OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2023: Question 21
10 marks · Hard difficulty · Structured Questions
Determine the concentration of iron(II) ions in a river water sample using titration with potassium manganate(VII) and explain the steps in modifying the experiment to determine total iron using redox electrode potentials.
Practise this questionQuestion
Question text
21 Some grass fertilisers contain compounds of iron.
During heavy rain, a fertiliser is washed into a nearby river causing the water to be polluted with a
mixture of iron(II) and iron(III) ions.
(a) A student determines the concentration of iron(II) ions in a sample of river water by titration
with potassium manganate(VII).
25.0 cm3 portions of river water are acidified with dilute sulfuric acid. Each portion is titrated
with 0.00250 mol dm–3 potassium manganate(VII) until a colour change is seen.
MnO –(aq) + 8H+(aq) + 5Fe2+(aq) Mn2+(aq) + 4H O(l) + 5Fe3+(aq)
(i) State the colour change seen at the end point of the titration.
from … to … [1]
(ii) The student’s titration results are shown in the table below.
The trial titre has been omitted.
12 3
Final volume / cm3 12.65 25.60 38.35
Initial volume / cm3 0.00 12.65 25.60
Titre volume / cm3
Complete the table above and calculate the mean titre that the student should use to
determine the concentration of iron(II) ions in the river water.
mean titre = … cm3 [2]
(iii) Determine the concentration, in mol dm–3, of iron(II) ions in the river water.
concentration = … mol dm–3 [3]
(b) The student modifies the experiment in (a) to determine the combined concentration of
iron(II) and iron(III) ions in the river water.
The student’s method is shown below.
Step 1 Add excess zinc to a 250.0 cm3 sample of river water and warm gently.
Step 2 Cool the solution and remove excess zinc by filtration.
Step 3 Acidify 25.0 cm3 portions of the filtrate from Step 2. Then titrate each portion
with 0.00250 mol dm–3 potassium manganate(VII) until a colour change is seen.
The table below shows information about three redox systems.
Redox system Half-equation E ө/ V
1 Zn2+(aq) + 2e– Zn(s) –0.76
2 Fe3+(aq) + e– Fe2+(aq) +0.77
3 MnO –(aq) + 8H+(aq) + 5e– Mn2+(aq) + 4H O(l) +1.51
Use the information in the table above to explain the reasons for Step 1 and Step 2.
Reason(s) for Step 1 …
Reason(s) for Step 2 …
[4]
Mark scheme
Show the mark scheme
AO
Question Answer Marks Guidance
element
21 (a) (i) Colourless to (pale) pink 1 AO1.1 ALLOW Pale purple
DO NOT ALLOW purple
(ii) 2 AO2.8 x2
12.65 12.95 12.75
✓
12.65+12.75 3
2 = 12.7(0) cm ✓
(iii) FIRST CHECK THE ANSWER ON ANSWER LINE 3 ALLOW ECF from incorrect titre in 21 a ii)
if answer = 6.35 × 10–3 award 3 marks for 3 marks
e.g. Titre of 12.78 cm3 gives 6.39 x 10-3
------------------------------------------------------------------------- AO2.8 --------------------------------------------------------
n(MnO –) in titration ALLOW 3 SF or more throughout
12.7
= (0.00250 1000 ) ALLOW ECF throughout
= 3.175 × 10–5 ✓
ALLOW
2+ 3 n(Fe2+) in 250 cm3 = 1.5875 × 10–3 (mol)
n(Fe ) in 25.0 cm
–5 so [Fe2+] in 25 cm3
= (3.175 × 10 5)
–4 = 1.5875 × 10–3 0.25 = 6.35 × 10–3
= 1.5875 × 10 (mol) ✓
Common errors for 2 marks
[Fe2+] = (1.5875 × 10–4 0.025)
2.46 x 10-2 (volumes transposed)
OR (1.5875 × 10–4 x 40)
1.25 x 10-2 (same volume used twice)
= 6.35 × 10–3 (mol dm–3) ✓
1.27 x 10-3 (no x 5)
2.54 x 10-4 ( 5)
(b) 4 AO3.1 ALLOW ORA throughout
26 1 IGNORE larger/smaller/greater/less
throughout
System 1/E⦵(Zn) is more negative/less positive than ALLOW Eo = (+)1.53(V)
system 2/ E⦵(Fe3+) ✓ AO3.4 ALLOW comparison if Fe system is
1 identified
Eqm 2 shifts to right AND Eqm 1 shifts to left
OR AO3.1
Zinc reduces iron(III) ions (to iron(II)) 1
OR
Zn + 2Fe3+ → Zn2+ + 2Fe2+ ✓ AO3.4
System 1/E⦵(Zn) is more negative than system 3/ ALLOW Eo = (+) 2.27(V)
E⦵(MnO –) ✓ ALLOW comparison if MnO - is identified
Eqm 3 shifts to right AND Eqm 1 shifts to left
OR
(If unfiltered), MnO – oxidise zinc
OR
2MnO – + 5Zn +16H+ → 2Mn2+ + 5Zn2+ + 8H O ✓
How to answer it
Redox Titrations and Electrode Potentials Study Guide
What this question tests
This multi-step question assesses your mastery of redox titrations, processing titration data to find mean titres, stoichiometry calculations involving moles and concentration, and applying standard electrode potentials (E°) to predict the feasibility of redox reactions and justify practical procedures.
Colour Change at the End Point
✅ Correct Answer
From colourless to (pale) pink
💡 Key Knowledge
Manganate(VII) ions (MnO₄⁻) are deep purple in solution. At the start of the titration, they react immediately with Fe²⁺ ions to form colourless Mn²⁺ and Fe³⁺. At the end point, a single drop of excess KMnO₄ provides an unreacted hint of purple/pink colour.
❌ Common Errors
Do not write "purple" as the final colour, as this indicates a massive over-titration. Do not write "clear" instead of "colourless".
Processing Titration Data
✅ Correct Answer
Completed Table: Titre 2 = 12.95 cm³
Mean Titre: 12.70 cm³ (using concordant titres 1 and 3: 12.65 and 12.75)
🧠 Exam Technique
Always calculate individual titres first: Final minus Initial.
• Titre 1: 12.65 - 0.00 = 12.65
• Titre 2: 25.60 - 12.65 = 12.95
• Titre 3: 38.35 - 25.60 = 12.75
Select only concordant titres (within 0.10 cm³ of each other) to calculate the mean. Titres 1 and 3 average to 12.70 cm³.
Calculating Concentration of Iron(II) Ions
📐 Step-by-Step Calculation
- Moles of MnO₄⁻ in titration:
0.0250 mol dm⁻³ × (12.70 ÷ 1000) = 3.175 × 10⁻⁵ mol - Moles of Fe²⁺ in 25.0 cm³ portion:
Using the 1:5 stoichiometry from the equation: 3.175 × 10⁻⁵ × 5 = 1.5875 × 10⁻⁴ mol - Concentration of Fe²⁺ in mol dm⁻³:
Concentration = Moles ÷ Volume (in dm³) = 1.5875 × 10⁻⁴ ÷ 0.0250 = 6.35 × 10⁻³ mol dm⁻³
❌ Common Calculation Traps
- Forgetting to multiply by the 1:5 stoichiometric ratio.
- Dividing by 1000 incorrectly or messing up volume scale factors.
- Using the wrong volume for the final concentration step.
Explaining Experimental Modifications Using E° Values
💡 Key Knowledge: Redox Feasibility
A reaction is feasible if the system with the more negative E° goes backwards (oxidation/loss of electrons) and the system with the more positive E° goes forwards (reduction/gain of electrons).
✅ Correct Answer: Reason for Step 1
System 1 (Zn) has a more negative E° (−0.76 V) than System 2 (Fe³⁺/Fe²⁺, +0.77 V). Therefore, zinc reduces iron(III) to iron(II) ( Zn + 2Fe³⁺ → Zn²⁺ + 2Fe²⁺ ).
✅ Correct Answer: Reason for Step 2
Unreacted solid zinc must be removed by filtration so it does not subsequently react with or reduce the potassium manganate(VII) added in Step 3 (or reduce Fe³⁺ further, or consume the titrant).
🧠 Top-Level Response Tips
Always explicitly compare the specific E° values by referencing their numerical values or stating clearly which system is "more negative / less positive" before stating the direction of equilibrium shifts.
Topics
Module 2: Foundations in chemistry · Module 5: Physical chemistry and transition elements · Practical Activity Groups · PAG 2: Acid-base titration · 2.1 Atoms and reactions · 5.2 Energy
Question and mark scheme from the OCR A-Level Chemistry examination, Periodic table, elements and physical chemistry (01), June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.