OCR A-Level Chemistry Unified chemistry (03), November 2021: Question 2

9 marks · Hard difficulty · Practical Questions

Determine the correct ionic equation for the reaction of europium with dilute aqueous acid by analyzing experimental gas collection results, evaluating temperature effects, and predicting precipitate masses.

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Question

Question text

2 Europium (Ar = 152.0) reacts with dilute aqueous acid in a redox reaction, forming a solution and

hydrogen gas.

A student proposed three possible ionic equations for this reaction, forming europium ions with

different charges:

Equation 1 2Eu(s) + 2H+(aq) 2Eu+(aq) + H (g)

Equation 2 Eu(s) + 2H+(aq) Eu2+(aq) + H (g)

Equation 3 2Eu(s) + 6H+(aq) 2Eu3+(aq) + 3H (g)

The student plans to carry out an investigation to determine which equation is correct.

Hydrochloric acid is used as the source of H+(aq) ions.

The student’s method is outlined below.

Step 1 Using a 3 decimal place balance, weigh a suitable container for the reaction.

Add about 1 g of europium to the container and reweigh.

Step 2 Set up apparatus for gas collection.

Add an excess of dilute hydrochloric acid to the europium.

Step 3 Measure the volume of gas produced.

Results

Mass of container = 32.795 g

Mass of container + europium = 33.783 g

Volume of gas collected = 152 cm3

(a) Draw a labelled diagram of suitable apparatus for this investigation.

[2]

(b) Analyse the student’s results to conclude which of Equation 1 or 2 or 3, is supported by the

experimental results.

Assume that the conditions in the laboratory are ‘room temperature and pressure’ (RTP).

correct equation (1, 2, or 3) = … [3]

hydrochloric acid. The apparatus gets hot during the reaction.

Predict how the hot apparatus would change the student’s results and the conclusion in (b).

Explain your answer.

… [2]

(d) The student modifies their method as outlined below:

• 1.52 g (0.01 mol) of europium is reacted with an excess of dilute hydrochloric acid.

• An excess of aqueous sodium hydroxide is added to the reaction mixture.

• A precipitate forms which is collected, dried and weighed.

Explain how the mass of precipitate formed would allow the student to conclude which of

Equation 1 or 2 or 3 is correct.

… [2]

Mark scheme

Show the mark scheme

Question Answer Marks element Guidance

2 (a) Closed system that would work (Labels not required) 2 AO3.3

Reaction apparatus with tube/side arm × 2 ALLOW small gaps provided there is an attempt

AND gas collection apparatus to show closed system

AND closed system

DO NOT ALLOW delivery tube below reaction

Labels mixture

Reaction apparatus, e.g.:

Conical flask, Buchner flask/conical flask with For reaction apparatus,

side arm, test-tube, boiling tube. • DO NOT ALLOW flask, volumetric flask,

AND beaker, measuring cylinder

Gas collection apparatus: • Delivery tube, bung does NOT need a label

(gas) syringe

OR gas collection over water with labelled ALLOW labels for diagram without closed system

measuring cylinder / burette (e.g. bung missing),

i.e. 2nd mark but not 1st mark

ALLOW any of these diagrams.

ALLOW a single line for the tube

IGNORE Sealed end of delivery tube

IGNORE size of syringe/measuring

cylinder/burette

(b) 152 –3 3 AO2.8

n(H2) = 24000 OR 6.33…. × 10 (mol)

×2

0.988 –3

n(Eu) = OR 6.5(0) × 10 (mol) 152 6.5(0) × 10–3 (mol)

Ratio H2 : Eu 1 : 1 ALLOW 0.97(4) : 1

AND

Equation 2 is correct AO3.2 ALLOW ECF from incorrect n(Eu)

×1 OR/AND n (H2)

Only ALLOW if n(H2) AND n(Eu) are approximately ----------------------------------------------------------------

equal ALLOW approach that calculates mass Eu from

6.33…. × 10–3 mol H for each equation, e.g.

ALLOW use of ideal gas equation at a reasonable Equation 1: 2 × 6.33 × 10–3 × 152

temperature and pressure. = 1.9.. g

Equation 2: 1 × 6.33 × 10–3 × 152

e.g. Using 100 kPa and 298 K, n(H ) = 6.14 × 10–3 mol

2 = 0.96.. g

Equation 3: 2/3 × 6.33 × 10–3 × 152

= 0.64.. g

0.988 matched to 0.96 g and Equation 2

Use judgment

----------------------------------------------------------------

ALLOW approach that calculates volume H2 from

6.50 × 10–3 mol Eu for each equation, e.g.

Equation 1: 0.5 × 24000 × 6.50 × 10–3

= 78 cm3

Equation 2: 1 × 24000 × 6.50 × 10–3

= 156 cm3

Equation 3: 1.5 × 24000 × 6.50 × 10–3

= 234 cm3

152 matched to 156 cm3 and Equation 2

Use judgment

Question Answer 10 Marks element Guidance

×2 IGNORE effect of rate, e.g. rate increases

Ratio H2 : Eu would be larger

IGNORE gas equation should be used to find n(H2)

ALLOW Equation 3 linked to H2 : Eu > 1

(d) Qual 2 AO3.4

Precipitates have different molar masses ×2

Precipitates have different formulae ALLOW precipitates are EuOH, Eu(OH)2 Eu(OH)3

OR precipitates have different number of OH– ions

Quant

Equation 2 forms precipitate with M = 186

OR with formula Eu(OH)2

Equation 2 forms 1.86 g precipitate

mass of precipitate – mass of precipitate – mass of Eu

Molar mass M of precipitate = moles precipitate ALLOW Moles OH = molar mass of OH–

mass of precipitate – mass of precipitate –1.52

OR OR Moles OH = 17

moles Eu

mass of precipitate

OR 0.01

How to answer it

Europium Redox & Stoichiometry Investigation

What this question tests

This multi-part exam question assesses practical skills in gas collection apparatus design, quantitative mole calculations involving molar gas volume at RTP, application of reacting ratios to deduce stoichiometry, evaluation of experimental errors, and gravimetric analysis using precipitation reactions.

Part (a): Gas Collection Apparatus

Designing and drawing a functional experimental setup

✅ Correct Answer Requirements

A closed system containing reaction apparatus connected to gas collection apparatus.
Reaction vessel: Conical flask, round-bottom flask, test tube, or boiling tube fitted with a bung and delivery tube.
Gas collection method: A gas syringe OR an inverted measuring cylinder/burette filled with water in a trough.

🧠 Exam Technique & Tips

Keep diagrams simple and 2D. Clearly show a bung sealing the reaction container so no gas escapes.
Ensure the delivery tube enters the collection vessel properly without dipping below the reaction mixture liquid level.

❌ Common Errors & Penalties

Using open containers like beakers or volumetric flasks (loses the closed system mark).
Drawing delivery tubes dipping under the acid level inside the reaction vessel, which would force acid back up or block gas flow.

Marks available: 2 marks (1 for functional closed apparatus setup, 1 for correct component labels).

Part (b): Stoichiometry Calculation & Analysis

Using experimental data to determine the correct ionic equation

📐 Step-by-Step Calculation

Step 1: Calculate moles of hydrogen gas collected.

n(H₂) = Volume / 24000 = 152 / 24000 = 6.333 × 10⁻³ mol

Step 2: Calculate moles of europium reacted.

n(Eu) = Mass / Ar = 0.988 / 152.0 = 6.50 × 10⁻³ mol

Step 3: Determine the reacting mole ratio (H₂ : Eu).

Ratio = 6.333 × 10⁻³ : 6.50 × 10⁻³ ≈ 1 : 1

✅ Correct Answer

Equation 2 is supported by the results.

Reason: The experimental mole ratio of H₂ to Eu is approximately 1 : 1, which directly matches the stoichiometric coefficients in Equation 2: Eu(s) + 2H⁺(aq) → Eu²⁺(aq) + H₂(g) .

💡 Alternative Approach

You can also calculate theoretical expected values:

Equation 1 expects 1.90 g Eu
Equation 2 expects 0.96 g Eu (very close to the 0.988 g used)
Equation 3 expects 0.64 g Eu

Marks available: 3 marks (1 for calculating n(H₂), 1 for calculating n(Eu) or ratio, 1 for identifying Equation 2 with justification).

Part (c): Temperature Effects on Gas Volume

Evaluating thermal expansion impacts on experimental gas collection

✅ Correct Answers

Gas volume: Would be larger than at RTP.
Ratio: H₂ : Eu ratio would appear larger.

💡 Scientific Explanation

Using concentrated acid is exothermic and heats the apparatus. According to Charles's Law and the Ideal Gas Equation ( pV = nRT ), gas expands as temperature increases at constant pressure, resulting in a artificially inflated gas volume reading for the same number of moles.

❌ Common Errors

Students often incorrectly discuss changes in reaction rate rather than focusing strictly on gas volume expansion caused by higher temperature.

Marks available: 2 marks (1 for larger gas volume prediction, 1 for explaining the temperature/expansion effect on the ratio).

Part (d): Gravimetric Analysis via Precipitation

Using stoichiometry and mass of precipitates to confirm ion charge

✅ Correct Answers & Qualitative Logic

Europium ions with different charges form precipitates with different chemical formulae and molar masses (e.g., EuOH, Eu(OH)₂, Eu(OH)₃).
Quantitative check: Equation 2 forms Eu(OH) ₂ with M = 186 g mol⁻¹, producing exactly 1.86 g of precipitate from 0.01 mol of Eu.

🧠 Exam Technique & Calculation Guidance

To prove Equation 2 using mass data:

Molar mass of precipitate = (Mass of precipitate) / 0.01

If Equation 2 is correct, molar mass = 1.86 / 0.01 = 186 g mol⁻¹, matching Eu(OH)₂ .

❌ Common Pitfalls

Failing to link the mass of the precipitate directly back to the moles of europium added initially (0.01 mol), or confusing hydroxide stoichiometry.

Marks available: 2 marks (1 for qualitative point on different molar masses/formulae, 1 for quantitative calculation or target mass like 1.86 g).

Topics

Module 2: Foundations in chemistry · Module 3: Periodic table and energy · Practical Activity Groups · PAG 1: Moles determination · 2.1 Atoms and reactions

Question and mark scheme from the OCR A-Level Chemistry examination, Unified chemistry (03), November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.