ChemistryISC Class 11 · Redox Reactions

Balancing Redox Equations: Oxidation-Number and Ion-Electron Methods, with Titrations

Updated 17 October 2026 · Fact-checked

In the oxidation-number method, balance the total increase and decrease in oxidation number, then balance the rest by inspection. In the ion-electron method, write and balance the two half-reactions (atoms, then O with H₂O, then H with H⁺, then charge with electrons), equalise the electrons and add. In basic medium, convert H⁺ into H₂O by adding OH⁻.

Follow the order: atoms, then O with H₂O, H with H⁺, charge with e⁻, then equalise electrons; in base add OH⁻.

Illustration drawn for QuestionBuddy.

Understand Balancing Redox Equations and Redox Titrations

Oxidation-number method. (1) Write the skeleton equation. (2) Assign oxidation numbers and find the elements oxidised and reduced. (3) Find the total increase and total decrease and equalise them with suitable coefficients. (4) Balance the remaining atoms: first other atoms, then H and O with H₂O (and H⁺ or OH⁻ in ionic equations). (5) Check atoms and charge.

Ion-electron method (acidic medium). (1) Split into oxidation and reduction half-reactions. (2) Balance each for atoms other than O and H. (3) Balance O by adding H₂O. (4) Balance H by adding H⁺. (5) Balance charge by adding electrons. (6) Multiply to equalise electrons, add, and cancel common species.

Basic medium. Balance first as if acidic, then add as many OH⁻ to both sides as there are H⁺, combine H⁺ + OH⁻ into H₂O, and cancel excess water.

Redox titrations. In permanganate titrations in acid, MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O acts as its own indicator (pink end point). The dichromate half-reaction is Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O. In iodometry, I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻. Use the balanced equation as the mole ratio: moles of titrant × (coefficient of analyte/coefficient of titrant) = moles of analyte.

Common balanced equations. 5Fe²⁺ + MnO₄⁻ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O; 6Fe²⁺ + Cr₂O₇²⁻ + 14H⁺ → 6Fe³⁺ + 2Cr³⁺ + 7H₂O; 2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O.

Balancing Redox Equations and Redox Titrations: formulas and key facts

Permanganate (acid): $M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$; Electrons gained: 5.
Dichromate (acid): $C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 C r^{3 +} + 7 H_{2} O$; Electrons gained: 6.
Iodometry: $I_{2} + 2 S_{2} O_{3}^{2 -} \to 2 I^{-} + S_{4} O_{6}^{2 -}$; 1 mol I₂ needs 2 mol thiosulphate.
Titration: $m o l eso f ana l y t e = m o l eso f t i t r an t \times (ana l y t ecoe f f i c i e n t / t i t r an t coe f f i c i e n t)$; From the balanced equation.

Balancing Redox Equations and Redox Titrations in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Titrating iron in a tablet

Suppose 25.0 mL of an Fe²⁺ solution is titrated with 0.0200 M K₂Cr₂O₇ in acid and needs 20.0 mL.

Write the balanced equation and find the concentration of Fe²⁺.

Try it first, then show the solutionSolution

Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O.
Moles of Cr₂O₇²⁻ = 0.0200 × 0.0200 = 4.00 × 10⁻⁴ mol.
Fe²⁺ = 6 × 4.00 × 10⁻⁴ = 2.40 × 10⁻³ mol in 25.0 mL, so 0.0960 M.

Answer: 0.0960 M

Memory hook: The coefficient ratio 6 : 1 comes from the balanced equation.

Tricks to remember Balancing Redox Equations and Redox Titrations

Short memory aids. Say each one aloud once, then use it on a question the same day.

Atoms, O, H, charge
In the ion-electron method balance in that order.
Use it when: Half-reactions.
Acid first, then OH⁻
For basic medium, balance as for acid and then neutralise H⁺ with OH⁻.
Use it when: Basic-medium equations.

Balancing Redox Equations and Redox Titrations: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Electrons in a half-reaction · 1 mark · MCQBoard level

In acidic medium, the number of electrons gained by one MnO₄⁻ ion to form Mn²⁺ is:
(a) 1
(b) 3
(c) 5
(d) 7

Try it first, then show the solutionSolution

Mn goes from +7 to +2: a gain of 5 electrons.

Answer: (c) 5

Example 2: Oxidation-number method · 3 marks · Short answerBoard level

Balance the equation for zinc and very dilute nitric acid, Zn + HNO₃ → Zn(NO₃)₂ + NH₄NO₃ + H₂O, by the oxidation-number method.

Try it first, then show the solutionSolution

Zn: 0 → +2 (increase of 2); N: +5 → −3 in NH₄⁺ (decrease of 8). (1 mark)
Equalise: 4 Zn (increase 8) and 1 NH₄NO₃ (decrease 8), giving 4 Zn(NO₃)₂. (1 mark)
N total = 8 + 1 (in NH₄⁺) + 1 (in NO₃⁻) = 10 HNO₃; H balance gives 3 H₂O: 4Zn + 10HNO₃ → 4Zn(NO₃)₂ + NH₄NO₃ + 3H₂O (O: 30 = 24 + 3 + 3). (1 mark)

Answer: 4Zn + 10HNO₃ → 4Zn(NO₃)₂ + NH₄NO₃ + 3H₂O

Example 3: Ion-electron method in acid · 3 marks · Short answerHarder board level

Balance MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺ in acidic solution.

Try it first, then show the solutionSolution

Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. (1 mark)
Oxidation: Fe²⁺ → Fe³⁺ + e⁻ (× 5). (1 mark)
Sum: MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O; charge 17 on each side. (1 mark)

Answer: MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O

Example 4: Basic medium · 4 marks · Short answerHarder board level

Balance MnO₄⁻ + I⁻ → MnO₂ + I₂ in basic medium.

Try it first, then show the solutionSolution

Reduction: MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻. (1 mark)
Oxidation: 2I⁻ → I₂ + 2e⁻. (1 mark)
Multiply the first by 2 and the second by 3 to equalise 6 electrons. (1 mark)
2MnO₄⁻ + 6I⁻ + 4H₂O → 2MnO₂ + 3I₂ + 8OH⁻ (charge −8 on both sides; O: 12 = 12). (1 mark)

Answer: 2MnO₄⁻ + 6I⁻ + 4H₂O → 2MnO₂ + 3I₂ + 8OH⁻

Example 5: Dichromate titration · 5 marks · Long answerHOTS / competency

Balance Cr₂O₇²⁻ + Fe²⁺ → Cr³⁺ + Fe³⁺ in acidic medium. If 25.0 mL of an Fe²⁺ solution needs 20.0 mL of 0.0200 M K₂Cr₂O₇, find the molarity of Fe²⁺ and the mass of iron in 250 mL of it (Fe = 56).

Try it first, then show the solutionSolution

Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O; Fe²⁺ → Fe³⁺ + e⁻ (× 6). (1 mark)
Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O. (1 mark)
Cr₂O₇²⁻: 0.0200 × 0.0200 = 4.00 × 10⁻⁴ mol; Fe²⁺: 6 × 4.00 × 10⁻⁴ = 2.40 × 10⁻³ mol in 25.0 mL. (1 mark)
Molarity = 2.40 × 10⁻³/0.0250 = 0.0960 M. (1 mark)
In 250 mL: 0.0960 × 0.250 = 0.0240 mol, mass = 0.0240 × 56 = 1.34 g. (1 mark)

Answer: 0.0960 M; 1.34 g of iron in 250 mL.

Where students lose marks in Balancing Redox Equations and Redox Titrations

The slips examiners see most often, and what to write instead.

Mistake: Not balancing charge
Leaving the equation with different total charges.
Do this: Check atoms and total charge on both sides.
Usually costs: 1 to 2 marks.
Mistake: Adding H⁺ in basic medium
Writing H⁺ in the final equation.
Do this: Convert to H₂O using OH⁻ in basic solution.
Usually costs: 2 marks.
Mistake: Wrong mole ratio in titration
Using 1 : 1 for MnO₄⁻ and Fe²⁺.
Do this: Use the ratio in the balanced equation (1 : 5).
Usually costs: 2 marks.

Exam tips

Write both half-reactions before combining them.
Verify atoms and charge on both sides.
State the medium (acidic or basic) in the answer.
For titrations, give the balanced equation and the mole ratio before calculating.

Practice questions on Balancing Redox Equations and Redox Titrations

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
In the ion-electron method, the oxygen atoms in an acidic medium are balanced by adding:
(a) OH⁻
(b) H⁺
(c) H₂O
(d) electrons
Hint
Then H⁺ balances the hydrogen.
Answer
(c) H₂O
Q2 · 2 marks · Short answerBoard level
Balance Fe₂O₃ + CO → Fe + CO₂.
Hint
Each CO takes one O.
Answer
Fe₂O₃ + 3CO → 2Fe + 3CO₂.
Q3 · 3 marks · Short answerHarder board level
Balance Cl₂ + OH⁻ → Cl⁻ + ClO₃⁻ + H₂O (hot, concentrated alkali).
Hint
Chlorine is both oxidised and reduced.
Answer
3Cl₂ + 6OH⁻ → 5Cl⁻ + ClO₃⁻ + 3H₂O (Cl: 6 = 5 + 1; charge: −6 = −6).
Q4 · 3 marks · Short answerHarder board level
Balance H₂O₂ + Fe²⁺ → Fe³⁺ + H₂O in acidic solution.
Hint
H₂O₂ gains 2 electrons.
Answer
H₂O₂ + 2H⁺ + 2e⁻ → 2H₂O; Fe²⁺ → Fe³⁺ + e⁻ (× 2). H₂O₂ + 2Fe²⁺ + 2H⁺ → 2Fe³⁺ + 2H₂O.
Q5 · 4 marks · Short answerHarder board level
Balance MnO₄⁻ + C₂O₄²⁻ → Mn²⁺ + CO₂ in acid and find the volume of 0.10 M KMnO₄ needed for 25.0 mL of 0.25 M oxalate.
Hint
Mole ratio from the balanced equation: 2 : 5.
Answer
2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O. Oxalate = 6.25 × 10⁻³ mol, so MnO₄⁻ = 2.5 × 10⁻³ mol; V = 2.5 × 10⁻³/0.10 = 0.025 L = 25 mL.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a chemist determines copper in a sample. Cu²⁺ is treated with excess iodide: 2Cu²⁺ + 4I⁻ → 2CuI + I₂, and the iodine is titrated with thiosulphate: I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻. The titration needs 20.0 mL of 0.100 M thiosulphate (Cu = 63.5).
(i) Find the moles of thiosulphate. (1 mark)
(ii) Find the moles of I₂. (1 mark)
(iii) Find the moles of Cu²⁺. (1 mark)
(iv) Find the mass of copper. (1 mark)
Hint
Follow the chain thiosulphate → I₂ → Cu²⁺.
Answer
(i) 0.0200 × 0.100 = 2.00 × 10⁻³ mol. (ii) 1.00 × 10⁻³ mol. (iii) 2 × 1.00 × 10⁻³ = 2.00 × 10⁻³ mol. (iv) 2.00 × 10⁻³ × 63.5 = 0.127 g.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Electrons gained by Cr₂O₇²⁻ in acid?Show answer
6.
2.Mole ratio MnO₄⁻ : Fe²⁺?Show answer
1 : 5.
3.How is O balanced in the ion-electron method?Show answer
By adding H₂O.
4.Which titration is self-indicating?Show answer
Permanganate.

Balancing Redox Equations and Redox Titrations: frequently asked questions

Which method should I use in the exam?

Either method is accepted. The oxidation-number method is quicker for molecular equations; the ion-electron method is better for ionic equations and for basic medium.

Why must electrons be equal?

Electrons lost by the reducing agent are exactly those gained by the oxidising agent.

Why is acid added in permanganate titrations?

In acid, MnO₄⁻ is reduced to colourless Mn²⁺ (5 electrons); in neutral or basic solution it gives MnO₂ (3 electrons), and the stoichiometry changes.