Edexcel International A Level Chemistry: Redox equilibria (YCH11)

This guide teaches Topic 16: Redox Equilibria from the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017. It covers learning outcomes 16.1 to 16.19, including Core Practicals 12, 13a and 13b. Pearson places it in Unit 5: Transition Metals and Organic Nitrogen Chemistry; this is International A2 (IA2) content. The Unit 6 paper (Practical Skills in Chemistry II) tests procedures from Units 4 and 5, so expect the core practicals there as well.

Course links: the Edexcel International A Level Chemistry hub and the printable Edexcel Chemistry checklist. Then use the redox equilibria revision notes and the redox equilibria practice questions. This topic builds on the Unit 2 redox chemistry guide and the entropy guide. For weak spots, take a free diagnostic.

The outcomes in Topic 16

Outcome	What you must be able to do
16.1	Oxidation and reduction by electrons and oxidation number, s-, p- and d-block
16.2–16.4	E⦵, its standard conditions, the standard hydrogen electrode, why a reference is needed
16.5, 16.6	Measuring E⦵ for element/ion and ion/ion systems; Core Practical 12
16.7, 16.8	Calculate E⦵cell and write cell diagrams
16.9	Explain why the conditions matter when measuring E
16.10, 16.11	Feasibility; E⦵cell, ΔStotal and ln K
16.12	Explain the limits of these predictions
16.13, 16.14	Electrochemical series; feasibility of disproportionation
16.15–16.17	Redox titrations, uncertainty and validity; Core Practicals 13a and 13b
16.18, 16.19	Fuel cells; hydrogen-oxygen electrode reactions in acid and alkali

16.1 Oxidation and reduction

Oxidation is loss of electrons, or an increase in oxidation number. Reduction is the reverse. Both definitions work in every block.

Worked example 1: three blocks in one reaction

In 2Al + 3Cu²⁺ → 2Al³⁺ + 3Cu, aluminium (p-block) goes from 0 to +3: it is oxidised and is the reducing agent. Copper (d-block) goes from +2 to 0: it is reduced. In Ca + Cl₂ → CaCl₂, calcium (s-block) goes from 0 to +2.

16.2–16.4 Standard electrode potentials and the hydrogen electrode

The standard electrode potential, E⦵, is the voltage a half-cell produces when connected to a standard hydrogen electrode, with both under standard conditions. The conditions are:

• temperature 298 K
• pressure of any gases 100 kPa
• concentration of ions 1.00 mol dm⁻³

In the standard hydrogen electrode, platinum coated with platinum black sits in 1.00 mol dm⁻³ H⁺ ions (hydrochloric acid, say) while H₂ at 100 kPa is bubbled past it, all at 298 K. By definition, E⦵ = 0.00 V. The platinum is inert and conducts; the porous coating speeds up the H₂/H⁺ equilibrium.

A reference is needed because a single electrode potential cannot be measured: a voltmeter needs two connections, so it reads a difference. Fixing one half-cell at zero puts every other half-cell on a common scale.

16.5 and Core Practical 12: measuring and investigating cells

(i) Metal or non-metal with its ions. For a metal, dip a clean strip of the metal into a 1.00 mol dm⁻³ solution of its ions. For a non-metal such as chlorine, bubble the gas at 100 kPa over a platinum electrode in a 1.00 mol dm⁻³ chloride solution.

(ii) Two ions of one element. Put a platinum electrode into a solution containing both ions, each at 1.00 mol dm⁻³, for example Fe²⁺ and Fe³⁺. Platinum is needed because neither ion is a solid.

Join the half-cell to the hydrogen electrode (or another half-cell of known E⦵) with a salt bridge and a high-resistance voltmeter. A salt bridge of filter paper soaked in potassium nitrate lets ions move without the solutions mixing; nitrate ions do not precipitate metal ions, whereas a chloride bridge would precipitate silver chloride in a silver half-cell. The voltmeter draws almost no current, so it reads the emf.

In Core Practical 12 you build several cells and compare readings with values calculated from E⦵ data. Clean metals with emery paper first. Gaps between measured and calculated values usually come from non-standard conditions or an oxide layer.

16.7 and 16.8 Cell emf and cell diagrams

E⦵cell = E⦵(right) − E⦵(left), where the right-hand half-cell is the positive electrode (more positive E⦵) and the left-hand half-cell is the negative electrode.

In a cell diagram, a single line is a phase boundary, a comma separates species in one phase, and a double line is the salt bridge. Electrodes go at the outer ends; the left half-cell is written as an oxidation, the right as a reduction.

Worked example 2: a nickel–bromine cell

Use these values: Ni²⁺(aq) + 2e⁻ ⇌ Ni(s), E⦵ = −0.25 V; Br₂(aq) + 2e⁻ ⇌ 2Br⁻(aq), E⦵ = +1.07 V.

1. Bromine has the more positive E⦵, so it is the positive (right) electrode and needs platinum.
2. Cell diagram: Ni(s) | Ni²⁺(aq) || Br₂(aq), Br⁻(aq) | Pt(s)
3. E⦵cell = +1.07 − (−0.25) = +1.32 V
4. Overall: Ni + Br₂ → Ni²⁺ + 2Br⁻.

16.9 Why conditions matter

A half-cell is an equilibrium, such as Ag⁺(aq) + e⁻ ⇌ Ag(s). Changing the concentration of an ion shifts that equilibrium and changes E. Raising [Ag⁺] shifts it to the right, so E becomes more positive; lowering [Ag⁺] does the opposite. Temperature and gas pressure also matter, so a measured E matches E⦵ only under standard conditions.

16.10 and 16.11 Feasibility, entropy and K

A reaction is thermodynamically feasible if its E⦵cell is positive: E⦵(reduction) − E⦵(oxidation).

Worked example 3: air and iron(II) solutions

Use these values: O₂ + 4H⁺ + 4e⁻ ⇌ 2H₂O, E⦵ = +1.23 V; Fe³⁺ + e⁻ ⇌ Fe²⁺, E⦵ = +0.77 V.

For oxygen oxidising Fe²⁺ in acid: E⦵cell = +1.23 − (+0.77) = +0.46 V. This is positive, so O₂ + 4H⁺ + 4Fe²⁺ → 2H₂O + 4Fe³⁺ is feasible. Iron(II) solutions for titration are therefore made up fresh.

The specification states that E⦵cell is directly proportional to the total entropy change and to ln K. So:

• E⦵cell positive means ΔStotal positive, and K greater than 1.
• E⦵cell negative means ΔStotal negative, and K less than 1.
• A larger positive E⦵cell means a larger K, so the reaction goes further towards completion.

Worked example 4: comparing two reactions

Reaction P has E⦵cell = +0.20 V and reaction R +0.60 V, with the same number of electrons transferred. As ln K is proportional to E⦵cell, ln K for R is 0.60/0.20 = 3 times ln K for P, so K for R equals K for P cubed.

16.12 Limits of the predictions

E⦵ values say whether a reaction can happen, not how fast. A feasible reaction with a high activation energy may not be seen: the system is kinetically stable. E⦵ values also assume standard conditions. Concentrated solutions, heat or other gas pressures can change E enough to reverse a prediction, especially when E⦵cell is close to zero.

16.13 and 16.14 The electrochemical series and disproportionation

Standard electrode potentials are also called standard reduction potentials, as each half-equation is written as a reduction. In order, they form the electrochemical series: the strongest oxidising agents are on the left of the most positive half-equations, and the strongest reducing agents on the right of the most negative.

Disproportionation is one species being oxidised and reduced at once. Find the half-equation where the species is reduced and the one where it is the product (reverse this for its oxidation). E⦵cell = E⦵(species reduced) − E⦵(species as product).

Worked example 5: hydrogen peroxide

Use these values: H₂O₂ + 2H⁺ + 2e⁻ ⇌ 2H₂O, E⦵ = +1.77 V; O₂ + 2H⁺ + 2e⁻ ⇌ H₂O₂, E⦵ = +0.68 V.

1. E⦵cell = +1.77 − (+0.68) = +1.09 V, so 2H₂O₂ → 2H₂O + O₂ is feasible.
2. Yet the solution keeps for months: without a catalyst the reaction is slow (16.12).

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

Iron(II) with manganate(VII): 5Fe²⁺ + MnO₄⁻ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O. MnO₄⁻ : Fe²⁺ is 1 : 5. Acidify with dilute sulfuric acid, not hydrochloric acid, whose chloride ions manganate(VII) would oxidise. It is self-indicating: once all the Fe²⁺ has reacted, one extra drop of manganate(VII) leaves a pale pink colour that persists.

Iodine with thiosulfate: 2S₂O₃²⁻ + I₂ → S₄O₆²⁻ + 2I⁻. I₂ : S₂O₃²⁻ is 1 : 2. An oxidising agent releases iodine from excess potassium iodide. Titrate until the brown turns pale yellow, add starch (blue-black), then continue until the blue-black just disappears. Starch added early binds a lot of iodine, which is released slowly and blurs the end point.

Worked example 6 (structured): iron in a moss killer

5.00 g of moss killer was dissolved in dilute sulfuric acid, giving 250.0 cm³ of solution. Titrating 25.0 cm³ portions needed a mean of 18.45 cm³ of 0.0200 mol dm⁻³ KMnO₄. Find the percentage by mass of iron(II).

1. Moles MnO₄⁻ = 0.0200 × 18.45/1000 = 3.69 × 10⁻⁴ mol
2. Moles Fe²⁺ in 25.0 cm³ = 5 × 3.69 × 10⁻⁴ = 1.845 × 10⁻³ mol
3. Moles Fe²⁺ in 250.0 cm³ = 1.845 × 10⁻² mol
4. Mass Fe = 1.845 × 10⁻² × 55.8 = 1.03 g
5. Percentage = 1.03/5.00 × 100 = 20.6%

Worked example 7 (unstructured): copper in brass

1.20 g of brass was dissolved in acid, neutralised and diluted to 250.0 cm³. Excess potassium iodide was added to 25.0 cm³ of this: 2Cu²⁺ + 4I⁻ → 2CuI + I₂. The iodine needed 12.60 cm³ of 0.100 mol dm⁻³ sodium thiosulfate. Find the percentage of copper.

Plan the ratio first: 2Cu²⁺ gives 1 I₂, which needs 2S₂O₃²⁻, so Cu²⁺ : S₂O₃²⁻ = 1 : 1.

• Moles S₂O₃²⁻ = 0.100 × 12.60/1000 = 1.26 × 10⁻³ mol = moles Cu²⁺ in 25.0 cm³
• Moles Cu in 250.0 cm³ = 1.26 × 10⁻² mol; mass = 1.26 × 10⁻² × 63.5 = 0.800 g
• Percentage = 0.800/1.20 × 100 = 66.7%

Worked example 8: uncertainty and validity (16.16)

For worked example 6, take each burette reading as ±0.05 cm³, pipette ±0.06 cm³, flask ±0.30 cm³, and each of two balance readings ±0.005 g.

Measurement	Uncertainty	Percentage
Titre 18.45 cm³	±0.10 cm³ (two readings)	0.54%
Pipette 25.0 cm³	±0.06 cm³	0.24%
Flask 250.0 cm³	±0.30 cm³	0.12%
Mass 5.00 g	±0.01 g	0.20%
Total		1.10%

1.10% of 20.6% is about ±0.23 percentage points. A label claim of 22.0% differs by 1.4 percentage points, far more than this. The apparatus cannot explain it, so look for a systematic cause, such as air oxidising some iron(II) first. The titre contributes most, so a larger titre (more dilute KMnO₄) would cut the uncertainty most.

16.18 and 16.19 Fuel cells

A fuel cell uses the energy released when a fuel reacts with oxygen to generate a voltage. Fuel and oxygen are fed in continuously at separate electrodes. Hydrogen, methanol and other hydrogen-rich fuels are used.

Electrolyte	Negative electrode	Positive electrode
Acidic	H₂ → 2H⁺ + 2e⁻	O₂ + 4H⁺ + 4e⁻ → 2H₂O
Alkaline	H₂ + 2OH⁻ → 2H₂O + 2e⁻	O₂ + 2H₂O + 4e⁻ → 4OH⁻

Both give 2H₂ + O₂ → 2H₂O overall.

Common errors

• Subtracting E⦵ values the wrong way round, giving a negative emf for a cell that works.
• Multiplying E⦵ when a half-equation is scaled up. E⦵ never changes.
• Leaving platinum out of an ion/ion half-cell.
• Forgetting the ×5 for Fe²⁺, the ×2 for thiosulfate, or the scale-up to the full flask.
• Adding starch too early.

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.