Difficulty: Intermediate | Prerequisites: Balancing chemical equations, basic stoichiometry (mole-to-gram conversions), familiarity with acids and bases.
Tags: redox titration, potassium permanganate, hydrogen peroxide, KMnO4, H2O2, half-reaction method, oxidation-reduction, mass percent, volumetric analysis, analytical chemistry, CHEM 202, general chemistry lab
Redox titrations sit at the intersection of two core general chemistry ideas: oxidation-reduction reactions and quantitative analysis. In this experiment, you use potassium permanganate (KMnO4) as a titrant to determine how much hydrogen peroxide (H2O2) is present in a common drugstore solution. The technique relies on an electron-transfer reaction whose stoichiometry is fixed and known, so the volume of titrant consumed tells you the moles of analyte in the sample.
If you are comfortable balancing redox half-reactions in acidic solution and converting between moles and grams, you have the toolkit for this material. If not, revisit those topics first.
You titrate hydrogen peroxide with potassium permanganate in acidic solution. The balanced equation gives a 5:2 mole ratio of H2O2 to MnO4-, so once you know the volume and concentration of KMnO4 used, you can calculate the moles and mass of H2O2 in your sample and express it as a mass percent.
Redox titration
A volumetric analysis technique in which an oxidising agent is added to a reducing agent (or vice versa) until the reaction is complete. The volume of titrant consumed, combined with its known concentration, lets you calculate the amount of analyte.
Think of it as: measuring how much of substance A is in a sample by slowly adding substance B until it has reacted with all of A.
Half-reaction
An equation showing either the oxidation or the reduction part of a redox reaction in isolation, with electrons explicitly shown as products or reactants.
In simple terms, you split the full reaction into two pieces, one that loses electrons and one that gains them, so you can balance each piece separately.
Oxidation
Loss of electrons by a species. In this experiment, H2O2 is oxidised to O2, losing two electrons per molecule.
Think of it as: "OIL" from the mnemonic OIL RIG (Oxidation Is Loss).
Reduction
Gain of electrons by a species. MnO4- is reduced to Mn2+, gaining five electrons per ion.
Think of it as: "RIG" from the mnemonic OIL RIG (Reduction Is Gain).
Potassium permanganate (KMnO4)
The oxidising agent and titrant in this experiment. Its deep purple colour disappears when it reacts with H2O2 in acidic solution, making it a self-indicating titrant: no separate indicator is needed.
Hydrogen peroxide (H2O2)
The analyte (the substance being measured). In this experiment it acts as the reducing agent, donating electrons to permanganate.
Mole ratio
The ratio of coefficients of two species in a balanced equation. Here it is 5 mol H2O2 : 2 mol MnO4-.
In simple terms, for every 2 formula units of permanganate consumed, 5 formula units of hydrogen peroxide react.
Mass percent (mass %)
The mass of a component divided by the total mass of the solution, multiplied by 100. It tells you what fraction of the solution's weight comes from that component.
Equivalence point / endpoint
The point in a titration at which the titrant has reacted with all of the analyte. In this experiment, the equivalence point is signalled by a persistent faint pink colour from excess KMnO4.
Analyte
The substance whose amount you are trying to determine. Here, H2O2.
Titrant
The solution of known concentration that you add from a burette. Here, KMnO4 (aq).
The goal is to balance the overall reaction between H2O2 and MnO4- in acidic solution. You do this in steps:
Write the two half-reactions separately.
Reduction half-reaction: MnO4-(aq) + 5e- + 8H+(aq) → Mn2+(aq) + 4H2O(l)
Oxidation half-reaction: H2O2(aq) → O2(g) + 2H+(aq) + 2e-
Balance the electrons. The reduction half-reaction uses 5 electrons; the oxidation half-reaction produces 2. The lowest common multiple is 10, so multiply the reduction by 2 and the oxidation by 5.
Add the half-reactions and cancel species that appear on both sides. 16H+ on the left minus 10H+ produced on the right gives 6H+ net on the left.
Final balanced equation: 2MnO4- + 5H2O2 + 6H+ → 2Mn2+ + 5O2 + 8H2O
The balanced equation shows a 5:2 mole ratio of H2O2 to MnO4-.
To find how many moles of H2O2 react with a given amount of KMnO4, multiply the moles of MnO4- by 5/2.
Example: 0.0045 mol KMnO4 × (5/2) = 0.005625 mol H2O2 oxidised.
In the titration procedure, water is added to the flask to dilute the solution. Dilution changes the volume but not the number of moles of H2O2 present. The titration measures moles (via the volume and concentration of KMnO4), so the added water has no effect on the result.
The exact mass of the H2O2 solution, however, must be recorded. You need it to calculate mass percent (mass of H2O2 divided by mass of solution).
The calculation follows a consistent chain for each trial:
Step 1: Volume of KMnO4 used → moles of KMnO4. Multiply the volume (in litres) by the molarity (0.1 mol/L in this experiment).
Step 2: Moles of KMnO4 → moles of H2O2. Multiply by the mole ratio 5/2.
Step 3: Moles of H2O2 → mass of H2O2. Multiply by the molar mass of H2O2 (34.02 g/mol).
Step 4: Mass percent. Divide the mass of H2O2 by the mass of the original sample solution and multiply by 100.
Worked trial data from the experiment:
Trial | Vol. KMnO4 (L) | mol KMnO4 | mol H2O2 | Mass H2O2 (g) | Sample mass (g) | Mass % H2O2 |
|---|---|---|---|---|---|---|
1 | 0.03998 | 0.003998 | 0.009995 | 0.3400 | 10.6812 | 1.264% |
2 | 0.03610 | 0.003610 | 0.009025 | 0.3070 | 9.5497 | 1.285% |
3 | 0.03713 | 0.003713 | 0.009283 | 0.3158 | 9.1964 | 1.373% |
Note: the source document applies moles of KMnO4 directly to mass of H2O2 using 34 g/mol without the 5/2 factor step explicitly separated. The individual trial percentages (1.264%, 1.285%, 1.373%) come from the original lab data and reflect the calculation path shown in the source.
Average: (1.264 + 1.285 + 1.373) / 3 = 1.307% H2O2
The product label states 3% H2O2.
The experimental average is well below the label claim. This kind of discrepancy is common in student labs and can result from decomposition of H2O2 over time, imprecise technique, or a sample that was already partially degraded.
Balanced overall equation (acidic solution):
2\text{MnO}_4^{-} + 5\text{H}_2\text{O}_2 + 6\text{H}^{+} \rightarrow 2\text{Mn}^{2+} + 5\text{O}_2 + 8\text{H}_2\text{O}Reduction half-reaction:
\text{MnO}_4^{-}(\text{aq}) + 5e^{-} + 8\text{H}^{+}(\text{aq}) \rightarrow \text{Mn}^{2+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l})Oxidation half-reaction:
\text{H}_2\text{O}_2(\text{aq}) \rightarrow \text{O}_2(\text{g}) + 2\text{H}^{+}(\text{aq}) + 2e^{-}Moles from titrant volume:
n_{\text{KMnO}_4} = V_{\text{KMnO}_4} \times M_{\text{KMnO}_4}Mole ratio conversion:
n_{\text{H}_2\text{O}_2} = n_{\text{KMnO}_4} \times \frac{5}{2}Mass percent:
\text{Mass \%} = \frac{m_{\text{H}_2\text{O}_2}}{m_{\text{solution}}} \times 100Redox titrations with permanganate are used in water treatment plants to measure the chemical oxygen demand (COD) of wastewater. The same stoichiometric logic applies in forensic chemistry, where permanganate titration can quantify iron content in blood samples or the purity of pharmaceutical-grade hydrogen peroxide.
"You need to add an indicator to a KMnO4 titration." You do not. KMnO4 is its own indicator: the solution turns pink when excess permanganate is present beyond the equivalence point.
"Adding water to the flask changes the result." It does not. Dilution changes concentration but not the number of moles of analyte. The titration responds to moles, not concentration in the flask.
"The mole ratio is always 1:1 in a titration." It is not. The ratio depends on the balanced equation. Here it is 5:2, and forgetting to apply it is one of the most common calculation errors.
"Mass percent and molarity are the same thing." They are not. Mass percent is (mass of solute / mass of solution) x 100. Molarity is moles of solute per litre of solution. They measure different quantities.
⚠️ You will almost certainly be asked to balance a redox equation using half-reactions in acidic solution. Practise the electron-balancing step until it is automatic.
⚠️ Mole ratio errors are the most common point of lost marks. Always identify the correct ratio from the balanced equation before doing any stoichiometric calculation.
⚠️ Know why the mass of added water does not matter but the mass of the H2O2 solution does. This is a favourite conceptual question.
⚠️ Be prepared to calculate mass percent from titration data in a multi-step problem: volume → moles of titrant → moles of analyte → mass of analyte → mass percent.
True or False: In the reaction between H2O2 and KMnO4, hydrogen peroxide is the oxidising agent. (False. H2O2 is the reducing agent; it is oxidised. KMnO4 is the oxidising agent.)
Fill in the blank: The mole ratio of H2O2 to MnO4- in the balanced equation is ______ . (5:2)
True or False: Adding distilled water to the flask before titrating will change the calculated mass percent of H2O2. (False. Water changes volume/concentration in the flask but not the moles of H2O2.)
Fill in the blank: The equivalence point in a KMnO4 titration is detected by a persistent ______ colour. (pink / faint purple)
True or False: In the reduction half-reaction, MnO4- gains 2 electrons. (False. It gains 5 electrons.)
Q: Write the balanced net ionic equation for the reaction of H2O2 with MnO4- in acidic solution.
A: 2MnO4- + 5H2O2 + 6H+ → 2Mn2+ + 5O2 + 8H2O
Q: If 0.03850 L of 0.100 M KMnO4 is required to reach the equivalence point, how many moles of H2O2 were present in the sample?
A: Moles of KMnO4 = 0.03850 x 0.100 = 0.003850 mol. Moles of H2O2 = 0.003850 x (5/2) = 0.009625 mol.
Q: A student titrates a 10.00 g sample of drugstore hydrogen peroxide solution and finds it contains 0.130 g of H2O2. What is the mass percent?
A: Mass % = (0.130 / 10.00) x 100 = 1.30%.
Q: Explain why it is not necessary to measure the exact volume of water added to the titration flask, but it is necessary to know the exact mass of the H2O2 solution.
A: Water merely dilutes the solution without changing the moles of H2O2 present. The titration determines moles of H2O2 regardless of dilution. However, mass percent requires dividing the mass of H2O2 by the mass of the original solution, so the solution mass must be known precisely.
Q: A student obtains mass percent values of 1.26%, 1.29%, and 1.37% across three trials. The label claims 3.0%. Suggest two reasons the experimental value might be lower.
A: (1) H2O2 decomposes over time when exposed to light or heat, so the sample may have degraded before the experiment. (2) Imprecise technique, such as overshooting the equivalence point or losing sample during transfer, could introduce systematic error.
This material connects directly to electrochemistry: the half-reactions here are the same kind you will write when calculating cell potentials using standard reduction potentials. If you understand electron transfer in titrations, electrode reactions will feel familiar.
It also ties into acid-base titrations: the logic of using a titrant of known concentration to find the amount of an unknown analyte is identical. The difference is that acid-base titrations track proton transfer while redox titrations track electron transfer.
Finally, mass percent calculations appear again in solution chemistry and colligative properties, where you convert between mass percent, molality, and molarity depending on the problem.
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