Edexcel A-Level Chemistry Paper 3, June 2025: Question 7
14 marks · Hard difficulty · Practical Techniques and Data Analysis
Analyse the concentration of ethanol in an alcoholic drink using a back titration with acidified sodium dichromate(VI), potassium iodide, and sodium thiosulfate.
Practise this questionQuestion
Question text
7 A laboratory technician analysed a sample of an alcoholic drink to determine the
concentration of ethanol, in g dm−3.
20.00 cm3 of the alcoholic drink was added to a 250 cm3 volumetric flask and made up
to the mark with deionised water.
The ethanol in 5.00 cm3 of the diluted sample was oxidised in the equipment shown.
The sample was suspended in an open beaker in a sealed conical flask over 10.00 cm3
of acidified sodium dichromate(VI), of concentration 0.0780 mol dm−3.
5.00 cm3 of the
diluted sample
10.00 cm3 of 0.0780 mol dm–3 acidified
sodium dichromate(VI) solution
The apparatus was left in a warm place for 24 hours for the
acidified sodium dichromate(VI) to oxidise the ethanol.
Excess potassium iodide solution was then added to the remaining
acidified sodium dichromate(VI) in the conical flask, forming iodine.
All of this iodine was then titrated using 0.0220 mol dm−3 sodium thiosulfate solution,
Na2S2O3(aq).
The volume of Na S2O3(aq) required in the titration was 32.70 cm3.
The equations for the reactions that take place during the experiment are shown.
3CH CH OH(l) + 2Cr O2−(aq) + 16H+(aq) → 3CH COOH(aq) + 4Cr3+(aq) + 11H O(l)
32 2 7 3 2
Cr O2−(aq) + 14H+(aq) + 6I−(aq) → 2Cr3+(aq) + 3I (aq) + 7H O(l)
27 2 2
2S O2−(aq) + I (aq) → 2I−(aq) + S O2−(aq)
23 2 4 6
(a) Explain why the apparatus was left in a warm room and for 24 hours.
(2)
(b) (i) Calculate the number of moles of ethanol in 5 cm3 of the diluted drink.
(5)
*P77834A02036*
(ii) Calculate the concentration of ethanol in the alcoholic drink in g dm−3.
(3)
(c) It was suggested that the titration as described might lead to an unreliable result.
State two ways the procedure could be amended to counter this suggestion.
(2)
(d) Another technician suggested that it was unnecessary to suspend the drink
sample above the acidified sodium dichromate(VI) solution and proposed a
change to the procedure.
The proposal was that the drink sample should be mixed with the acidified*P77834A02136*
sodium dichromate(VI) solution and then heated under reflux to allow the
ethanol to be oxidised.
Give one reason to support this change and one reason to leave the method
unchanged. In each case justify your answer.
(2)
Reason to support this change
Reason to leave the method unchanged
(Total for Question 7 = 14 marks)
Mark scheme
Show the mark scheme
Question
Answer Additional Guidance Mark
Number
7(a) An explanation that makes reference to two of the following points: (2)
• (warmth) to allow for (all) the ethanol to evaporate (1) Ignore to speed the reaction up
Ignore activation energy
Do not award to evaporate the acidified
dichromate(VI)
• (24 hr) to allow (time for) the ethanol to fully react / fully (1) Allow the reaction goes to completion
oxidise Allow all of ethanol reacts
• (To allow full oxidation of the ethanol) to ethanoic acid (1) Allow to ensure there is no ethanol /
ethanal left
‘To allow time for all the ethanol to
react to form ethanoic acid’, scores P2
and P3
Question
Answer Additional Guidance Mark
Number
7(b)(i) Example of calculation (5)
• calculation of moles of thiosulfate (1) (32.70 ÷ 1000) × 0.0220 = 7.194 × 10–4 (mol)
• deduction of moles of iodine (1) 7.194 × 10–4 ÷ 2 = 3.597 × 10–4
• deduction of moles of acidified dichromate(VI) (1) 3.597 × 10–4 ÷ 3 = 1.199 × 10–4 (mol)
that reacted to form iodine
• calculation of moles of acidified dichromate(VI) (1) [(10.0 ÷ 1000) × 0.078] = 7.80 × 10–4
that reacted with ethanol 7.80 × 10–4 – 1.199 × 10–4 = 6.601 × 10–4 (mol)
• deduction of moles of ethanol in 5 cm3 sample (1) 6.601 × 10–4 × (3 ÷ 2)
= 9.9015 × 10–4 / 9.90 × 10–4 (mol)
Allow TE throughout Ignore SF except 1SF
Correct answer without working scores 5
Question
Answer Additional Guidance Mark
Number
7(b)(ii) Example of calculation (3)
• calculation of moles of ethanol in 250 cm3 of (1) 9.9015 × 10–4 × (250 ÷ 5) = 0.049508 (mol)
diluted drink
• calculation of concentration of alcoholic drink (1) 0.049508 ÷ (20 ÷ 1000) = 2.4754 (mol dm–3)
in mol dm–3
(1) 2.4754 × 46 = 113.87 / 114 (g dm–3)
• calculation of concentration of alcoholic drink
in g dm–3 Allow TE throughout including from part bi
Ignore SF except 1SF
Correct answer without working scores 3
Question
Answer Additional Guidance Mark
Number
7(c) An answer that makes reference to the following points: (2)
• repeat the whole experiment / (1) Do not allow just repeat the titration
set up more than one version of the
experiment and run concurrently
• take more than one sample / (1) If no other mark is scored allow repeat the titration
smaller samples (so more than one and calculate a mean / repeat the titration until
titration can be carried out) concordant results are obtained
Ignore starch / indicator
Question
Answer Additional Guidance Mark
Number
7(d) An answer that makes reference to the following points: (2)
M1 reason to support change
• heating under reflux will speed up the reaction (1)
or
• less chance of incomplete reaction with heating under
reflux
M2 reason to remain unchanged
• might be other (less volatile substances) that could also be (1) Ignore the ethanol and the ethanoic acid
oxidised by the dichromate(VI) produced might react to form an ester
or
• greater risk of uncontrolled reaction associated with Ignore greater risk of flammability
heating under reflux
or
• greater risk of some of the ethanol evaporating with reflux Allow some ethanol may escape from the
reflux condenser before it is oxidised
(Total for Question 7 = 14 marks)
How to answer it
Determination of Ethanol via Redox Back Titration
This question assesses your mastery of complex multi-step redox titrations and practical experimental design, specifically:
- Understanding indirect/back-titration mechanics involving dichromate(VI), iodide, and thiosulfate ions.
- Tracking stoichiometric reacting ratios across three linked redox equations.
- Calculating dilution factors (aliquot to volumetric flask to initial sample) and unit conversions ( mol dm⁻³ to g dm⁻³ ).
- Evaluating experimental procedures, reliability, repeatability, and method modifications (reflux vs. vapour diffusion).
Part (a): Purpose of Warmth and 24-Hour Duration
Explaining experimental conditions [2 Marks]
✅ Correct Answer (Award 1 mark per bullet)
- Warmth: To allow the ethanol to evaporate (volatilise) out of the suspended beaker into the flask.
- 24 hours: To allow sufficient time for all the ethanol to fully react / fully oxidise into ethanoic acid (ensures reaction goes to completion).
🧠 Exam Technique
Notice the physical setup: the ethanol sample is physically separated from the oxidising solution in a suspended inner beaker. Therefore, ethanol must vaporise to contact the dichromate(VI).
❌ Common Errors & Lost Marks
- Stating warmth is "to speed up the reaction" or "provide activation energy" — Ignored by examiners because warmth specifically enables evaporation here.
- Claiming the warmth is to "evaporate the acidified dichromate" — Explicitly rejected.
- Saying "to ensure equilibrium is reached" — this is an irreversible redox oxidation, not an equilibrium system.
Part (b)(i): Moles of Ethanol in 5.00 cm³ Diluted Drink
Five-step stoichiometric calculation [5 Marks]
📐 Step-by-Step Back Titration Calculation
- Calculate moles of thiosulfate, S₂O₃²⁻:
n(S₂O₃²⁻) = (32.70 ÷ 1000) × 0.0220 = 7.194 × 10⁻⁴ molAward 1 Mark - Determine moles of I₂ liberated:
From equation (3): 2S₂O₃²⁻ ≡ 1I₂
n(I₂) = 7.194 × 10⁻⁴ ÷ 2 = 3.597 × 10⁻⁴ molAward 1 Mark - Determine moles of remaining Cr₂O₇²⁻:
From equation (2): 1Cr₂O₇²⁻ ≡ 3I₂
n(Cr₂O₇²⁻ unreacted) = 3.597 × 10⁻⁴ ÷ 3 = 1.199 × 10⁻⁴ molAward 1 Mark - Calculate moles of Cr₂O₇²⁻ that reacted with ethanol:
Initial n(Cr₂O₇²⁻) = (10.00 ÷ 1000) × 0.0780 = 7.800 × 10⁻⁴ mol
n(Cr₂O₇²⁻ reacted) = 7.800 × 10⁻⁴ - 1.199 × 10⁻⁴ = 6.601 × 10⁻⁴ molAward 1 Mark - Calculate moles of ethanol in 5.00 cm³ sample:
From equation (1): 3 ethanol ≡ 2Cr₂O₇²⁻
n(ethanol) = 6.601 × 10⁻⁴ × (3 ÷ 2) = 9.9015 × 10⁻⁴ mol (or 9.90 × 10⁻⁴ mol)Award 1 Mark
❌ Common Calculation Traps
- Wrong mole ratios: Missing the ÷ 3 ratio between I₂ and Cr₂O₇²⁻, or missing the 3/2 ratio between ethanol and Cr₂O₇²⁻.
- Forgetting it's a back titration: Attempting to link thiosulfate directly to ethanol without subtracting unreacted dichromate from initial dichromate.
💡 Examiner Guidance
Full marks are awarded for the correct final answer even without working shown. Transfer of Error (TE) is applied across all steps if an early arithmetic slip occurs.
Part (b)(ii): Concentration of Ethanol in Drink
Dilution scaling and mass concentration conversion [3 Marks]
📐 Step-by-Step Scaling & Unit Conversion
- Scale up to the 250 cm³ volumetric flask:
The 5.00 cm³ portion was taken from 250 cm³.
n(ethanol in 250 cm³) = 9.9015 × 10⁻⁴ × (250 ÷ 5) = 0.049508 molAward 1 Mark - Calculate concentration in original 20.00 cm³ sample (mol dm⁻³):
The 0.049508 mol came entirely from 20.00 cm³ of original alcoholic drink.
c(mol dm⁻³) = 0.049508 ÷ (20.00 ÷ 1000) = 2.4754 mol dm⁻³Award 1 Mark - Convert to g dm⁻³ using Mᵣ:
Mᵣ(C₂H₅OH) = (2 × 12.0) + (6 × 1.0) + 16.0 = 46.0 g mol⁻¹
Concentration = 2.4754 × 46.0 = 113.87 g dm⁻³ (rounds to 114 g dm⁻³)Award 1 Mark
❌ Dilution Inversion Trap
Students frequently confuse the dilution factor: remember the original drink is concentrated and the flask solution is dilute. Multiplying by 250/5 gives total moles; dividing by (20/1000) yields original concentration.
Part (c): Improving Titration Reliability
Experimental repeatability [2 Marks]
✅ Accepted Improvements (State two)
- Repeat the whole experiment / set up replicate flasks concurrently (to calculate a mean).
- Use larger total volumes or take smaller aliquots from the reaction mixture so that multiple titrations can be carried out to obtain concordant titres.
❌ Insufficient Responses
- Just saying "repeat the titration" scores 0 unless qualified by explaining how you obtain enough sample to do so, because in the given procedure, all the iodine in the flask is consumed in a single titration!
- Adding starch indicator (starch is already a standard indicator for iodine-thiosulfate titrations; it does not solve the reliability issue of a single run).
Part (d): Evaluating the Reflux Alternative
Weighing advantages and disadvantages of direct reflux [2 Marks]
✅ Reason to Support Change (1 mark)
- Heating under reflux will significantly speed up the reaction rate (saving 24 hours).
- Reduces the risk of incomplete oxidation.
✅ Reason to Leave Unchanged (1 mark)
- Other non-volatile / less volatile compounds in the drink (e.g. sugars, flavourings, other alcohols) could be oxidised if mixed directly with Cr₂O₇²⁻, giving an erroneously high ethanol value.
- Risk of volatile ethanol vapour escaping past the condenser before reacting.
- Direct mixing can lead to a vigorous/uncontrolled exothermic reaction upon heating.
🧠 Why the Suspended Beaker Method is Clever
Suspending the sample above the oxidant acts as an in-situ vapour distillation. Only volatile components (like ethanol) evaporate and transfer across into the dichromate solution, leaving non-volatile sugars or additives behind.
Topics
Physical Chemistry · Organic Chemistry · Topic 5: Formulae, Equations and Amounts of Substance · Topic 6: Organic Chemistry I · Topic 3: Redox I
Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 3, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.