Edexcel International A Level Chemistry: Redox equilibria (YCH11) – Revision Notes

These notes condense Topic 16: Redox Equilibria of the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017, outcomes 16.1 to 16.19 with Core Practicals 12, 13a and 13b. The topic is in Unit 5: Transition Metals and Organic Nitrogen Chemistry, at International A2 (IA2) level. For full explanations and worked examples, read the redox equilibria study guide first.

When you are ready, try the redox equilibria practice questions. Track outcomes on the Edexcel Chemistry checklist and find every topic on the course hub. Oxidation number rules are in the revision notes on Unit 2 redox. A free diagnostic shows where to focus.

Definitions (16.1–16.4, 16.13)

• Oxidation: electron loss; oxidation number goes up. Reduction: electron gain; oxidation number goes down. True for s-, p- and d-block elements alike.
• Standard electrode potential, E⦵: the voltage of a half-cell measured against a standard hydrogen electrode, both under standard conditions.
• Standard conditions: 298 K; 100 kPa for any gas; 1.00 mol dm⁻³ for each ion.
• Standard hydrogen electrode: platinum coated with platinum black; H₂ at 100 kPa; H⁺(aq) at 1.00 mol dm⁻³; 298 K. E⦵ = 0.00 V by definition.
• Why a reference? Only a potential difference between two electrodes can be measured.
• Standard reduction potential: another name for E⦵, since each half-equation is written as a reduction.
• Electrochemical series: half-equations listed in order of E⦵. At the positive end, left-hand species are the strongest oxidising agents. At the negative end, right-hand species are the strongest reducing agents.

Building half-cells (16.5, Core Practical 12)

System	Electrode	Solution
Metal / metal ion	the metal itself, cleaned	its ion at 1.00 mol dm⁻³
Non-metal / ion	platinum, with gas at 100 kPa	the ion at 1.00 mol dm⁻³
Two ions, one element	platinum	both ions, each 1.00 mol dm⁻³

Connect to the reference with a potassium nitrate salt bridge and a high-resistance voltmeter. The salt bridge completes the circuit by ion movement without mixing the solutions; nitrate forms no precipitates. The high-resistance meter draws almost no current, so it reads the true emf.

Cell emf and diagrams (16.7, 16.8)

Method in steps

1. Find both E⦵ values.
2. More positive E⦵ = positive electrode, written on the right, as a reduction.
3. E⦵cell = E⦵(right) − E⦵(left). It must be positive for the cell as written.
4. Balance electrons to get the overall equation. Never multiply E⦵ values.

Notation: | phase boundary; , same phase; || salt bridge; electrodes at the ends.

Worked reminder. Use these values: Al³⁺/Al −1.66 V; Cu²⁺/Cu +0.34 V. Al(s) | Al³⁺(aq) || Cu²⁺(aq) | Cu(s); E⦵cell = +0.34 − (−1.66) = +2.00 V; 2Al + 3Cu²⁺ → 2Al³⁺ + 3Cu.

Conditions, feasibility and their limits (16.9–16.12)

• A half-cell is an equilibrium. Raising the concentration of the oxidised form moves it right and makes E more positive. Temperature and pressure also change E.
• Feasible if E⦵cell > 0.
• E⦵cell ∝ ΔStotal ∝ ln K (same number of electrons):

E⦵cell	ΔStotal	K	Meaning
Large, positive	Large, positive	Very large	Goes almost to completion
Small, positive	Small, positive	Greater than 1, not huge	Equilibrium mixture
Negative	Negative	Less than 1	Not feasible as written

• Doubling E⦵cell doubles ln K, so K is squared.
• Limits: E⦵ predicts thermodynamics, not rate. A feasible reaction can still be imperceptibly slow when its activation energy is large (kinetic stability). Non-standard conditions, such as concentrated reagents or heat, can change E and reverse a prediction, most often when E⦵cell is close to zero.

Disproportionation (16.14)

Take E⦵ for the half-equation where the species is reduced, minus E⦵ for the half-equation where it is the product.

Worked reminder. Use these values: Cu⁺ + e⁻ ⇌ Cu, +0.52 V; Cu²⁺ + e⁻ ⇌ Cu⁺, +0.15 V. 2Cu⁺ → Cu + Cu²⁺: E⦵cell = +0.52 − 0.15 = +0.37 V, so copper(I) in aqueous solution is expected to disproportionate.

Redox titrations (16.15–16.17, Core Practicals 13a and 13b)

	Iron(II) / manganate(VII)	Iodine / thiosulfate
Ionic equation	5Fe²⁺ + MnO₄⁻ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O	2S₂O₃²⁻ + I₂ → S₄O₆²⁻ + 2I⁻
Ratio	Fe²⁺ : MnO₄⁻ = 5 : 1	S₂O₃²⁻ : I₂ = 2 : 1
Indicator	none (self-indicating)	starch, added near the end
End point	first permanent pale pink	blue-black to colourless
Conditions	excess dilute sulfuric acid	excess KI to release I₂ from the oxidising agent

Why sulfuric acid? Hydrochloric acid would be oxidised to chlorine by manganate(VII). Nitric acid is itself an oxidising agent. A weak acid cannot supply enough H⁺.

Method in steps (any redox titration)

1. Moles of titrant = concentration × titre in dm³.
2. Use the equation ratio (chain the ratios if there are two steps, such as oxidant → I₂ → S₂O₃²⁻).
3. Scale from the pipetted portion to the whole solution.
4. Convert to mass, then to a percentage or concentration.

Useful chain ratios: 2Cu²⁺ gives one I₂, so Cu²⁺ : S₂O₃²⁻ = 1 : 1. IO₃⁻ gives 3I₂, so IO₃⁻ : S₂O₃²⁻ = 1 : 6.

Uncertainty and validity (16.16)

• Percentage uncertainty = (uncertainty ÷ measured value) × 100.
• A titre is the difference of two burette readings, so its uncertainty is twice that of one reading.
• Total percentage uncertainty = the sum of the percentage uncertainties of all measurements.
• If the gap between your result and an accepted value is bigger than the total uncertainty, apparatus alone cannot explain it. Suggest a systematic error, such as air oxidising Fe²⁺.
• A bigger titre or bigger mass reduces percentage uncertainty.

Fuel cells (16.18, 16.19)

A fuel cell turns the energy released when a fuel reacts with oxygen into a voltage. Fuel and oxygen are supplied continuously. Hydrogen, methanol and other hydrogen-rich fuels are used.

Electrolyte	Negative electrode (oxidation)	Positive electrode (reduction)
Acid	H₂ → 2H⁺ + 2e⁻	O₂ + 4H⁺ + 4e⁻ → 2H₂O
Alkali	H₂ + 2OH⁻ → 2H₂O + 2e⁻	O₂ + 2H₂O + 4e⁻ → 4OH⁻

Overall in both: 2H₂ + O₂ → 2H₂O.

Must-know distinctions

• E vs E⦵: E is the measured value under any conditions; E⦵ only under standard conditions.
• Thermodynamic vs kinetic: feasible (E⦵cell > 0) does not mean fast.
• Positive vs negative electrode: in a cell, reduction happens at the positive electrode.
• Metal vs platinum electrode: platinum only when no solid in the half-equation can act as the electrode.

Quick self-test

1. State the three standard conditions for E⦵.
2. Why is platinum black used in the hydrogen electrode?
3. Use these values: Mg²⁺/Mg −2.37 V; Fe³⁺/Fe²⁺ +0.77 V. Calculate E⦵cell and write the cell diagram.
4. Use these values: Br₂/Br⁻ +1.07 V; Sn⁴⁺/Sn²⁺ +0.15 V. Can bromine oxidise Sn²⁺? Give E⦵cell.
5. A reaction has E⦵cell = −0.40 V. State the sign of ΔStotal and whether K is above or below 1.
6. Give two reasons a reaction with positive E⦵cell might not be seen.
7. How many moles of Fe²⁺ react with 22.40 cm³ of 0.0150 mol dm⁻³ KMnO₄?
8. How many moles of thiosulfate react with 0.0100 mol of I₂?
9. A titre of 12.50 cm³ comes from two burette readings, each ±0.05 cm³. What is its percentage uncertainty?
10. Write the positive electrode reaction in an alkaline hydrogen-oxygen fuel cell.
11. Why is starch not added at the start of an iodine titration?

Answers

1. 298 K; 100 kPa for gases; 1.00 mol dm⁻³ for ions.
2. It gives a large surface area where the H₂/H⁺ equilibrium is set up quickly.
3. E⦵cell = +0.77 − (−2.37) = +3.14 V; Mg(s) | Mg²⁺(aq) || Fe³⁺(aq), Fe²⁺(aq) | Pt(s).
4. E⦵cell = +1.07 − 0.15 = +0.92 V, so yes: Br₂ + Sn²⁺ → 2Br⁻ + Sn⁴⁺ is feasible.
5. ΔStotal is negative; K is less than 1.
6. High activation energy (slow rate); non-standard conditions changing E.
7. 5 × 0.0150 × 0.02240 = 1.68 × 10⁻³ mol.
8. 0.0200 mol.
9. ±0.10 ÷ 12.50 × 100 = 0.80%.
10. O₂ + 2H₂O + 4e⁻ → 4OH⁻.
11. A lot of iodine binds to starch and is released slowly, so the colour change is less sharp.

Where marks are usually lost

• Writing E⦵cell = E⦵(left) − E⦵(right) and getting a negative value for a working cell.
• Multiplying E⦵ by the factor used to balance electrons.
• Giving only two of the three standard conditions, or “room temperature” instead of 298 K.
• Omitting platinum from an ion/ion half-cell, or from the hydrogen electrode.
• Describing a salt bridge as carrying electrons. It carries ions.
• Stating a reaction “will happen” from E⦵ alone when the question asks about limitations.
• Using a 1 : 1 ratio for Fe²⁺ with manganate(VII), or 1 : 1 for iodine with thiosulfate.
• Forgetting to scale up from the pipetted portion to the volumetric flask.
• Counting a titre as one burette reading when working out uncertainty.
• Writing H⁺ in the alkaline fuel cell equations, or OH⁻ in the acid ones.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017, published by Pearson Education Limited. Unit 5: Transition Metals and Organic Nitrogen Chemistry, Topic 16: Redox Equilibria, outcomes 16.1 to 16.19.