amazon.co.ukEquilibria and Redox
A* revision for Kc, Kp, Le Chatelier's principle, oxidation states, half-equations and electrochemical cells with E° calculations, for AQA, Edexcel, OCR, WJEC and CCEA A-Level Chemistry.
🎯 Learning Objectives
- Apply Le Chatelier's principle to concentration, pressure and temperature changes, including industrial compromise conditions
- Write Kc and Kp expressions, calculate equilibrium constants from ICE tables and partial pressures, and assign units
- Explain why only temperature changes K, and why catalysts do not affect equilibrium position
- Assign oxidation states and construct balanced half-equations and full redox equations in acidic and alkaline conditions
- Calculate E°cell from standard electrode potentials and predict feasibility and reaction direction
- Describe electrochemical cells, salt bridges and cell notation, and evaluate fuel cells and rechargeable batteries
📋 Key Definitions and Core Concepts
Dynamic equilibrium: Forward and reverse reactions occur at equal rates in a closed system, so concentrations remain constant.
Le Chatelier's principle: When a change in concentration, pressure or temperature is applied, the equilibrium shifts to oppose the change.
Kc: Equilibrium constant in concentrations, products over reactants, each raised to its stoichiometric coefficient.
Kp: Equilibrium constant in partial pressures; pA = mole fraction × total pressure.
Partial pressure: The pressure a gas would exert if it alone occupied the volume: mole fraction × total pressure.
Oxidation number: A measure of electron loss/gain assigned by rules; increase = oxidation, decrease = reduction.
Standard electrode potential (E°): The potential of a half-cell measured against the standard hydrogen electrode under standard conditions (298 K, 1 mol dm⁻³, 100 kPa).
Disproportionation: A reaction in which one element is simultaneously oxidised and reduced, e.g. Cl₂ in water.
Fuel cell: A cell that generates electricity continuously from the reaction of a fuel (e.g. H₂) with oxygen; products at point of use are water.
Dynamic Equilibria, Kc and Le Chatelier's Principle
A dynamic equilibrium exists in a closed system when forward and reverse reactions occur at equal rates: concentrations stay constant, but both reactions continue. Le Chatelier's principle: when a change in concentration, pressure or temperature is applied, the equilibrium position shifts to oppose the change. Concentration: adding a species shifts equilibrium to consume it; removing shifts to replace it. Pressure (gases only): increasing pressure shifts to the side with fewer gas moles. Temperature: heating shifts in the endothermic direction — for an exothermic forward reaction, yield falls as temperature rises. Only temperature changes the value of Kc or Kp; concentration and pressure change the position but not K, because the system re-establishes the same ratio. A catalyst speeds both directions equally — faster equilibrium, no change in position or K.
Kc = [products]/[reactants] at equilibrium, each concentration raised to the power of its stoichiometric coefficient. For aA + bB ⇌ cC + dD: Kc = [C]^c[D]^d / [A]^a[B]^b. Units: count powers — e.g. for esterification CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O, Kc has no units (equal moles each side). Homogeneous equilibria have all species in the same state; heterogeneous equilibria involve different states — pure solids and pure liquids are omitted from K expressions because their concentrations are constant.
ICE-table calculations: given initial moles and the equilibrium amount of one species, fill the change row (x per coefficient), deduce equilibrium moles, divide by volume for concentrations, substitute into Kc. Exam tip: state the change as ±(coefficient × x) explicitly — that row carries the method marks. Industrial context: Haber process (450 °C compromise temperature, ~200 atm, Fe catalyst — low temperature favoured by equilibrium but too slow, so a compromise is chosen); Contact process for SO₃ (~2 atm because yield is already high; V₂O₅ catalyst).
Kp, Partial Pressures and Gas Equilibria
Kp is the equilibrium constant for gas equilibria expressed in partial pressures. Partial pressure of gas A: pA = mole fraction of A × total pressure, where mole fraction = moles of A / total moles of gas. Kp = product of p( products )^coefficients / p( reactants )^coefficients. Example for 2SO₂(g) + O₂(g) ⇌ 2SO₃(g): Kp = p(SO₃)² / [p(SO₂)² × p(O₂)], units: (kPa)²/(kPa)³ = kPa⁻¹.
Worked example: 2.00 mol SO₂ and 1.00 mol O₂ reach equilibrium with 1.20 mol SO₃ formed at total pressure P. Change row: SO₂ −1.20, O₂ −0.60, SO₃ +1.20; equilibrium moles: 0.80, 0.40, 1.20; total = 2.40. Mole fractions: 1/3, 1/6, 1/2. Partial pressures: P/3, P/6, P/2. Kp = (P/2)² / [(P/3)² × (P/6)] = (P²/4) / (P³/54) = 13.5/P kPa⁻¹. Method marks: change row, mole fractions, substitution, units.
Temperature is the only variable that alters Kp. Raising pressure at constant temperature leaves Kp unchanged but shifts position to fewer gas moles. The relationship Kp = Kc(RT)^Δn where Δn = moles of gaseous products − moles of gaseous reactants links the two constants (more heavily used by some boards, e.g. WJEC/CCEA). When asked to predict the effect of a change on yield, first compare gas moles; when asked about K, first ask whether temperature changed.
Redox: Oxidation States and Half-Equations
Oxidation is loss of electrons, reduction is gain (OIL RIG); an oxidising agent gains electrons (is itself reduced), a reducing agent loses electrons. Oxidation state rules: free element = 0; Group 1 = +1; Group 2 = +2; Al = +3; O = −2 except in peroxides (−1) and OF₂ (+2); H = +1 except in metal hydrides (−1); F always −1. The sum of oxidation states equals the overall charge. Nomenclature: oxidation states in Roman numerals — iron(III) means Fe³⁺, manganate(VII) Mn in +7.
Half-equations — method: (1) balance the main species; (2) balance O with H₂O; (3) balance H with H⁺ (acidic conditions); (4) balance charge with electrons on the correct side. Example: MnO₄⁻ → Mn²⁺ gives MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. In alkaline conditions, balance as if acidic then add OH⁻ to both sides to cancel H⁺ (forming water) — e.g. for ClO⁻ → Cl₂ in alkaline solution.
Combining half-equations: multiply so electrons cancel, then add and cancel common species. Classic redox titration equation: MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺ (manganate(VII) oxidising Fe(II), self-indicating — purple to colourless at the endpoint). Work out oxidation state changes: Mn +7 → +2 (gain 5e⁻), Fe +2 → +3 (lose 1e⁻), so 5 Fe²⁺ per MnO₄⁻. Disproportionation: one species is both oxidised and reduced — chlorine in water: Cl₂ + H₂O ⇌ HClO + HCl (0 → +1 and −1), the basis of water treatment.
Electrochemical Cells and Standard Electrode Potentials
An electrochemical cell converts chemical energy to electrical energy from two redox half-cells connected by a wire (external circuit) and a salt bridge (usually KNO₃ — mobile ions complete the circuit without reacting). In cell notation, the more negative electrode is written on the left: Zn | Zn²⁺ || Cu²⁺ | Cu. Oxidation occurs at the anode (negative in a galvanic cell), reduction at the cathode. The standard hydrogen electrode (Pt | H₂(g, 100 kPa) | H⁺(1 mol dm⁻³), E° = 0.00 V) is the reference against which all standard electrode potentials are measured under standard conditions.
E°cell = E°(reduction at cathode) − E°(reduction at anode) — always subtract the more negative (or less positive) reduction potential. Example: Zn | Zn²⁺ || Cu²⁺ | Cu with E°(Zn²⁺/Zn) = −0.76 V, E°(Cu²⁺/Cu) = +0.34 V: E°cell = 0.34 − (−0.76) = +1.10 V > 0 → spontaneous; Zn is oxidised, Cu²⁺ reduced, overall Zn + Cu²⁺ → Zn²⁺ + Cu. A reaction is feasible when E°cell > 0; the more positive E°, the stronger the oxidising agent (more readily reduced).
Limitations: E° values are standard (298 K, 1 mol dm⁻³, 100 kPa) and predict thermodynamic feasibility only — they say nothing about rate (activation energy) or the actual conditions. Under non-standard conditions the Nernst equation applies: E = E° − (0.059/n) log Q at 298 K (Q = reaction quotient; for the half-cell Mⁿ⁺/M, Q = 1/[Mⁿ⁺]). Increasing the concentration of the metal ion makes E more positive. At equilibrium E°cell = 0 and E°cell = (0.059/n) log K links cell potential to the equilibrium constant.
Commercial cells: fuel cells — H₂ + ½O₂ → H₂O; alkaline version has anode H₂ + 2OH⁻ → 2H₂O + 2e⁻, cathode ½O₂ + H₂O + 2e⁻ → 2OH⁻; advantages: higher efficiency and only water produced at point of use; limitation: H₂ storage/transport, energy cost of producing H₂. Rechargeable cells (Li-ion, NiMH, lead-acid): reversible redox — charging reverses the reaction by forcing electrons the other way. Non-rechargeable: irreversible reactions (e.g. zinc-carbon). Environmental arguments (recycling, lithium mining, hydrogen from renewable electricity) are standard 3-mark evaluation points.
Non-standard condition example: a Cu²⁺/Cu half-cell with [Cu²⁺] = 0.010 mol dm⁻³ (standard 1.0): E = E° − (0.059/2) log(1/[Cu²⁺]) = +0.34 − (0.0295 × 2) = +0.28 V — the more dilute ion solution gives a less positive potential. Concentration cells (same electrode, different concentrations) generate a voltage purely from the concentration difference; the spontaneous direction increases the concentration of the dilute half-cell until E reaches 0 at equilibrium. When predicting direction from E° values with non-standard conditions, compare the actual E values rather than E° — a reaction feasible under standard conditions can be reversed by concentrating products or diluting reactants. Exam technique: quote the Nernst equation with n (electrons transferred) identified for the specific half-cell — MnO₄⁻/Mn²⁺ has n = 5, Cu²⁺/Cu n = 2 — then substitute carefully, taking the log of the correct quotient.
Industrial Equilibria: Compromise Conditions and Evaluations
Industrial processes are the applied 6-marker of this topic: predict the conditions Le Chatelier's principle would favour, then explain why industry chooses a compromise. Haber process (N₂ + 3H₂ ⇌ 2NH₃, ΔH = −92 kJ/mol): equilibrium favours high pressure (4 → 2 gas moles) and low temperature (exothermic forward reaction), but low temperature is too slow and very high pressure is expensive and dangerous. Industry uses ~450 °C (compromise between yield and rate) and ~200 atm with an iron catalyst; the ammonia is cooled and liquefied as it forms, and unreacted N₂/H₂ recycled — recycling shifts the effective position right without changing K. Contact process (2SO₂ + O₂ ⇌ 2SO₃, ΔH = −196 kJ/mol): ~450 °C, ~2 atm (yield already ~99% at low pressure, so higher pressure is not worth the cost), V₂O₅ catalyst. Methanol production (CO + 2H₂ ⇌ CH₃OH, ΔH = −90 kJ/mol): similar compromise logic; ethanol hydration (C₂H₄ + H₂O ⇌ C₂H₅OH, ΔH = −42 kJ/mol): excess steam shifts position right, ~300 °C, 60 atm, H₃PO₄ catalyst.
Writing the evaluation: (1) state the equilibrium-favoured condition with the mole-count or exothermic-direction reasoning; (2) state the kinetic problem (too slow at the equilibrium-favourable temperature); (3) state the cost/safety factor (high pressure expensive, containment); (4) conclude the compromise condition and catalyst. Each step earns a mark — '450 °C because it is a compromise' alone does not. Catalyst evaluation: does not change yield or K, but allows a lower temperature for the same rate, saving energy and reducing the yield penalty of higher temperature.
Exam traps: confusing 'yield' with 'rate' in answers (pressure increases both here, but for exothermic reactions temperature raises rate while lowering yield); forgetting that only temperature changes K; omitting that a catalyst speeds both directions equally. When asked 'explain the effect on K of increasing temperature', answer with the direction of the endothermic change: for an exothermic forward reaction, K decreases as temperature rises because the equilibrium shifts left. Link back to ΔG = ΔH − TΔS for the thermodynamic perspective: K falls as T rises for exothermic reactions because the equilibrium constant relates to ΔG by ΔG = −RT ln K.
💡 Worked Example
Exam-style: Calculate E° for the cell Zn | Zn²⁺ || Cu²⁺ | Cu given E°(Zn²⁺/Zn) = −0.76 V and E°(Cu²⁺/Cu) = +0.34 V. State whether the cell reaction is spontaneous and write the overall equation.
Mark-scheme logic: E°cell = E°(cathode) − E°(anode) = +0.34 − (−0.76) = +1.10 V. Positive E°cell → reaction is feasible/spontaneous. Zinc is oxidised at the anode (more negative potential), Cu²⁺ is reduced at the cathode. Overall: Zn + Cu²⁺ → Zn²⁺ + Cu. Mark points: correct subtraction; sign; identify anode/cathode processes; overall equation. (AQA 3.1.11.2)
❓ Practice Questions & Mark Schemes
Balance the equation MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺ in acidic solution, showing the two half-equations.
Answer: Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O (balance O with H₂O, H with H⁺, charge with electrons). Oxidation: Fe²⁺ → Fe³⁺ + e⁻. Multiply iron half-equation by 5 and add: MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺.
E°(Ag⁺/Ag) = +0.80 V, E°(Cu²⁺/Cu) = +0.34 V. Which species is the stronger oxidising agent? Justify and write the overall cell reaction.
Answer: Ag⁺ is the stronger oxidising agent — more positive standard reduction potential (+0.80 > +0.34 V), so more readily reduced. E°cell = 0.80 − 0.34 = +0.46 V > 0, feasible: 2Ag⁺ + Cu → 2Ag + Cu²⁺.
For 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), write the Kp expression. If total pressure is doubled at constant temperature, what happens to the yield of SO₃ and why?
Answer: Kp = p(SO₃)² / [p(SO₂)² × p(O₂)]. Kp unchanged (temperature constant) but position shifts to fewer gas moles (3 → 2), so SO₃ yield increases. Exploited in the Contact process.
Why can't standard electrode potentials predict reaction rate?
Answer: E° values describe thermodynamic feasibility under standard conditions only. Rate depends on activation energy, concentration and temperature; a positive E°cell reaction can still be immeasurably slow if Ea is large (e.g. kinetic inhibition in H₂/O₂ mixtures without a spark or catalyst).
Write the half-equations for each electrode in an alkaline hydrogen-oxygen fuel cell and the overall equation. State one advantage over internal combustion engines.
Answer: Anode: H₂ + 2OH⁻ → 2H₂O + 2e⁻. Cathode: ½O₂ + H₂O + 2e⁻ → 2OH⁻. Overall: H₂ + ½O₂ → H₂O. Advantages: greater efficiency (more energy as useful work, less heat) and only water at the point of use — no CO₂ or pollutants.
2.00 mol of PCl₅ dissociates in a 10.0 dm³ container at constant T; at equilibrium 1.40 mol of Cl₂ is present. Calculate Kc for PCl₅ ⇌ PCl₃ + Cl₂.
Answer: Equilibrium moles: PCl₅ 0.60, PCl₃ 1.40, Cl₂ 1.40. Concentrations: 0.060, 0.140, 0.140 mol dm⁻³. Kc = (0.140 × 0.140)/0.060 = 0.327 (units: mol dm⁻³).