Oxidising ethanol to ethanoic acid and vinegar
Compare the specified chemical and bacterial oxidation routes, keeping them distinct from oxygen-free fermentation and complete combustion.
Extended only (Supplement) · Chemistry 0620, 2026–2028.
Before you begin
- Read ethanol and ethanoic-acid structures and define oxidation.
- Use moles and molar masses in a stated idealised quantitative model.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which group distinguishes ethanoic acid from ethanol?
What you will learn
- Describe ethanol oxidation with acidified aqueous potassium manganate(VII).
- Describe bacterial oxidation during vinegar production and distinguish the resulting acid from combustion products.
Two specified oxidation routes give the acid
Ethanol CH₃CH₂OH can be oxidised to ethanoic acid CH₃COOH using acidified aqueous potassium manganate(VII). State acidified, aqueous and the reagent's full name; plain water or a nickel catalyst does not supply the specified oxidising route. In the conventional organic shorthand, CH₃CH₂OH+2[O]→CH₃COOH+H₂O, where [O] denotes oxygen supplied by the oxidising agent rather than a bottled reactant called [O].
Ethanoic acid is also formed by bacterial oxidation of ethanol during vinegar production. This involves oxygen, unlike the specified oxygen-free yeast fermentation that first makes ethanol from glucose. An overall bacterial-oxidation model is C₂H₅OH+O₂→CH₃COOH+H₂O. Vinegar is an aqueous mixture containing ethanoic acid, not a claim of pure isolated acid.
Follow the product structure and the stated model
In this oxidation, the two-carbon framework is retained while the alcohol end becomes a carboxylic-acid group: ethanol C₂H₆O becomes ethanoic acid C₂H₄O₂. This is different from complete combustion, which forms CO₂ and H₂O rather than retaining the organic acid. Both involve oxidation, so the word alone is insufficient to identify the product.
For a supplied ideal model with complete conversion, one mole of ethanol gives one mole of ethanoic acid. With C=12, H=1 and O=16, their molar masses are 46 and 60 g/mol. The acid can have greater mass than the ethanol used because oxidation includes an external oxygen source; conservation applies across all reactants and products, including water. Real yield and purity require additional data.
Pause and explain
Which statement correctly separates fermentation and vinegar oxidation?
Worked example
In the stated ideal bacterial-oxidation model, 0.20 mol ethanol is fully converted. Find the theoretical ethanoic-acid mass, using C=12, H=1, O=16.
Show the worked solution
- C₂H₅OH+O₂→CH₃COOH+H₂O gives a 1:1 ethanol-to-acid ratio.
- CH₃COOH totals C₂H₄O₂, molar mass 2×12+4+2×16=60 g/mol.
- 0.20×60=12 g theoretical ethanoic acid; this assumes the stated complete conversion.
Answer 12 g theoretical ethanoic acid in the supplied complete-conversion model.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
State both required routes for forming ethanoic acid from ethanol and distinguish their agents.
Give me a hint
Name the complete chemical reagent and the bacterial context.
Compare my reasoning
- Chemical oxidation uses acidified aqueous potassium manganate(VII).
- Bacterial oxidation of ethanol occurs during vinegar production with oxygen involved.
- Both form ethanoic acid; neither is simply the earlier oxygen-free glucose fermentation.
Acidified aqueous potassium manganate(VII), or bacterial oxidation during vinegar production; both oxidise ethanol to ethanoic acid.
Look for these in your work
- Both specified routes and the chemical conditions appear.
- Oxidation is distinguished from making ethanol by fermentation.
Calculate the theoretical acid mass from 4.6 g ethanol in the stated complete-conversion model, using molar masses 46 and 60 g/mol.
Give me a hint
The ethanol-to-acid mole ratio is one to one.
Compare my reasoning
- Ethanol amount is 4.6÷46=0.10 mol.
- Complete conversion gives 0.10 mol ethanoic acid.
- Mass is 0.10×60=6.0 g; external oxygen explains why the acid mass can exceed the original ethanol mass.
6.0 g theoretical ethanoic acid, assuming complete conversion and the stated 1:1 ratio.
Look for these in your work
- The mole conversion and ratio are correct.
- The ideal-yield assumption accompanies the answer.
For the ideal bacterial model, 46 g ethanol reacts with 32 g O₂ to give 60 g acid and water. Find the water mass and assess a claim that the acid's extra mass violates conservation.
Give me a hint
Compare whole-system masses, not ethanol alone with acid alone.
Compare my reasoning
- Total reactant mass is 46+32=78 g.
- The remaining product mass is 78−60=18 g water, agreeing with one mole H₂O.
- External oxygen is included in the reactants; 60+18=78 g conserves total mass.
18 g water; 46+32=60+18=78 g, so the acid's greater mass does not violate conservation.
Look for these in your work
- The external oxygen and water are included.
- The whole-system mass balance is explicit.
Common mistakes
- Omitting acidified or aqueous from the required manganate(VII) reagent.
- Calling oxygen-free yeast fermentation the same step as bacterial vinegar oxidation.
- Predicting CO₂ instead of the stated retained two-carbon acid.
- Treating a theoretical complete-conversion mass as a measured real yield.
What can you explain now?
Which chemical reagent is specified for oxidising ethanol to ethanoic acid?
Compare with the explanation
Acidified aqueous potassium manganate(VII).
The acidification and aqueous condition are part of the stated reagent description.
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Make it stick
Recall both oxidation routes and their agents, draw the alcohol-to-acid change and check a complete-conversion mass with external oxygen included.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.