen from a metal ore. Modern definitions extend this concept to all electron transfer processes, regardless of whether oxygen is present.

📘 Definition

Oxidation

A process that results in the loss of electrons from a species, or an increase in the oxidation number of an atom.

Example:

$N a \to N a^{+} + e^{-}$ : sodium is oxidised

📘 Definition

Reduction

A process that results in the gain of electrons by a species, or a decrease in the oxidation number of an atom.

Example:

$C l_{2} + 2 e^{-} \to 2 C l^{-}$ : chlorine is reduced

📐 Worked Example

Identify which species is oxidised and which is reduced in the reaction: $Z n (s) + C u S O_{4} (a q) \to Z n S O_{4} (a q) + C u (s)$

1
Write the net ionic equation by removing the spectator sulfate ion:
2
$Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$
3
Track electron change: Zinc loses 2 electrons to form a zinc ion:
4
$Zn \rightarrow Zn^{2+} + 2e^-$
5
Copper(II) ion gains 2 electrons to form solid copper:
6
$Cu^{2+} + 2e^- \rightarrow Cu(s)$
7
Conclusion: Zinc lost electrons so it is oxidised. Copper ion gained electrons so it is reduced.

2. Oxidation Numbers: Rules and Assignment★★☆☆☆⏱ 7 min

Oxidation number (also called oxidation state) is a formal number assigned to an atom to track its degree of oxidation or reduction. It follows a consistent set of rules that work for all compounds and ions.

📘 Definition

Oxidising and Reducing Agents

An oxidising agent causes oxidation of another species by accepting electrons, so it is itself reduced. A reducing agent causes reduction of another species by donating electrons, so it is itself oxidised.

Example:

In the zinc-copper reaction above, $C u^{2 +}$ is the oxidising agent, and $Z n$ is the reducing agent.

The oxidation number of an uncombined element is always 0
The oxidation number of a monatomic ion equals its ionic charge
Oxygen has an oxidation number of -2 in most compounds (-1 in peroxides, positive in compounds with fluorine)
Hydrogen has +1 when bonded to non-metals, -1 when bonded to metals (hydrides)
Sum of oxidation numbers in a neutral compound = 0; sum in a polyatomic ion = ion charge

📐 Worked Example

Calculate the oxidation number of sulfur in the sulfate ion $S O_{4}^{2 -}$

1
Apply the rules: Oxygen has oxidation number -2, and the sum of oxidation numbers equals the ion charge (-2).
2
Let $x$ = oxidation number of S. Set up the balance equation:
3
$x + 4(-2) = -2$
4
Solve for $x$ :
5
$x = -2 + 8 = +6$
6
Conclusion: Sulfur has an oxidation number of +6 in sulfate ions.

✓ Quick check

Test your understanding of oxidation number rules:

What is the oxidation number of manganese in $M n O_{4}^{-}$ ?
- +4
- +7
- -1
- +2
Reveal answer
B —
Correct. $x + 4 (- 2) = - 1 \to x = + 7$

3. Balancing Redox Reactions with Oxidation Numbers★★★☆☆⏱ 8 min

Oxidation number change can be used to quickly balance simple redox equations. The total increase in oxidation number from oxidation must equal the total decrease from reduction, because electrons lost equal electrons gained.

Assign oxidation numbers to all atoms to identify changing oxidation states
Calculate the change in oxidation number per atom for both oxidation and reduction
Find the simplest whole number ratio that makes total increase equal to total decrease
Add coefficients to balance the redox changes, then balance remaining atoms and charge

📐 Worked Example

Balance the reaction: $C u + H N O_{3} \to C u (N O_{3})_{2} + N O + H_{2} O$

1
Assign oxidation numbers: Cu goes from 0 → +2 (change = +2 per Cu). N goes from +5 → +2 (change = -3 per N).
2
Find the least common multiple of 2 and 3 = 6. Total change must be +6 and -6, so 3 Cu atoms and 2 N atoms are changed.
3
Add coefficients: 3 Cu → 3 $C u (N O_{3})_{2}$ , 2 N reduced → 2 NO. Total N = 3(2) + 2 = 8 → 8 $H N O_{3}$ .
4
Balance H: 8 H from $H N O_{3}$ → 4 $H_{2} O$ .
5
Check oxygen: Left = 8×3 = 24, Right = 3×6 + 2×1 + 4×1 = 24. Balanced.
6
$3Cu + 8HNO_3 \rightarrow 3Cu(NO_3)_2 + 2NO + 4H_2O$

4. Common Pitfalls

Wrong move:

Assuming oxygen always has an oxidation number of -2

Why:

This ignores common exceptions like peroxides (O = -1) and oxygen fluorides (O has positive oxidation number)

Correct move:

Always check if the compound is a peroxide or contains fluorine before assigning an oxidation number to oxygen

Wrong move:

Confusing an oxidising agent with the species that is oxidised

Why:

Oxidising agents cause oxidation of other species, so they accept electrons and are themselves reduced

Correct move:

Memorise: oxidising agent = reduced, reducing agent = oxidised

Wrong move:

Treating oxidation number as actual charge for covalent compounds

Why:

Oxidation number is a formal accounting tool, not a measurement of actual charge in covalent species

Correct move:

Follow the oxidation number rules regardless of bonding type to get the correct value

Wrong move:

Forgetting to balance total oxidation number change, only balancing per atom

Why:

Unmatched total change means electrons lost do not equal electrons gained, leading to an unbalanced equation

Correct move:

Always match total increase in oxidation number to total decrease before balancing other atoms

5. Quick Reference Cheatsheet

Rule	Oxidation Number / Outcome
Uncombined element	0
Monatomic ion	Equal to ion charge
Oxygen (most compounds)	-2
Oxygen (peroxides)	-1
Hydrogen (bonded to non-metals)	+1
Hydrogen (bonded to metals)	-1
Sum (neutral compound)	0
Sum (polyatomic ion)	Equal to ion charge
Oxidation	Loss of e⁻, ON increase
Reduction	Gain of e⁻, ON decrease

6. Frequently Asked

Is oxidation number the same as ionic charge?

No. Oxidation number is a formal accounting tool for electron transfer, even for covalent compounds where atoms do not have full ionic charge. Only for monatomic ions does oxidation number equal the actual ionic charge.

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2025 · Paper 1
Assign oxidation number to Mn
2024 · Paper 2
Identify reducing agent in reaction
2023 · Paper 1
Balance simple redox reaction

Going deeper

referenceIB Chemistry Data BookletUse for standard electrode potential values in later topics

What's Next

This sub-topic is the foundation for all advanced electrochemistry topics in IB HL Chemistry. Mastery of oxidation numbers and redox definitions is required to write half-equations for cells, calculate cell potentials, solve redox titration problems, and understand electrolysis. Redox reactions also appear in organic chemistry, where they are used to synthesise key functional groups like alcohols and carboxylic acids. Build on this foundation with the following topics: