“My answer does not match, even though I followed the acid-base method.” — In a redox titration it is the amount of electrons (e⁻), not H⁺ and OH⁻, that balances. Three steps solve it:
- Read the electron count (the e⁻ coefficient) from the half-reactions
- Set up the equality using electron count × moles
- Solve for the concentration or volume
How it differs from an acid-base titration
The apparatus is similar; what balances is not.
- Acid-base: H⁺ = OH⁻ (valence × concentration × volume)
- Redox: electrons accepted = electrons released (electron count × moles)
For acid-base, see the acid-base titration calculator and the molarity guide.
The electron count comes from the half-reaction
Do not memorise the numbers; read the coefficient of e⁻.
| Substance | Role | Half-reaction | Electrons |
|---|---|---|---|
| Potassium permanganate (acidic) | Oxidizer | MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O | 5 |
| Potassium dichromate | Oxidizer | Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O | 6 |
| Hydrogen peroxide (as oxidizer) | Oxidizer | H₂O₂ + 2H⁺ + 2e⁻ → 2H₂O | 2 |
| Oxalic acid | Reducer | H₂C₂O₄ → 2CO₂ + 2H⁺ + 2e⁻ | 2 |
| Hydrogen peroxide (as reducer) | Reducer | H₂O₂ → O₂ + 2H⁺ + 2e⁻ | 2 |
| Iron(II) ion | Reducer | Fe²⁺ → Fe³⁺ + e⁻ | 1 |
| Sodium thiosulfate | Reducer | 2S₂O₃²⁻ → S₄O₆²⁻ + 2e⁻ | 1 (per S₂O₃²⁻) |
The same substance can have different half-reactions depending on its role. Thiosulfate releases two electrons per two ions, so one ion carries one. See OpenStax for balancing half-reactions.
The procedure
- Moles on the known side: concentration (mol/L) × volume (L)
- Electron moles: electron count × moles
- Solve the equality for the unknown side’s moles
- Convert back to a concentration or volume
Convert mL to litres first.
Example 1: standardising permanganate
A 0.0500 mol/L oxalic acid solution (20.0 mL) reaches the end point at 16.0 mL of unknown KMnO₄.
- Oxalic acid: 0.0500 × 20.0/1000 = 1.00 × 10⁻³ mol
- Electrons released: 1.00 × 10⁻³ × 2 = 2.00 × 10⁻³ mol
- For KMnO₄, 5x = 2.00 × 10⁻³, so x = 4.00 × 10⁻⁴ mol
- Concentration: 4.00 × 10⁻⁴ ÷ (16.0/1000) = 0.0250 mol/L
The oxidizer’s moles must be smaller than the reducer’s because its electron count is larger.
Example 2: the volume needed
A 0.10 mol/L hydrogen peroxide solution (10.0 mL) is titrated with 0.020 mol/L KMnO₄.
- H₂O₂: 0.10 × 10.0/1000 = 1.0 × 10⁻³ mol
- Electrons released: 1.0 × 10⁻³ × 2 = 2.0 × 10⁻³ mol
- KMnO₄ moles: 2.0 × 10⁻³ ÷ 5 = 4.0 × 10⁻⁴ mol
- Volume: 4.0 × 10⁻⁴ ÷ 0.020 × 1000 = 20.0 mL
Example 3: indirect iodometry
When an oxidizer is hard to titrate directly, excess potassium iodide liberates iodine (I₂), which is titrated with thiosulfate. Electrons pass along the chain, so electrons accepted by the oxidizer = moles of thiosulfate used (1 e⁻ per S₂O₃²⁻).
For 20.0 mL of 0.010 mol/L dichromate titrated with 0.100 mol/L thiosulfate:
- Electrons: 6 × 0.010 × 20.0/1000 = 1.20 × 10⁻³ mol
- Thiosulfate moles: 1.20 × 10⁻³ mol
- Volume: 1.20 × 10⁻³ ÷ 0.100 × 1000 = 12.0 mL
If 24.0 mL of thiosulfate were used instead, the dichromate concentration would be 0.0200 mol/L.
Common mistakes
- Using the acid-base “valence”: it counts H⁺, not electrons
- Reversing the ratio: a larger electron count means fewer moles are needed
- Counting thiosulfate as divalent: each ion carries one electron
- Using 5 electrons outside acidic conditions: neutral or basic conditions stop at MnO₂ with 3 e⁻
- Forgetting the mL-to-L conversion, or rounding too early (see significant figures)
Conditions and limits
- The reaction must go to completion under the stated conditions. Permanganate is its own indicator: its colour marks the end point
- Sulfuric acid is used because hydrochloric acid acts as a reducer and nitric acid as an oxidizer, which would break the stoichiometry
- Back-titration and COD, which titrate the remaining oxidizer in a second step, are outside the three basic patterns here
- Round once at the end to the digits given in the problem (often three)
Check your work with the tool
The Redox Titration Calculator on Tools Hub fills in the half-reaction and electron count when you pick a substance.
Pick the substances
Choose oxalic acid as the reducer: the tool fills in H₂C₂O₄ → 2CO₂ + 2H⁺ + 2e⁻ and the electron count 2. Choose acidic potassium permanganate as the oxidizer to fill in 5.
Enter Example 1
For the known reducer enter 0.0500 and 20.0; for the oxidizer enter 16.0. Press Calculate to see "Concentration = 0.0250 mol/L" with the electron working.
Switch patterns
Use "Volume needed for the reaction" for Example 2, and "Indirect iodine titration" for Example 3, which can solve for either the thiosulfate volume or the oxidizer concentration.
Tool used in this guide
Redox Titration Calculator
Electron counts, concentrations, volumes and indirect iodometry with the working shown. Free, browser-only, no sign-up.
FAQ
How is this different from an acid-base titration?
Acid-base balances H⁺ and OH⁻ (valence × concentration × volume); redox balances electron moles (electron count × moles). The “valence” now counts electrons.
Where do I find the electron count?
It is the coefficient of e⁻ in the half-reaction: 5 for MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O, 6 for dichromate. Writing the half-reaction each time is safer than memorising numbers.
Why does thiosulfate count as one electron?
The half-reaction 2S₂O₃²⁻ → S₄O₆²⁻ + 2e⁻ releases two electrons per two ions, so each carries one — a common trap.
Why is sulfuric acid used?
It is neither an oxidizer nor a reducer, so it only supplies H⁺. Hydrochloric acid would act as a reducer and nitric acid as an oxidizer.
What was checked
On October 3, 2026, the electron counts, moles, electron moles, concentrations and volumes were recalculated by hand and compared with the listed references. The examples are study calculations, not experimental results. For hands-on work with reagents, follow your institution’s procedures and supervisor’s instructions.
