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Balancing Redox Reactions: The Half-Equation Method

Learn to balance redox equations using oxidation numbers and the half-equation method. Covers electron transfer, acidic and basic solutions, worked examples, and exam strategies.

V
Vectora Team
STEM Education
9 min read
2026-01-10

What Are Redox Reactions?

Redox stands for Reduction-Oxidation — reactions where electrons are transferred between species. They are among the most important reactions in chemistry, from rusting to respiration to batteries.

Remember with the mnemonic OIL RIG:

  • Oxidation Is Loss (of electrons)
  • Reduction Is Gain (of electrons)

Learning Goals: By the end of this guide, you should be able to:

  1. Assign oxidation numbers to any element in a compound.
  2. Identify which species is oxidised and which is reduced.
  3. Write and balance half-equations.
  4. Combine half-equations into overall balanced redox equations.
  5. Balance equations in acidic and basic solutions.

Oxidation Numbers — The Rules

RuleExample
Elements in their natural state = 0Fe=0Fe = 0, O2=0O_2 = 0
Monoatomic ions = their chargeNa+=+1Na^+ = +1, Cl−=−1Cl^- = -1
Oxygen is usually -2Exception: peroxides (−1-1)
Hydrogen is usually +1Exception: metal hydrides (−1-1)
Fluorine is always -1Most electronegative element
Sum of oxidation numbers = overall chargee.g., in SO42−SO_4^{2-}: S+4(−2)=−2S + 4(-2) = -2, so S=+6S = +6

The Half-Equation Method

Step-by-Step (Acidic Solution)

  1. Assign oxidation numbers to identify what's oxidised and what's reduced.
  2. Write two separate half-equations — one for oxidation, one for reduction.
  3. Balance atoms other than O and H.
  4. Balance O by adding H2OH_2O.
  5. Balance H by adding H+H^+.
  6. Balance charge by adding electrons (e−e^-).
  7. Equalise electrons in both half-equations (multiply if needed).
  8. Add the half-equations and cancel common species.

Example: Fe2+Fe^{2+} and MnO4−MnO_4^- in Acidic Solution

Oxidation half: Fe2+→Fe3++e−Fe^{2+} \rightarrow Fe^{3+} + e^-

Reduction half (balance step by step):

  1. MnO4−→Mn2+MnO_4^- \rightarrow Mn^{2+}
  2. Balance O: MnO4−→Mn2++4H2OMnO_4^- \rightarrow Mn^{2+} + 4H_2O
  3. Balance H: MnO4−+8H+→Mn2++4H2OMnO_4^- + 8H^+ \rightarrow Mn^{2+} + 4H_2O
  4. Balance charge: MnO4−+8H++5e−→Mn2++4H2OMnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O

Combine: Multiply oxidation by 5 to equalise electrons:

5Fe2++MnO4−+8H+→5Fe3++Mn2++4H2O5Fe^{2+} + MnO_4^- + 8H^+ \rightarrow 5Fe^{3+} + Mn^{2+} + 4H_2O

Redox Equation Balancer

Enter any redox reaction and watch the half-equations being balanced step by step. See oxidation numbers change and electrons transfer in real time.
Launch Redox Balancer

Oxidising and Reducing Agents

TermDefinitionWhat Happens to It
Oxidising agentCauses oxidation in another speciesGets reduced itself (gains electrons)
Reducing agentCauses reduction in another speciesGets oxidised itself (loses electrons)

In the example above: MnO4−MnO_4^- is the oxidising agent (it gets reduced), and Fe2+Fe^{2+} is the reducing agent (it gets oxidised).

Worked Examples

Example 1: Assign Oxidation Numbers in Cr2O72−Cr_2O_7^{2-}

2(Cr)+7(−2)=−2⇒2Cr−14=−2⇒Cr=+62(Cr) + 7(-2) = -2 \quad \Rightarrow \quad 2Cr - 14 = -2 \quad \Rightarrow \quad Cr = +6

Chromium is in oxidation state +6.

Example 2: Balance Cu+HNO3→Cu(NO3)2+NO2+H2OCu + HNO_3 \rightarrow Cu(NO_3)_2 + NO_2 + H_2O

Oxidation: Cu→Cu2++2e−Cu \rightarrow Cu^{2+} + 2e^-

Reduction: NO3−+2H++e−→NO2+H2ONO_3^- + 2H^+ + e^- \rightarrow NO_2 + H_2O

Multiply reduction by 2: 2NO3−+4H++2e−→2NO2+2H2O2NO_3^- + 4H^+ + 2e^- \rightarrow 2NO_2 + 2H_2O

Overall: Cu+2NO3−+4H+→Cu2++2NO2+2H2OCu + 2NO_3^- + 4H^+ \rightarrow Cu^{2+} + 2NO_2 + 2H_2O

Full equation: Cu+4HNO3→Cu(NO3)2+2NO2+2H2OCu + 4HNO_3 \rightarrow Cu(NO_3)_2 + 2NO_2 + 2H_2O

Example 3: Identifying Redox in Displacement

Question: Zn+CuSO4→ZnSO4+CuZn + CuSO_4 \rightarrow ZnSO_4 + Cu. Identify what is oxidised and reduced.

Solution:

  • ZnZn: 0→+20 \rightarrow +2 → oxidised (loses 2 electrons, reducing agent)
  • Cu2+Cu^{2+}: +2→0+2 \rightarrow 0 → reduced (gains 2 electrons, oxidising agent)

Common Mistakes

  1. Not checking that electron counts match — Before combining half-equations, the number of electrons lost must equal the number gained. If they don't, multiply one or both half-equations.

  2. Forgetting to balance O and H — In acidic solution: use H2OH_2O for oxygen, H+H^+ for hydrogen. In basic solution: also add OH−OH^- at the end.

  3. Assigning wrong oxidation numbers — Always check oxygen (-2 normally) and hydrogen (+1 normally) before solving for the unknown element.

  4. Mixing up oxidising and reducing agents — The oxidising agent is reduced (gains electrons). This is counterintuitive. Remember: "the oxidising agent gets reduced."

Exam Tips (A-Level / AP / IB)

  • Show all steps explicitly: assign oxidation numbers → write half-equations → balance → combine.
  • When balancing, check your final equation: same number of each element on both sides, same total charge on both sides.
  • On the AP exam, you often need to identify the oxidising and reducing agents AND write the net ionic equation.
  • For colour changes involving transition metals, know the common ones: MnO4−MnO_4^- (purple) → Mn2+Mn^{2+} (colourless), Cr2O72−Cr_2O_7^{2-} (orange) → Cr3+Cr^{3+} (green).

Frequently Asked Questions

What is the difference between oxidation and reduction?

Oxidation is the loss of electrons (increase in oxidation number). Reduction is the gain of electrons (decrease in oxidation number). They always occur together — you cannot have one without the other.

How do you identify a redox reaction?

Check if any element changes its oxidation number. If at least one element is oxidised and one is reduced, it's a redox reaction. If no oxidation numbers change, it's not redox.

Why does the half-equation method work?

It ensures conservation of both mass and charge. By balancing each half separately and then combining, you guarantee that electrons lost equal electrons gained, and all atoms and charges are balanced.