# Oxidation numbers and redox reactions

> Chemistry · CIE A-Level
> Source: https://www.owlsprep.com/study/cie-9701-u6-oxidation-numbers-and-redox-reactions/

This core module covers rules for assigning oxidation numbers, identifying redox processes, distinguishing oxidising/reducing agents, and balancing full redox equations from half-equations for CIE 9701 A-Level Chemistry.

**Prerequisites:** [Basic chemical equation balancing and ionic bonding](https://www.owlsprep.com/study/cie-9701-u2-ionic-compounds/)

## Learning objectives

- Assign oxidation numbers to atoms in any compound or ion
- Distinguish oxidation and reduction using oxidation number changes
- Identify oxidising and reducing agents in redox reactions
- Write balanced half-equations for acidic/alkaline conditions
- Combine half-equations to form balanced full redox equations

## Rules for Assigning Oxidation Numbers

**Oxidation Number** — The hypothetical charge an atom would have if all bonds in the compound/ion were fully ionic (also called oxidation state)

*Example:* The oxidation number of Na in NaCl is +1, Cl is -1

- An element in its elemental state has an oxidation number of 0
- A monatomic ion has an oxidation number equal to its charge
- Fluorine always has oxidation number -1 in all compounds
- Oxygen is usually -2, except *peroxides* (-1) and when bonded to F (positive)
- Hydrogen is usually +1, except *metal hydrides* (-1)
- Sum of oxidation numbers = 0 for neutral compounds, equals overall charge for ions

**Worked example:** Assign oxidation numbers to all elements in (a) $\text{H}_2\text{SO}_4$ (b) $\text{Cr}_2\text{O}_7^{2-}$

1. For (a) $\text{H}_2\text{SO}_4$: Let oxidation number of S = $x$. Use rules: H = +1, O = -2. Sum to 0 for neutral compound:
2. $$2(+1) + x + 4(-2) = 0 \\ 2 + x - 8 = 0 \\ x = +6$$
3. Result: H = +1, S = +6, O = -2. For (b) $\text{Cr}_2\text{O}_7^{2-}$: Let Cr = $x$. Sum equals overall charge -2:
4. $$2x + 7(-2) = -2 \\ 2x - 14 = -2 \\ 2x = 12 \\ x = +6$$
5. Result: Cr = +6, O = -2

> **Exam tip:** Always confirm if the species is a neutral compound or a charged ion when calculating the sum of oxidation numbers

## Oxidation, Reduction and Redox Agents

Oxidation is an *increase* in oxidation number, corresponding to loss of electrons. Reduction is a *decrease* in oxidation number, corresponding to gain of electrons. An oxidising agent (oxidant) is reduced and oxidises another species, while a reducing agent (reductant) is oxidised and reduces another species.

> **Memory Hook**
>
> OIL RIG: Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons)

**Worked example:** Identify which element is oxidised, which is reduced, and name the oxidising/reducing agents in the reaction: $\text{Zn}(s) + \text{CuSO}_4(aq) \rightarrow \text{ZnSO}_4(aq) + \text{Cu}(s)$

1. Assign oxidation numbers to all species:
2. Zn(s): 0, Cu$^{2+}$ (CuSO$_4$): +2, Zn$^{2+}$ (ZnSO$_4$): +2, Cu(s): 0
3. Compare changes: Zn goes from 0 → +2 (increase, so oxidised). Cu goes from +2 → 0 (decrease, so reduced).
4. Conclusion: Zn is the reducing agent (it is oxidised, reduces Cu$^{2+}$), Cu$^{2+}$ from CuSO$_4$ is the oxidising agent (it is reduced, oxidises Zn).

## Writing Balanced Half-Equations

Half-equations separate oxidation and reduction processes, showing electron transfer. They are used to build full balanced redox equations, and require balancing for both atoms and charge. The method differs slightly for acidic vs alkaline conditions.

**Worked example:** Write a balanced half-equation for the reduction of $\text{MnO}_4^-$ to $\text{Mn}^{2+}$ in acidic solution

1. 1. Write unbalanced equation for the species:
2. $$MnO_4^- \rightarrow Mn^{2+}$$
3. 2. Balance Mn: already balanced (1 Mn each side). Balance O by adding H$_2$O to the O-deficient side:
4. $$MnO_4^- \rightarrow Mn^{2+} + 4H_2O$$
5. 3. Balance H by adding H$^+$ (for acidic conditions) to the H-deficient side:
6. $$8H^+ + MnO_4^- \rightarrow Mn^{2+} + 4H_2O$$
7. 4. Balance charge by adding electrons to the more positive side. Left charge = +8 -1 = +7, right charge = +2. Add 5e$^-$ to left:
8. $$5e^- + 8H^+ + MnO_4^- \rightarrow Mn^{2+} + 4H_2O$$
9. 5. Check: all atoms balanced, total charge on both sides = +2, so half-equation is correct

> **tip**
>
> For alkaline conditions: after balancing with H+, add an equal number of OH$^-$ to both sides, then combine H$^+$ + OH$^-$ into H$_2$O and simplify

*Calculator:* forbidden

## Combining Half-Equations for Full Redox Equations

To form a full balanced redox equation, the total number of electrons lost in oxidation must equal the total number gained in reduction. Adjust coefficients to equalise electrons, then add the half-equations and cancel common species.

**Worked example:** Combine the oxidation half-equation $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$ with the MnO$_4^-$ reduction half-equation from the previous example to form a full balanced equation

1. 1. Oxidation loses 1 e-, reduction gains 5 e-. Multiply oxidation half-equation by 5 to equalise electrons:
2. $$5Fe^{2+} \rightarrow 5Fe^{3+} + 5e^-$$
3. 2. Add the two half-equations together:
4. $$5Fe^{2+} + 5e^- + 8H^+ + MnO_4^- \rightarrow 5Fe^{3+} + 5e^- + Mn^{2+} + 4H_2O$$
5. 3. Cancel the 5 electrons on both sides, no other common species to cancel:
6. $$5Fe^{2+} + 8H^+ + MnO_4^- \rightarrow 5Fe^{3+} + Mn^{2+} + 4H_2O$$
7. 4. Check: Atoms are balanced, total charge left = +10 +8 -1 = +17, right = +15 +2 = +17, so equation is balanced

**Check your understanding**

Test your understanding:

1. What is the coefficient of $\text{I}_2$ when the reaction between $\text{I}^-$ and $\text{Cr}_2\text{O}_7^{2-}$ (forming $\text{I}_2$ and $\text{Cr}^{3+}$ in acidic solution) is fully balanced?

   - A) 1
   - B) 3
   - C) 6
   - D) 2

   *Why:* Oxidation half: $2I^- \rightarrow I_2 + 2e^-$, reduction half: $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$. Multiply oxidation by 3 to get 6 electrons lost, so 3 $I_2$ is produced.

*Calculator:* forbidden

## Common pitfalls

- **Wrong:** Assuming oxygen is always -2, even in peroxides
  - Why it fails: Common exception to the rule that is frequently tested in multiple choice questions
  - Correct: Check for peroxide (O-O) bonds or oxygen bonded to fluorine, and use -1 for oxygen in peroxides
- **Wrong:** Using 0 as the sum of oxidation numbers for polyatomic ions
  - Why it fails: Students forget to account for the overall charge of the ion, leading to incorrect oxidation number calculations
  - Correct: Sum of oxidation numbers always equals the overall charge of the species, only 0 for neutral compounds
- **Wrong:** Labelling the oxidised species as the oxidising agent
  - Why it fails: Common confusion between the process (oxidation/reduction) and the role of the species (agent)
  - Correct: Remember: Oxidising agents get reduced, reducing agents get oxidised
- **Wrong:** Adding half-equations without equalising the number of electrons first
  - Why it fails: Skipping this step leads to unbalanced charge in the final full equation
  - Correct: Always cross-multiply half-equations so total electrons lost = total electrons gained before adding
- **Wrong:** Leaving H+ in the final equation for reactions in alkaline solution
  - Why it fails: H+ does not exist in high concentration in alkaline conditions, so the equation is incorrect
  - Correct: Neutralise all H+ by adding equal OH- to both sides, then combine into water and simplify

## Cheatsheet

| Rule / Term | Key Value / Meaning |
| --- | --- |
| Elemental state | Oxidation number = 0 |
| Monatomic ion | ON = ion charge |
| F in all compounds | ON = -1 |
| O (not peroxide/F) | ON = -2 |
| O in peroxides | ON = -1 |
| H (not metal hydrides) | ON = +1 |
| H in metal hydrides | ON = -1 |
| Sum ON (neutral) | = 0 |
| Sum ON (ion) | = overall ion charge |
| Oxidation | Increase in ON, lose e⁻ |
| Reduction | Decrease in ON, gain e⁻ |
| Oxidising agent | Gets reduced, gains e⁻ |
| Reducing agent | Gets oxidised, loses e⁻ |

## What's next

Mastering oxidation numbers and redox balancing is the foundation for all further electrochemistry topics in CIE A-Level Chemistry. You will apply these core skills to calculate standard cell potentials, predict reaction feasibility, and solve quantitative problems involving electrolytic cells and Faraday's laws of electrolysis. Redox concepts also appear regularly in organic chemistry, for example to identify oxidation and reduction processes in reactions of alcohols and carbonyl compounds. Building on this topic will prepare you for both multiple choice and structured extended response questions that make up a large proportion of exam marks.

- [Electrolysis Principles and Applications](https://www.owlsprep.com/study/cie-9701-u6-electrolysis-principles-and-applications/)
- [Redox half-equations](https://www.owlsprep.com/study/cie-9701-u6-redox-half-equations/)
- [Chemical equilibria (AS)](https://www.owlsprep.com/study/cie-9701-u7-overview/)

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