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

This study guide teaches 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, covering outcomes 5.2.3(a) to (k). It sits in Module 5, which the specification assigns to Component 01 (Periodic table, elements and physical chemistry) and Component 03 (Unified chemistry). PAG8, electrochemical cells, is linked to 5.2.3(g).

See the OCR chemistry hub; tick off outcomes on the printable checklist. Oxidation number rules are in the Redox (2.1.5) study guide. Short on time? Try a free diagnostic first.

What 5.2.3 asks of you

Outcome	What you must be able to do	Guidance in the specification
5.2.3(a)	Explain and use oxidising agent and reducing agent	See also 2.1.5
5.2.3(b)	Build redox equations from half-equations and oxidation numbers	M0.2
5.2.3(c)	Interpret and predict electron-transfer reactions	–
5.2.3(d)	Techniques for redox titrations, including Fe²⁺/MnO₄⁻ and I₂/S₂O₃²⁻	HSW4
5.2.3(e)	Titration calculations, structured and non-structured: (i) Fe²⁺/MnO₄⁻ and I₂/S₂O₃²⁻, (ii) non-familiar systems	M0.1–M2.4
5.2.3(f)	Use standard electrode potential, E⦵, measured with a hydrogen electrode	E⦵ data provided on papers
5.2.3(g)	Measure cell potentials of (i) metals or non-metals with their ions, (ii) ions of one element in two oxidation states, with Pt	Same-element ions at 1 mol dm⁻³ or equimolar; PAG8
5.2.3(h)	Calculate a standard cell potential from two E⦵ values	–
5.2.3(i)	Predict feasibility from cell potentials; limits (kinetics, concentration)	M0.3; HSW6
5.2.3(j)	Apply electrode potentials to modern storage cells	Details and data supplied; HSW9 risks of Li-based cells
5.2.3(k)	Explain how a fuel cell creates a voltage and the change at each electrode	No recall of equations; data supplied

5.2.3(a): oxidising and reducing agents

An oxidising agent takes electrons from another species. It is reduced, so its oxidation number falls. A reducing agent gives electrons to another species. It is oxidised, so its oxidation number rises.

Worked example 1

Identify the agents in 2Fe³⁺(aq) + Sn²⁺(aq) → 2Fe²⁺(aq) + Sn⁴⁺(aq).

Iron goes from +3 to +2 (reduced); tin goes from +2 to +4 (oxidised). So Fe³⁺ is the oxidising agent and Sn²⁺ is the reducing agent.

5.2.3(b): building redox equations

Method 1: half-equations

1. Balance the changing atom, then O with H₂O, then H with H⁺.
2. Balance charge with electrons.
3. Equalise electrons, add, and cancel species on both sides.

Worked example 2: manganate(VII) and hydrogen peroxide

Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O Oxidation: H₂O₂ → O₂ + 2H⁺ + 2e⁻

Match ten electrons (×2 and ×5), add, and cancel 10H⁺: 2MnO₄⁻ + 6H⁺ + 5H₂O₂ → 2Mn²⁺ + 5O₂ + 8H₂O. Check: O is 18 on each side; charge is +4 on each side.

Method 2: oxidation numbers

Make the total increase equal the total decrease.

Worked example 3: iodate(V) and sulfite

IO₃⁻ oxidises SO₃²⁻ to sulfate and is reduced to iodide.

• I: +5 → −1, a decrease of 6.
• S: +4 → +6, an increase of 2.
• Three S atoms balance one I: IO₃⁻ + 3SO₃²⁻ → I⁻ + 3SO₄²⁻.

Oxygen (12) and charge (−7) already balance.

5.2.3(c): interpreting and predicting electron transfer

Electrons lost by the reducing agent equal electrons gained by the oxidising agent, so you can predict an unfamiliar product.

Worked example 4: how far does zinc reduce vanadium?

In acid, 3.00 × 10⁻³ mol of zinc reduces 2.00 × 10⁻³ mol of VO₂⁺ (vanadium +5). Assume vanadium gains every electron zinc loses.

1. Zn → Zn²⁺ + 2e⁻, so electrons released = 2 × 3.00 × 10⁻³ = 6.00 × 10⁻³ mol.
2. Electrons per vanadium = 6.00 × 10⁻³ ÷ 2.00 × 10⁻³ = 3.
3. New oxidation number = +5 − 3 = +2, so the product is V²⁺.

5.2.3(d): redox titration techniques

Manganate(VII) and iron(II)

• Put potassium manganate(VII) in the burette and the Fe²⁺ solution, measured by pipette, in the conical flask.
• Acidify the flask with excess dilute sulfuric acid. Do not use hydrochloric acid: manganate(VII) oxidises chloride ions (see worked example 9), so the titre would be too large.
• No indicator is added. Purple MnO₄⁻ is decolourised as it is reduced to almost colourless Mn²⁺. The end point is the first permanent pale pink.
• The dark solution hides the bottom of the meniscus, so read the top consistently.

Iodine and thiosulfate

• The oxidising agent is added to excess potassium iodide, releasing yellow-brown iodine.
• Titrate with sodium thiosulfate: I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻.
• At pale yellow, add starch (blue-black). The end point is when the blue-black colour just disappears.
• Adding starch near the end gives a sharper end point than adding it at the start.

Average only concordant titres (within 0.10 cm³).

5.2.3(e): titration calculations

Worked example 5 (structured, Fe²⁺/MnO₄⁻): iron tablets

Four iron tablets are dissolved in dilute sulfuric acid to give 250.0 cm³ of solution. Samples of 25.0 cm³ react with 0.00800 mol dm⁻³ KMnO₄. Concordant titres: 20.85, 20.95 and 20.90 cm³. The ionic equation is MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O.

1. Mean titre = 20.90 cm³.
2. n(MnO₄⁻) = 0.00800 × 20.90 ÷ 1000 = 1.672 × 10⁻⁴ mol.
3. n(Fe²⁺) in 25.0 cm³ = 5 × 1.672 × 10⁻⁴ = 8.36 × 10⁻⁴ mol.
4. In 250.0 cm³: 8.36 × 10⁻³ mol, so per tablet = 2.09 × 10⁻³ mol.
5. Mass of iron per tablet = 2.09 × 10⁻³ × 55.8 = 0.117 g, about 117 mg.

Worked example 6 (structured, I₂/S₂O₃²⁻): standardising thiosulfate

0.535 g of potassium iodate(V), KIO₃ (M = 214.0 g mol⁻¹), is made up to 250.0 cm³. A 25.0 cm³ portion is added to excess acidified KI:

IO₃⁻ + 5I⁻ + 6H⁺ → 3I₂ + 3H₂O

The iodine needs a mean titre of 24.60 cm³ of sodium thiosulfate.

1. n(KIO₃) in 25.0 cm³ = 0.535 ÷ 214.0 ÷ 10 = 2.50 × 10⁻⁴ mol.
2. IO₃⁻ gives 3I₂, which needs 6S₂O₃²⁻: ratio 1 : 6.
3. n(S₂O₃²⁻) = 6 × 2.50 × 10⁻⁴ = 1.50 × 10⁻³ mol.
4. Concentration = 1.50 × 10⁻³ ÷ 0.02460 = 0.0610 mol dm⁻³.

Worked example 7 (non-structured, non-familiar system)

10.0 cm³ of a hydrogen peroxide solution is diluted to 250.0 cm³. A 25.0 cm³ portion, acidified, needs a mean titre of 17.65 cm³ of 0.0200 mol dm⁻³ KMnO₄. Find the concentration of the original solution.

Plan first: equation (worked example 2 gives MnO₄⁻ : H₂O₂ = 2 : 5), moles, then dilution.

1. n(MnO₄⁻) = 0.0200 × 17.65 ÷ 1000 = 3.53 × 10⁻⁴ mol.
2. n(H₂O₂) in 25.0 cm³ = 3.53 × 10⁻⁴ × 5 ÷ 2 = 8.825 × 10⁻⁴ mol.
3. In 250.0 cm³, which all came from 10.0 cm³: 8.825 × 10⁻³ mol.
4. Original concentration = 8.825 × 10⁻³ ÷ 0.0100 = 0.883 mol dm⁻³.

5.2.3(f): standard electrode potential

Only the voltage between two half-cells can be measured. The reference is the standard hydrogen electrode: platinum in 1 mol dm⁻³ H⁺(aq), with hydrogen gas at 100 kPa passed over it, at 298 K. Its potential is defined as 0.00 V.

The standard electrode potential, E⦵, of a half-cell is the e.m.f. measured when that half-cell is connected to the standard hydrogen electrode under standard conditions: 298 K, 100 kPa and 1 mol dm⁻³ solutions. Half-equations are written as reductions, for example Ag⁺(aq) + e⁻ ⇌ Ag(s), E⦵ = +0.80 V.

More positive E⦵: the left-hand species is a stronger oxidising agent. More negative E⦵: the right-hand species is a stronger reducing agent.

5.2.3(g): measuring cell potentials (PAG8)

Connect two half-cells with a salt bridge (filter paper soaked in saturated KNO₃(aq)) and a high-resistance voltmeter, so almost no current flows.

• Metal with its ions: clean the metal with emery paper and dip it in a 1 mol dm⁻³ solution of its ions, for example lead in Pb²⁺(aq).
• Non-metal with its ions: use platinum, for example chlorine at 100 kPa bubbled over Pt in 1 mol dm⁻³ Cl⁻(aq).
• Ions of one element in two oxidation states: inert platinum dips into a solution of both ions, for example Fe³⁺ and Fe²⁺, each 1 mol dm⁻³ or equimolar.

The specification’s example PAG8 activity investigates how concentration affects a cell’s potential.

5.2.3(h): standard cell potential

E⦵cell = E⦵(positive electrode) − E⦵(negative electrode)

The half-cell with the more positive E⦵ is the positive electrode, where reduction happens.

Worked example 8

Fe³⁺(aq) + e⁻ ⇌ Fe²⁺(aq), E⦵ = +0.77 V; Pb²⁺(aq) + 2e⁻ ⇌ Pb(s), E⦵ = −0.13 V.

1. Positive electrode: Pt in the Fe³⁺/Fe²⁺ half-cell.
2. E⦵cell = +0.77 − (−0.13) = 0.90 V.
3. Fe³⁺ is reduced and lead is oxidised: 2Fe³⁺ + Pb → 2Fe²⁺ + Pb²⁺.

5.2.3(i): predicting feasibility and its limits

A reaction is feasible when the oxidising agent sits in a half-equation with a more positive E⦵ than the half-equation containing the reducing agent, so E⦵cell is positive.

Worked example 9: can acidified manganate(VII) oxidise chloride ions?

MnO₄⁻ + 8H⁺ + 5e⁻ ⇌ Mn²⁺ + 4H₂O, E⦵ = +1.51 V; Cl₂ + 2e⁻ ⇌ 2Cl⁻, E⦵ = +1.36 V.

E⦵cell = 1.51 − 1.36 = +0.15 V, so it is feasible: 2MnO₄⁻ + 16H⁺ + 10Cl⁻ → 2Mn²⁺ + 5Cl₂ + 8H₂O. This is why hydrochloric acid is not used in manganate(VII) titrations.

Limitations

• Kinetics: E⦵ values predict feasibility, not rate. A feasible reaction with a high activation energy can be too slow to notice. MnO₄⁻ (+1.51 V) should oxidise water (O₂/H₂O, +1.23 V), yet acidified manganate(VII) is stable enough for titrations because this reaction is very slow.
• Concentration: E⦵ applies only at 1 mol dm⁻³. Raising [Fe³⁺] shifts Fe³⁺ + e⁻ ⇌ Fe²⁺ to the right, making its potential more positive, so the cell in worked example 8 would exceed 0.90 V. When two E⦵ values are close, a concentration change can reverse the prediction.

5.2.3(j): storage cells

Storage cells hold their chemicals sealed inside; charging a rechargeable cell reverses the discharge reactions. You are given the half-equations and data.

Worked example 10: nickel–cadmium cell

Use these values: Cd(OH)₂ + 2e⁻ ⇌ Cd + 2OH⁻, E⦵ = −0.81 V; NiO(OH) + H₂O + e⁻ ⇌ Ni(OH)₂ + OH⁻, E⦵ = +0.49 V.

1. E⦵cell = 0.49 − (−0.81) = 1.30 V.
2. Cadmium is the negative electrode; on discharge Cd + 2OH⁻ → Cd(OH)₂ + 2e⁻.
3. Overall discharge: Cd + 2NiO(OH) + 2H₂O → Cd(OH)₂ + 2Ni(OH)₂. Charging runs this backwards.

Lithium-based cells are useful, but their materials can be toxic and damaged cells can catch fire.

5.2.3(k): fuel cells

A fuel cell uses the energy from a fuel reacting with oxygen to create a voltage. Fuel and oxygen are supplied continuously, so it needs no recharging.

In an acidic hydrogen fuel cell:

• Negative electrode (fuel oxidised): H₂ → 2H⁺ + 2e⁻ (0.00 V)
• Positive electrode (oxygen reduced): O₂ + 4H⁺ + 4e⁻ → 2H₂O (+1.23 V)
• Overall: 2H₂ + O₂ → 2H₂O; E⦵cell = 1.23 V

Common errors

• Calling MnO₄⁻ the reducing agent because it “is reduced”.
• Using the IO₃⁻ : I₂ ratio (1 : 3) instead of IO₃⁻ : S₂O₃²⁻ (1 : 6).
• Forgetting the ×10 for a 25.0 cm³ portion taken from 250.0 cm³.
• Doubling an E⦵ value when a half-equation is doubled.

Next steps

Condense this with the revision notes, then try the practice questions. Titration technique also appears in Acids (2.1.4).

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.