OCR A-Level Chemistry Unified chemistry (03), June 2018: Question 2
13 marks · Hard difficulty · Structured Questions
Investigate the decomposition of hydrogen peroxide, calculate initial concentration from gas volume, write redox equations using electrode potentials, and explain cobalt(II) complex equilibria and color changes.
Practise this questionQuestion
Question text
2 This question looks at reactions of hydrogen peroxide and of cobalt(II) ions.
(a) Aqueous hydrogen peroxide decomposes as shown in equation 2.1.
2H2O2(aq) → 2H2O(l) + O2(g) Equation 2.1
The reaction is catalysed by manganese(IV) oxide, MnO2.
A student investigates the decomposition of a hydrogen peroxide solution as outlined below.
• The student adds 50.00 cm3 of H O (aq) to a conical flask.
• The student adds a small spatula measure of MnO2 and quickly connects the flask
to a gas syringe.
• The student measures the volume of oxygen every 200 seconds.
Results
Time/s Volume of O /cm3
200 15
400 28
600 36
800 41
1000 46
1200 48
1400 50
(i) Process the results as outlined below.
• On page 5, plot a graph of volume of O2 against time.
• Use your graph to find the rate of the reaction, in cm3 s–1, at t = 500 s.
Show your working on the graph and in the space below.
rate = … cm3 s–1 [5]
(ii) The student allows the reaction in equation 2.1 to proceed until no more gas is evolved.
The volume of O in the syringe is now 55 cm3, measured at RTP.
Calculate the initial concentration of the H2O2.
Give your answer to two significant figures.
initial concentration of H O = … mol dm–3 [3]
(b) Hydrogen peroxide can act as an oxidising agent or as a reducing agent.
Some standard electrode potentials are shown below.
2H+(aq) + O (g) + 2e– H O (aq) E o = +0.68 V
22 2
H O (aq) + 2H+(aq) + 2e– 2H O(l) E o = +1.77 V
22 2
VO2+(aq) +2H+(aq) + e– V3+(aq) + H O(l) E o = +0.34 V
MnO –(aq) + 8H+(aq) + 5e– Mn2+(aq) + 4H O(l) E o = +1.51 V
Use this information to write an equation for a reaction in which hydrogen peroxide acts as a
reducing agent.
… [2]
(c) Cobalt(II) forms complex ions with water ligands and with chloride ligands.
• With water ligands, cobalt(II) forms a pink octahedral complex ion, [Co(H O) ]2+.
• With chloride ligands, cobalt(II) forms a blue tetrahedral complex ion.
A student dissolves cobalt(II) sulfate in water in a boiling tube. A pink solution forms.
Experiment 1
The student places the boiling tube in a water bath at 100 °C.
Concentrated hydrochloric acid is added dropwise.
The colour of the solution changes from pink to blue.
Experiment 2
The student places the boiling tube from experiment 1 in an ice/water bath at 0 °C.
The colour of the solution changes from blue to pink.
(i) Write the equilibrium equation for the reaction that takes place when the colour of the
solution changes.
… [1]
(ii) Explain the observations and predict whether the formation of the blue colour is
exothermic or endothermic.
… [2]
Mark scheme
Show the mark scheme
Question Answer Marks Guidance
2 (a) Graph 5
Graph of volume (y axis) against time (x axis)
AND Axes labelled with correct units
AND At least half graph paper in both directions
AND Linear scales
Points
7 points from 200–1400 s plotted
Point at 0,0 not required
Line
Curve drawn through origin (0,0)
AND
Curve not drawn with straight lines between points.
Rate
Attempted tangent on graph drawn to curve at
t = 500 ± 100 s
Rate calculated in range 0.037–0.047 (cm3 s–1)
50 – 11
e.g. for graph in guidance: = 0.042
920 – 0
-------------------------------------------------------------------------- ALLOW V OR Vol for volume
For tangents not drawn at 500 ± 100 s, ALLOW t for time
ALLOW ECF ONLY for a tangent drawn to the For ‘s’, ALLOW sec, seconds, etc
candidate’s line.
CARE:
Then calculate the gradient from candidate’s tangent.
Use of x and y coordinates at t = 500 s scores zero,
For inverse graphs of time against volume, 3
e.g. For volume = 33 cm and time = 500 s,
Graph mark will not be scored.
x and y coordinates gives 33/500 = 0.066
All other marks are available.
BUT rate = 1/ gradient = 0.037–0.047 (cm3 s–1)
(ii) FIRST CHECK THE ANSWER ON ANSWER LINE 8 3 ALLOW ECF throughout
If answer = 0.092 (mol dm–3) award 3 marks
------------------------------------------------------------------------- ALLOW 2 SF up to calculator value of
55 2.291666667 10–3
n(O ) = = 2.29 10–3 (mol)
2 24000
ALLOW calculation using ideal gas equation provided that
p = ~ 105 Pa and T in range 293–298 K.
ALLOW use of 8.31 for R (gives same answer)
1 105 55 10–6
e.g. n(O ) = = 2.22 10–3 (mol)
2 8.314 298
n(H O ) = 2.29 10–3 2 = 4.58 10–3 (mol)
n(H O ) = 2.22 10–3 2 = 4.44 10–3 (mol)
4.58 10–3 1000
[H O ] = = 0.092 (mol dm–3)
22 50.0 –3
4.44 10 1000 –3
(2 SF) [H2O2] = = 0.089 (mol dm )
50.0
(2 SF)
NOTE: 293 K gives 0.090 (mol dm–3)
Common errors
0.046 2 marks no 2 for n(H2O2)
(b) 2MnO – + 5H O + 6H+ → 2Mn2+ + 8H O + 5O 2 ALLOW multiples
42 2 2 2
ALLOW ⇌ instead of sign
Correctly balanced equation for MnO –/H O reaction but no
42 2
cancelling of H+ and/or e–
ALLOW 1 mark for final equation with correct balancing
numbers AND
Overall equation correct with all species cancelled ONE small slip in a formula OR charge
IGNORE annotations around equations,
i.e. treat as rough working
ALLOW 1 mark for: 2H2O2 2H2O + O2
(H2O2 is acting as both reducing and oxidising agent)
(c) (i) Equation 1 ALLOW reverse equation:
[Co(H O) ]2+ + 4Cl– ⇌ [CoCl ]2– + 6H O [CoCl ]2– + 6H O ⇌ [Co(H O) ]2+ + 4Cl–
26 4 2 4 2 2 6
OR [Co(H O) ]2+ + 4HCl ⇌ [CoCl ]2– + 6H O + 4H+ 9 but take care for subsequent explanations
26 4 2
IGNORE state symbols (even if wrong)
For [CoCl ]2– ,
ALLOW CoCl 2–, (CoCl )2–
For other representations, contact TL
(ii) 2 Mark independently
Equilibrium shift
equilibrium (shifts) to right at high temperature/100ºC ALLOW suitable alternatives for ‘to right’ e.g. towards
OR equilibrium shifts to left at low temperature/0ºC products OR in forward direction OR ‘favours the right’
ORA for ‘to left’
CARE: Direction of shift depends on direction of equilibrium
equation from 2c(i). Either look back or see the equation Temperature required but
copied at bottom of 2c(ii) marking zone. ALLOW ‘in ice for low temperature
OR ‘in boiling/hot water’ for high temperature
IGNORE shift to blue side or pink side
------------------------------------------------------------------- -------------------------------------------------
Enthalpy change
Endothermic
Total 13
How to answer it
Kinetics, Redox Potentials & Transition Elements Study Guide
This comprehensive multi-topic question assesses your ability to process experimental rate data by constructing graphs and drawing tangents, calculate concentrations using molar gas volumes, combine half-equations to form overall redox equations, and apply Le Chatelier's principle to transition metal ligand substitution equilibria and enthalpy changes.
Processing Experimental Results into a Rate Value
✅ Mark Scheme Requirements (5 Marks)
- Graph layout: Volume of O₂ (y-axis) against time (x-axis), axes correctly labelled with units ( cm³ and s ).
- Scale & Size: Linear scales covering at least half of the grid in both directions; 7 points plotted accurately.
- Line: Smooth curve passing through the origin ( 0, 0 ) without connecting data points with straight lines.
- Tangent: A clear tangent drawn precisely at t = 500 s (tolerance ± 100 s ).
- Calculation: Gradient calculated correctly, yielding a final rate value between 0.037 and 0.047 cm³ s⁻¹ .
🧠 Exam Technique & Tangent Construction
- Drawing Tangents: Use a sharp pencil and a transparent ruler. Ensure the line touches the curve at exactly t = 500 s and has a balanced slope matching the curve.
- Large Triangles: Pick coordinates far apart along your tangent line to minimise percentage errors when calculating the gradient ( Change in y / Change in x ).
❌ Common Errors to Avoid
- Reading coordinates directly off a single point on the curve at t = 500 s (doing y / x ) instead of calculating the gradient of a tangent.
- Connecting plotted points with point-to-point straight lines instead of drawing a smooth curve of best fit.
- Failing to start the curve at the origin ( 0, 0 ).
Calculating Initial Concentration of Hydrogen Peroxide
📐 Step-by-Step Calculation
- Find moles of O₂ gas evolved:
Using molar gas volume at RTP ( 24000 cm³ mol⁻¹ ):
n(O₂) = 55 / 24000 = 2.2917 × 10⁻³ mol - Use reacting ratios from Equation 2.1:
Equation shows 2H₂O₂(aq) → 2H₂O(l) + O₂(g ). Ratio of H₂O₂ : O₂ is 2 : 1 .
n(H₂O₂) = 2.2917 × 10⁻³ × 2 = 4.5833 × 10⁻³ mol - Calculate concentration in mol dm⁻³:
Volume of solution = 50.0 cm³ ( = 0.050 dm³ ).
[H₂O₂] = (4.5833 × 10⁻³ / 50.0) × 1000 = 0.09166... mol dm⁻³ - Apply Significant Figures:
Rounded to 2 significant figures as requested: 0.092 mol dm⁻³ .
Writing Combined Redox Equations
✅ Correct Answer (2 Marks)
2MnO₄⁻(aq) + 5H₂O₂(aq) + 6H⁺(aq) → 2Mn²⁺(aq) + 8H₂O(l) + 5O₂(g)
(Accept reverse equations or valid multiples, and equilibrium arrows instead of forward arrows).
💡 Key Knowledge: Using Electrode Potentials
- To find a reaction where hydrogen peroxide acts as a reducing agent, look for a half-equation where H₂O₂ is oxidised to O₂ (producing electrons on the right).
- Pair this with a species having a more positive E° value that acts as the oxidising agent (here, manganate(VII), MnO₄⁻ ).
Cobalt Complex Equilibria and Enthalpy Changes
✅ (i) Equilibrium Equation (1 Mark)
[Co(H₂O)₆]²⁺(aq) + 4Cl⁻(aq) ⇌ [CoCl₄]²⁻(aq) + 6H₂O(l)
State symbols are ignored. Reverse equations are fully accepted.
✅ (ii) Observations & Enthalpy (2 Marks)
- Equilibrium shift: Shifts to the right at high temperature ( 100 °C / boiling water) to form the blue complex.
- Enthalpy change: The forward reaction is endothermic. Raising the temperature favours the endothermic direction.
🧠 Examiner Strategy & Consistency
- Make sure your explanation in part (ii) matches the exact equation written in part (i)! If you wrote the reverse equation, your explanation of "right" vs "left" must align logically with the products you defined.
Topics
Module 3: Periodic table and energy · Module 5: Physical chemistry and transition elements · Practical Activity Groups · 3.2 Physical chemistry · 5.1 Rates, equilibrium and pH · 5.2 Energy · 5.3 Transition elements · PAG 9: Rates of reaction – continuous monitoring method
Question and mark scheme from the OCR A-Level Chemistry examination, Unified chemistry (03), June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.