OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), November 2020: Question 19
9 marks · Hard difficulty · Structured Questions
Use standard electrode potentials to draw an electrochemical cell, explain redox observations involving methanal and manganate(VII), and calculate the standard electrode potential for an oxygen fuel cell half-equation.
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Question text
19 Standard electrode potentials for four redox systems are shown in Table 19.1.
Redox ө
Half-equation E / V
system
1 CO (g) + 2H+(aq) + 2e− HCOOH(aq) −0.11
2 HCOOH(aq) + 2H+(aq) + 2e− HCHO(aq) + H O(l) −0.03
3 Ag+(aq) + e− Ag(s) +0.80
4 MnO −(aq) + 8H+(aq) + 5e− Mn2+(aq) + 4H O(l) +1.51
Table 19.1
(a) A student sets up a standard cell in the laboratory based on redox systems 3 and 4.
Draw a labelled diagram to show how this cell could be set up to measure its standard cell
potential at 298 K.
[3]
(b) A student warms a mixture of methanal, HCHO, and acidified potassium manganate(VII).
The student observes gas bubbles.
Explain this observation in terms of electrode potentials and equilibria.
Include overall equations in your answer.
… [4]
(c) Methanoic acid, HCOOH, can be used in a fuel cell. As with all fuel cells, the fuel (HCOOH) is
supplied at one electrode and the oxidant (oxygen) at the other electrode.
The standard cell potential for this fuel cell is 1.34 V.
The overall reaction is shown below.
HCOOH + ½O2 H2O + CO2
Using the information in Table 19.1, deduce the half-equation for the reaction at the oxygen
electrode, and calculate the standard electrode potential for the oxygen half-cell.
half-equation …
standard electrode potential = … V
[2]
Mark scheme
Show the mark scheme
AO
Question Answer Marks Guidance
element
19 (a) Circuit 3 Voltmeter must be shown AND salt bridge must be
Complete circuit AND voltmeter 3.4× 1 labelled
AND salt bridge linking two half-cells ALLOW small gaps in circuit
Half cells
Ag AND Ag+ AND 1 mol dm–3 solution 1.2× 1 If species in BOTH half cells are correct but
concentration of 1 mol dm–3 omitted,
Pt AND H+ AND MnO4- AND Mn2+ ALLOW 1 mark for BOTH half cells.
AND 1 mol dm–3 /equimolar solution 1.2×1
ALLOW acidified as an alternative for H+
IGNORE stated pressure
Not relevant here as no gas
(b) Comparison of E values 4 3.1× 2 IGNORE higher/lower
E of redox system 4 (MnO4- /Mn2+)is more
positive/less negative than E of redox systems 2 ALLOW Overall Ereaction = (+)1.54V OR (+)1.62V
(HCOOH/HCHO) OR 1 (CO2/HCOOH)
Equilibrium shift related to E values
More negative/less positive/system 2 For ‘shifts left’, ALLOW ‘is oxidised’
(HCOOH/HCHO) OR system 1 (CO2/HCOOH) OR ‘electrons are lost ’ OR ‘reducing agent’
shifts left
OR For ‘shifts right’, ALLOW ‘is reduced’
Less negative/more positive/system 4 (MnO4- /Mn2+) OR ‘electrons are gained’ OR ‘oxidising agent’
shifts right
IGNORE state symbols
• 2 and 4 ALLOW multiples
2MnO – + 5HCHO + 6H+ → 2Mn2+ + 5HCOOH + 3H O 3.2× 2 DO NOT ALLOW un-cancelled species, e.g. H+,
• 1 and 4 on both sides
2MnO4– + 5HCOOH + 6H+ → 2Mn2+ + 5CO2 + 8H2O ALLOW for 1 mark two balanced equations with
uncancelled species.
ALLOW combined equation for 2 marks:
4MnO4– + 5HCHO + 12H+ → 4Mn2+ + 5CO2 +11H2O
AO
element
(c) 2H+ + ½O2 +2e– → H2O 16 2 2.6 IGNORE state symbols
1.34 + (–0.11) = (+)1.23 (V) 2.2×1 ALLOW multiples
Total 9
How to answer it
OCR A-Level Chemistry: Standard Electrode Potentials
What this question tests
This multi-step question assesses your mastery of electrochemical cells and redox chemistry. You are tested on your ability to draw standard laboratory cells with correct components, apply standard electrode potential values (E°) to predict the direction of electron flow and equilibrium shifts, construct balanced overall redox equations, and manipulate cell potential calculations involving fuel cells.
Drawing a Standard Electrochemical Cell
A student sets up a standard cell based on redox systems 3 (Ag⁺/Ag, E° = +0.80 V) and 4 (MnO₄⁻/Mn²⁺, E° = +1.51 V).
✅ Correct Answer Requirements
- Circuit: Complete electrical circuit containing a high-resistance voltmeter.
- Salt Bridge: Clearly labelled salt bridge connecting the two half-cells (typically filter paper soaked in KNO₃(aq)).
- Half-cell 3 (Silver): Silver (Ag) metal electrode immersed in 1.0 mol dm⁻³ Ag⁺(aq) solution.
- Half-cell 4 (Manganate): Platinum (Pt) inert electrode in a solution containing 1.0 mol dm⁻³ MnO₄⁻, 1.0 mol dm⁻³ Mn²⁺, and H⁺(aq) (acidified).
💡 Key Knowledge
- Inert Electrodes: Platinum is mandatory when both the oxidized and reduced species are aqueous ions (e.g., MnO₄⁻ and Mn²⁺) because the ions cannot act as a solid electrical conductor.
- Concentrations: Standard conditions require all aqueous ions to be at 1.0 mol dm⁻³ .
❌ Common Errors
- Omitting the platinum electrode for system 4 and trying to dip a wire straight into the solution.
- Forgetting to state concentrations ( 1.0 mol dm⁻³ ) or labelling the salt bridge.
- Drawing the voltmeter bridging directly across the solutions instead of wiring it to both solid electrodes.
Predicting Redox Reactions & Gas Evolution
A student warms methanal (HCHO) with acidified potassium manganate(VII). Gas bubbles are observed. Explain this using electrode potentials and equilibria, including overall equations.
✅ Correct Answer Requirements
- Comparison: Note that E° of system 4 (MnO₄⁻/Mn²⁺ = +1.51 V) is more positive than systems 1 (CO₂/HCOOH = -0.11 V) and 2 (HCOOH/HCHO = -0.03 V).
- Equilibrium Shift: Systems 1 and/or 2 have more negative E° values, so their equilibria shift to the left (releasing electrons / getting oxidised). System 4 shifts to the right.
- Overall Equations:
With HCHO (system 2): 2MnO₄⁻ + 5HCHO + 6H⁺ → 2Mn²⁺ + 5HCOOH + 3H₂O
With HCOOH (system 1): 2MnO₄⁻ + 5HCOOH + 6H⁺ → 2Mn²⁺ + 5CO₂ + 8H₂O
(Gas bubbles = CO₂ gas evolved from complete oxidation).
🧠 Exam Technique
- Always explicitly state which E° value is more positive/negative before explaining equilibrium shifts.
- Make sure equations are balanced for atoms AND charges. Check that electrons cancel out completely (multiply half-equations to match electrons: 5 electrons for MnO₄⁻ vs 2 electrons for organic systems means a 2:5 scaling factor).
❌ Common Errors
- Using vague phrasing like "manganate has a higher potential" instead of "more positive E° value".
- Failing to cancel out spectator species or leaving un-cancelled H⁺ ions on both sides of the final equation.
Fuel Cell Half-Equations & Cell Calculations
Methanoic acid fuel cell has an overall cell potential of 1.34 V. Overall reaction: HCOOH + ½O₂ → H₂O + CO₂ . Deduce the oxygen half-equation and calculate the oxygen half-cell potential.
📐 Step-by-Step Calculation
- Deduce Oxygen Half-Equation: Oxygen acts as the oxidant in acidic conditions:
2H⁺ + ½O₂ + 2e⁻ ⇌ H₂O (or multiples). - Identify Fuel Half-Equation from Table 19.1:
System 1: CO₂ + 2H⁺ + 2e⁻ ⇌ HCOOH with E° = -0.11 V. - Apply Cell Potential Formula:
Cell Potential = E°(cathode) - E°(anode)
1.34 = E°(oxygen) - (-0.11)
E°(oxygen) = 1.34 - 0.11 = +1.23 V (Note: standard tabulated oxygen half-cell value confirms this).
❌ Calculation Traps & Sign Errors
- Sign Subtraction Trap: Subtracting a negative number becomes an addition. Watch your signs carefully: 1.34 = E° - (-0.11) rearranges to 1.34 - 0.11 = 1.23 V .
- Stoichiometry: Ensure half-equations balance electrons before combining potentials (do not multiply E° values when scaling half-equations!).
Topics
Module 5: Physical chemistry and transition elements · Module 6: Organic chemistry and analysis · Practical Activity Groups · 5.2 Energy · 6.1 Aromatic compounds, carbonyls and acids · PAG 8: Electrochemical cells
Question and mark scheme from the OCR A-Level Chemistry examination, Periodic table, elements and physical chemistry (01), November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.