OCR A Level Chemistry A: Redox and electrode potentials (H432) – Revision Notes

These revision notes condense section 5.2.3, Redox and electrode potentials, of the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017. They cover outcomes 5.2.3(a) to (k), all in Module 5. The specification assigns Module 5 to Component 01 (Periodic table, elements and physical chemistry) and Component 03 (Unified chemistry). For full explanations and longer worked examples, read the study guide.

Use the OCR chemistry hub for every H432 topic and the printable checklist to track outcomes. When you are ready, try the practice questions or a free diagnostic. Oxidation number rules sit in Redox (2.1.5).

Key terms (5.2.3(a) and (f))

Term	Meaning
Oxidising agent	Electron acceptor; it is reduced and its oxidation number falls
Reducing agent	Electron donor; it is oxidised and its oxidation number rises
Half-equation	One reduction or one oxidation, with electrons shown
Standard hydrogen electrode	H₂(g) at 100 kPa over Pt, in 1 mol dm⁻³ H⁺(aq), at 298 K; E⦵ = 0.00 V by definition
Standard electrode potential, E⦵	e.m.f. of a half-cell connected to the standard hydrogen electrode under standard conditions (298 K, 100 kPa, 1 mol dm⁻³)
Standard cell potential, E⦵cell	E⦵(positive electrode) − E⦵(negative electrode)
Storage cell	Sealed cell; rechargeable types reverse the discharge reaction when charged
Fuel cell	Uses energy from a fuel reacting with oxygen to create a voltage; reactants supplied continuously

Memory hook: the agent is named for what it does to the other species, not what happens to itself.

Building equations (5.2.3(b))

Method in steps: half-equations (acid)

1. Balance the atom that changes.
2. Add H₂O to balance O.
3. Add H⁺ to balance H.
4. Add e⁻ to balance charge.
5. Scale both half-equations to the same electron count, add them, then cancel H⁺, H₂O or e⁻ that appear on both sides.

Method in steps: oxidation numbers

1. Mark the element whose oxidation number rises and the one whose oxidation number falls.
2. Multiply by the number of atoms changing.
3. Choose coefficients so total increase = total decrease.
4. Balance O with H₂O and H with H⁺ if needed, then check charge.

Worked reminder: in acid, Cr₂O₇²⁻ → Cr³⁺ gives Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O. Check: charge +6 on each side.

Predicting electron transfer (5.2.3(c))

• Electrons lost = electrons gained. Use mole ratios to find a change in oxidation number, then the product.
• If one species is both oxidised and reduced, it is disproportionation (for example Cu⁺ forming Cu²⁺ and Cu, met again in 5.3.1).

Redox titrations (5.2.3(d))

Feature	Fe²⁺ / MnO₄⁻	I₂ / S₂O₃²⁻
Burette	KMnO₄(aq)	Na₂S₂O₃(aq)
Flask	Fe²⁺(aq) + excess dilute H₂SO₄	I₂ released by adding the oxidising agent to excess KI(aq)
Indicator	None: MnO₄⁻ is self-indicating	Starch, added when the iodine colour is pale yellow
End point	First permanent pale pink	Blue-black colour just disappears
Key ratio	MnO₄⁻ : Fe²⁺ = 1 : 5	I₂ : S₂O₃²⁻ = 1 : 2
Watch out	No HCl (Cl⁻ is oxidised); read top of meniscus	Too little KI leaves oxidising agent unreacted

Equations to know how to use:

• MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O
• I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻

Titration calculations (5.2.3(e))

Method in steps

1. Choose concordant titres (within 0.10 cm³) and average them.
2. Moles in the burette solution = c × V (V in dm³).
3. Use the ratio from the equation. For a two-step iodometric method, chain the ratios: e.g. IO₃⁻ → 3I₂ → 6S₂O₃²⁻.
4. Scale up for the volumetric flask (25.0 cm³ from 250.0 cm³ is × 10).
5. Convert to the quantity asked for: concentration, mass, percentage purity, or a formula value such as x in a hydrate.

Worked reminder: an oxidising agent releases 2.40 × 10⁻⁴ mol of I₂ from excess KI. This needs 2 × 2.40 × 10⁻⁴ = 4.80 × 10⁻⁴ mol of S₂O₃²⁻. With 0.0200 mol dm⁻³ thiosulfate, the expected titre is 4.80 × 10⁻⁴ ÷ 0.0200 = 0.0240 dm³, or 24.00 cm³.

For a non-familiar system, the question gives you a half-equation or an equation. Build the ratio from it before touching the numbers.

Measuring cell potentials (5.2.3(g), PAG8)

Half-cell type	Electrode	Solution
Metal / metal ion	The metal, cleaned with emery paper	1 mol dm⁻³ metal ions
Non-metal / ion	Platinum, with gas at 100 kPa if the element is a gas	1 mol dm⁻³ ions
Two ions of one element	Platinum	Both ions at 1 mol dm⁻³, or equimolar

Join half-cells with a salt bridge (e.g. filter paper soaked in saturated KNO₃(aq)) and a high-resistance voltmeter.

Cell potentials and feasibility (5.2.3(h) and (i))

• E⦵cell = E⦵(more positive) − E⦵(less positive). Reduction happens at the more positive half-cell.
• A reaction is feasible if the oxidising agent’s half-equation has the more positive E⦵, so E⦵cell is positive.
• Never multiply E⦵ when you scale a half-equation.

Must-know distinctions

Feasible	Actually observed
Decided by E⦵ values (thermodynamics)	Also depends on rate (activation energy)
Assumes 1 mol dm⁻³, 100 kPa, 298 K	Real concentrations shift each half-cell equilibrium

Concentration rule: raising the concentration of a species on the left of a half-equation makes that electrode potential more positive; raising one on the right makes it less positive.

Storage and fuel cells (5.2.3(j) and (k))

• All the half-equations and potentials you need are supplied. Identify the more negative half-cell as the negative electrode, where oxidation happens on discharge.
• Overall equation: reverse the more negative half-equation, balance electrons, add.
• Rechargeable cell: charging reverses the overall discharge reaction.
• Lithium-based cells: benefits weighed against toxicity and fire risk.
• Fuel cell: the fuel is oxidised at the negative electrode; oxygen is reduced at the positive electrode. The cell runs while fuel and oxygen are supplied.

Quick self-test

1. Define a reducing agent in terms of electrons.
2. Cl₂ + 2e⁻ ⇌ 2Cl⁻ is +1.36 V and Br₂ + 2e⁻ ⇌ 2Br⁻ is +1.09 V. Will chlorine oxidise bromide ions? Give E⦵cell.
3. Write the half-equation for Cr₂O₇²⁻ forming Cr³⁺ in acid.
4. 25.0 cm³ of Fe²⁺(aq) reacts with 18.40 cm³ of 0.0150 mol dm⁻³ KMnO₄. Find [Fe²⁺].
5. A titration uses 21.60 cm³ of 0.0500 mol dm⁻³ Na₂S₂O₃. How many moles of I₂ reacted?
6. Why is starch added near the end of an iodine–thiosulfate titration?
7. Al³⁺ + 3e⁻ ⇌ Al is −1.66 V and Cu²⁺ + 2e⁻ ⇌ Cu is +0.34 V. Calculate E⦵cell.
8. In the cell from question 7, what happens to the cell potential if [Cu²⁺] is increased?
9. I₂ + 2e⁻ ⇌ 2I⁻ is +0.54 V and Fe³⁺ + e⁻ ⇌ Fe²⁺ is +0.77 V. Does iodine oxidise Fe²⁺?
10. Cu²⁺ + e⁻ ⇌ Cu⁺ is +0.15 V and Cu⁺ + e⁻ ⇌ Cu is +0.52 V. Is the disproportionation of Cu⁺ feasible?
11. Why does an Fe³⁺/Fe²⁺ half-cell need a platinum electrode?
12. State one reason a feasible reaction might not be seen.

Answers

1. A species that donates electrons (and is itself oxidised).
2. Yes. E⦵cell = 1.36 − 1.09 = +0.27 V.
3. Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O.
4. n(MnO₄⁻) = 2.76 × 10⁻⁴ mol; n(Fe²⁺) = 1.38 × 10⁻³ mol; [Fe²⁺] = 0.0552 mol dm⁻³.
5. n(S₂O₃²⁻) = 1.08 × 10⁻³ mol, so n(I₂) = 5.40 × 10⁻⁴ mol.
6. With a lot of iodine present, the starch–iodine complex can be slow to break up, blurring the end point. Added late, it gives a sharp end point.
7. 0.34 − (−1.66) = 2.00 V.
8. Cu²⁺ + 2e⁻ ⇌ Cu shifts right, the copper electrode becomes more positive, so the cell potential increases.
9. No. E⦵cell would be 0.54 − 0.77 = −0.23 V. The reverse, Fe³⁺ oxidising I⁻, is feasible.
10. Yes. 2Cu⁺ → Cu²⁺ + Cu; E⦵cell = 0.52 − 0.15 = +0.37 V.
11. Neither species is a solid that conducts; inert platinum transfers electrons between the ions and the circuit.
12. High activation energy (slow rate), or non-standard concentrations change the electrode potentials.

Where marks are usually lost

• Naming the element (“manganese”) instead of the agent (MnO₄⁻ or “manganate(VII) ions”).
• Leaving electrons in a combined equation, or H⁺ on both sides.
• Using HCl to acidify a manganate(VII) titration, or saying “add indicator”.
• Describing the thiosulfate end point as “turns blue-black” (that is when starch is added).
• Forgetting the dilution factor when only part of a volumetric flask is titrated.
• Using a 1 : 2 or 1 : 3 ratio in an iodate(V) titration instead of chaining to 1 : 6.
• Giving a cell potential with a negative sign for the positive electrode, or omitting “V”.
• Saying a reaction “will happen” because E⦵cell is positive, with no mention of rate or concentration.
• Using a metal electrode (such as iron) for a half-cell of two iron ions.

Official syllabus

OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, published by OCR. Section 5.2.3, Redox and electrode potentials.