Redox half-equations

CIE A-Level Chemistry· Unit 6: Redox reactions and electrolysis· 15 min read

1. Introduction to half-equations★★☆☆☆⏱ 4 min

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.

📘 Definition

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: $ ext{Zn} ightarrow ext{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
1. 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.

2. Balancing half-equations in acidic conditions★★★☆☆⏱ 5 min

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

Balance all non-oxygen, non-hydrogen atoms first
Balance oxygen atoms by adding $e x t H_{2} O$ to the side that needs oxygen
Balance hydrogen atoms by adding $e x t H^{+}$ to the side that needs hydrogen
Balance total charge by adding electrons to the more positive side

📐 Worked Example

Balance the half-equation for the reduction of dichromate(VI) ( $Cr_{2} O_{7}^{2 -}$ ) to chromium(III) ( $Cr^{3 +}$ ) 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
$14H^+ + Cr_2O_7^{2-} + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$

3. Balancing half-equations in alkaline conditions★★★★☆⏱ 4 min

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

Complete all 4 steps for acidic balancing
Add $e x t O H^{-}$ ions to both sides equal to the number of $e x t H^{+}$ ions
Combine $e x t H^{+}$ + $e x t O H^{-}$ on the same side to form $e x t H_{2} O$
Cancel any excess $e x t H_{2} O$ molecules on both sides

📐 Worked Example

Balance the oxidation of sulfite ( $SO_{3}^{2 -}$ ) to sulfate ( $SO_{4}^{2 -}$ ) 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 $2 H^{+} + 2 O H^{-} = 2 H_{2} O$ , then simplify by canceling 1 $H_{2} O$ from both sides:
6
$2OH^- + SO_3^{2-} \rightarrow SO_4^{2-} + H_2O + 2e^-$

4. Combining half-equations to form full redox equations★★★☆☆⏱ 4 min

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

5. Common Pitfalls

Wrong move:

Writing electrons on the wrong side for oxidation/reduction

Why:

Confusion between electron loss and gain leads to wrong charge balance and lost marks

Correct move:

Oxidation = electrons out (right side, product), Reduction = electrons in (left side, reactant)

Wrong move:

Only multiplying electrons when scaling half-equations

Why:

Scaling only electrons leaves atoms and charge unbalanced, leading to a wrong full equation

Correct move:

Multiply every species in the half-equation by the scaling factor

Wrong move:

Balancing oxygen with $e x t H^{+}$ directly in acidic conditions

Why:

Skipping the step of balancing oxygen with water breaks the balancing order and leads to wrong atom counts

Correct move:

Follow the order: non-H/O atoms → O with H $_{2}$ O → H with H $^{+}$ → charge with electrons

Wrong move:

Leaving excess water in alkaline half-equations

Why:

Uncanceled excess water means the equation is not fully simplified, which loses an accuracy mark in CIE

Correct move:

After forming water from $e x t H^{+}$ and $e x t O H^{-}$ , cancel equal numbers of water molecules from both sides

6. Quick Reference Cheatsheet

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

7. Frequently Asked

Is balancing different for acidic vs alkaline conditions?

Yes. For acidic, use $e x t H^{+}$ to balance H and $e x t H_{2} e x t O$ to balance O. For alkaline, balance first as acidic, then add $e x t O H^{-}$ to neutralize $e x t H^{+}$ into $e x t H_{2} e x t O$ before simplifying.

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.

2022 · 12
Balance acidic half-equation
2023 · 22
Write alkaline half-equation
2024 · 11
Combine two half-equations

Going deeper

syllabusCIE 9701 A-Level Chemistry SyllabusVerify unit 6 learning outcomes

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.