# Redox half-equations

> CIE A-Level Chemistry · 9701
> Source: https://www.owlsprep.com/study/cie-9701-u6-redox-half-equations/

This module teaches you to construct, balance and combine redox half-equations, the core building blocks of all redox reactions. You will master balancing in both acidic and alkaline conditions, a required skill for almost all redox exam questions.

**Prerequisites:** [Oxidation number rules](https://www.owlsprep.com/study/cie-9701-u6-oxidation-numbers/); Basic ionic equation balancing

## Learning objectives

- Distinguish between oxidation and reduction half-equations
- Balance half-equations in acidic and alkaline conditions
- Combine half-equations to form full balanced redox equations
- Correctly place electrons in half-equations per exam conventions

## Introduction to half-equations

Any full redox reaction can be split into two separate half-equations: one for oxidation (loss of electrons) and one for reduction (gain of electrons). Half-equations isolate each process, making it much easier to balance complex redox reactions.

**Half-equation** — A balanced ionic equation that describes only one half of a redox reaction, with electrons explicitly shown to track charge transfer.

*Example:* Oxidation of zinc: $\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-$

CIE follows the universal convention: electrons are written on the product (right) side for oxidation (electrons lost) and on the reactant (left) side for reduction (electrons gained). Marks are always awarded for correct placement of electrons.

**Worked example:** Write the balanced half-equation for the reduction of silver(I) ions to solid silver.

1. 1. Write the unbalanced equation for reactant and product:
2. $$Ag^+ \rightarrow Ag$$
3. 2. Balance charge: silver goes from +1 to 0, so gains 1 electron. Add 1 $e^-$ to the left (reactant) side:
4. $$Ag^+ + e^- \rightarrow Ag$$

> **Exam tip:** Always check that total charge on the left equals total charge on the right after adding electrons.

## Balancing half-equations in acidic conditions

Most half-equation questions in CIE exams are for acidic conditions. Use this systematic 4-step method to guarantee a correct balanced equation:

1. Balance all non-oxygen, non-hydrogen atoms first
2. Balance oxygen atoms by adding $\text{H}_2\text{O}$ to the side that needs oxygen
3. Balance hydrogen atoms by adding $\text{H}^+$ to the side that needs hydrogen
4. Balance total charge by adding electrons to the more positive side

**Worked example:** Balance the half-equation for the reduction of dichromate(VI) ($\text{Cr}_2\text{O}_7^{2-}$) to chromium(III) ($\text{Cr}^{3+}$) in acidic solution.

1. Step 1: Balance chromium atoms: 2 Cr on left, so 2 Cr on right:
2. $$Cr_2O_7^{2-} \rightarrow 2Cr^{3+}$$
3. Step 2: Balance oxygen: 7 O on left, add 7 H$_2$O to the right:
4. $$Cr_2O_7^{2-} \rightarrow 2Cr^{3+} + 7H_2O$$
5. Step 3: Balance hydrogen: 14 H on right, add 14 H$^+$ to the left:
6. $$14H^+ + Cr_2O_7^{2-} \rightarrow 2Cr^{3+} + 7H_2O$$
7. Step 4: Balance charge: Left total charge = +12, right = +6. Add 6 e$^-$ to the left:
8. $$14H^+ + Cr_2O_7^{2-} + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$$

## Balancing half-equations in alkaline conditions

Balancing in alkaline conditions adds one extra step to the acidic method to convert $\text{H}^+$ to $\text{OH}^-$, this method is less error-prone than starting with $\text{OH}^-$ directly.

1. Complete all 4 steps for acidic balancing
2. Add $\text{OH}^-$ ions to both sides equal to the number of $\text{H}^+$ ions
3. Combine $\text{H}^+$ + $\text{OH}^-$ on the same side to form $\text{H}_2\text{O}$
4. Cancel any excess $\text{H}_2\text{O}$ molecules on both sides

**Worked example:** Balance the oxidation of sulfite ($\text{SO}_3^{2-}$) to sulfate ($\text{SO}_4^{2-}$) in alkaline solution.

1. After step 4 (acidic balance), we have:
2. $$H_2O + SO_3^{2-} \rightarrow SO_4^{2-} + 2H^+ + 2e^-$$
3. Add 2 OH$^-$ to both sides:
4. $$2OH^- + H_2O + SO_3^{2-} \rightarrow SO_4^{2-} + 2H^+ + 2OH^- + 2e^-$$
5. Combine $2H^+ + 2OH^- = 2H_2O$, then simplify by canceling 1 $H_2O$ from both sides:
6. $$2OH^- + SO_3^{2-} \rightarrow SO_4^{2-} + H_2O + 2e^-$$

> **tip**
>
> Always simplify fully: extra water or $\text{OH}^-$ left over will cost you a mark.

## Combining half-equations to form full redox equations

To get a full balanced redox equation, you combine one oxidation and one reduction half-equation. The total number of electrons lost in oxidation must equal the total number gained in reduction, so you scale half-equations by an integer if needed before adding.

**Worked example:** Combine the dichromate(VI) reduction half-equation from earlier with the oxidation of Fe$^{2+}$ to Fe$^{3+}$ to form a full acidic redox equation.

1. 1. Write the two balanced half-equations:
2. $$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$$
3. $$Fe^{2+} \rightarrow Fe^{3+} + e^-$$
4. 2. Equalize electrons: multiply the oxidation half-equation by 6:
5. $$6Fe^{2+} \rightarrow 6Fe^{3+} + 6e^-$$
6. 3. Add the equations and cancel the 6 electrons from both sides:
7. $$Cr_2O_7^{2-} + 6Fe^{2+} + 14H^+ \rightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O$$

## Common pitfalls

- **Wrong:** Writing electrons on the wrong side for oxidation/reduction
  - Why it fails: Confusion between electron loss and gain leads to wrong charge balance and lost marks
  - Correct: Oxidation = electrons out (right side, product), Reduction = electrons in (left side, reactant)
- **Wrong:** Only multiplying electrons when scaling half-equations
  - Why it fails: Scaling only electrons leaves atoms and charge unbalanced, leading to a wrong full equation
  - Correct: Multiply *every species* in the half-equation by the scaling factor
- **Wrong:** Balancing oxygen with $\text{H}^+$ directly in acidic conditions
  - Why it fails: Skipping the step of balancing oxygen with water breaks the balancing order and leads to wrong atom counts
  - Correct: Follow the order: non-H/O atoms → O with H$_2$O → H with H$^+$ → charge with electrons
- **Wrong:** Leaving excess water in alkaline half-equations
  - Why it fails: Uncanceled excess water means the equation is not fully simplified, which loses an accuracy mark in CIE
  - Correct: After forming water from $\text{H}^+$ and $\text{OH}^-$, cancel equal numbers of water molecules from both sides

## Cheatsheet

| Step | Acidic Conditions | Alkaline Conditions |
| --- | --- | --- |
| 1. Balance non-O/H | Balance atoms on both sides | Same as acidic |
| 2. Balance O atoms | Add $\text{H}_2\text{O}$ to deficit side | Same as acidic |
| 3. Balance H atoms | Add $\text{H}^+$ to deficit side | Same as acidic |
| 4. Balance charge | Add $e^-$ to more positive side | Same as acidic |
| 5. Convert to alkaline | N/A | Add equal $\text{OH}^-$ to both sides, combine $\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}$, cancel excess $\text{H}_2\text{O}$ |
| Combine half-equations | Equalize electrons, add, cancel electrons | Same as acidic |

## What's next

Redox half-equations are the foundation for all further redox topics in CIE A-Level Chemistry, including calculating electrode potentials, writing equations for electrolytic processes, and solving redox titration calculations. Almost every exam question on redox will require you to construct or use balanced half-equations at some step, so mastering this skill now makes all subsequent redox topics much easier. Next, you can explore full redox equation balancing, deepen your understanding of oxidation numbers, or move on to electrode potentials and electrochemical cells.

- [Chemical equilibria (AS)](https://www.owlsprep.com/study/cie-9701-u7-overview/)
- [Dynamic equilibrium characteristics](https://www.owlsprep.com/study/cie-9701-u7-dynamic-equilibrium-characteristics/)
- [Le Chatelier's principle](https://www.owlsprep.com/study/cie-9701-u7-le-chatelier-s-principle/)

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