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H2 Chemistry notes
Free H2 Chemistry notes on one page, for the A-Level H2 Chemistry syllabus (9476). Full notes for 5 of the 14 syllabus chapters, each with the key idea, the facts and formulas to know, a worked exam question, the common mistakes and exam tips. The other chapters have a checklist of what the syllabus asks for, so the page maps the whole syllabus.
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How to use these notes
- Pick one chapter. Read the key idea and the must-know points.
- Close the page and write down what you remember, then check and mark what you missed.
- Try the worked exam question before reading its steps, and read the common mistakes.
- Do questions on that chapter from a school paper and mark them. Redo the wrong ones a few days later.
Syllabus edition
These notes follow SEAB's H2 Chemistry syllabus 9476 for the A-Level examination in 2027. Source: SEAB, H2 Chemistry syllabus 9476 (2027).
Exam format
| Paper | What it is | Time | Marks | Weighting |
|---|---|---|---|---|
| Paper 1 | 30 compulsory multiple-choice questions, 5 to 8 of the multiple completion type | 1 h | 30 | 15% |
| Paper 2 | Compulsory structured questions, including data-based questions worth 20 to 25 marks | 2 h | 75 | 30% |
| Paper 3 | Section A, 55 marks: 3 to 4 compulsory structured questions. Section B, 20 marks: answer one of two questions. | 2 h | 75 | 35% |
| Paper 4 | Practical, including planning | 2 h 30 min | 50 | 20% |
- The Data Booklet, with the Periodic Table, is given for the theory papers. The Qualitative Analysis Notes are given in the practical.
- No books or laboratory notebooks are allowed in the practical.
The whole syllabus, chapter by chapter
All 14 chapters of the syllabus in order. 5 have full notes on this page. The rest have a checklist of what the syllabus asks you to do.
| Chapter | What it covers | On this page |
|---|---|---|
| 1. Atomic Structure | Sub-atomic particles, orbitals, electronic configuration and ionisation energy. | Checklist |
| 2. Chemical Bonding | Bond types, shapes of molecules, polarity, intermolecular forces and lattices. | Checklist |
| 3. The Gaseous State | Ideal gases, pV = nRT and partial pressures. | Checklist |
| 4. Theories of Acids and Bases | Arrhenius, Brønsted-Lowry and Lewis theories. | Checklist |
| 5. The Periodic Table | Trends across Period 3 and down Groups 2 and 17. | Checklist |
| 6. The Mole Concept and Stoichiometry | Relative masses, the mole, formulae and reacting quantities. | Checklist |
| 7. Chemical Energetics | Enthalpy changes, Hess' Law, Born-Haber cycles, entropy and Gibbs free energy. | Full notes |
| 8. Reaction Kinetics | Rate equations, orders, activation energy and catalysis. | Full notes |
| 9. Chemical Equilibria | Dynamic equilibrium, equilibrium constants and the Haber process. | Full notes |
| 10. Acid-Base Equilibria | Ka, Kb, Kw, pH, indicators and buffers. | Full notes |
| 11. Solubility Equilibria | Solubility product, the common ion effect and complex ions. | Checklist |
| 12. Organic Chemistry | Isomerism, mechanisms, and hydrocarbons through to nitrogen compounds and polymers. | Checklist |
| 13. Electrochemistry | Redox, electrode potentials, cells and electrolysis. | Full notes |
| 14. Transition Elements | Properties, redox systems, complexes, colour and catalysis. | Checklist |
H2 Chemistry at a glance
| Chapter | The one thing to remember |
|---|---|
| Chemical Energetics | q = mcΔT, then ΔH = −q ÷ n |
| Reaction Kinetics | rate = k[A]^(m)[B]^(n), with orders from experiment, not from the equation |
| Chemical Equilibria | Kc = [C]^(c)[D]^(d) ÷ ([A]^(a)[B]^(b)), leaving out pure solids and liquids |
| Acid and Base Equilibria | pH = −lg[H⁺], and pH + pOH = 14.00 at 25 °C |
| Electrochemistry | E°cell = E°(reduction) − E°(oxidation), feasible if positive |
Notes 1 of 5
Chemical Energetics
Key idea. Chemical energetics covers enthalpy changes and feasibility. Enthalpy changes are measured by calorimetry with q = mcΔT and combined using Hess's law, including Born-Haber cycles for lattice energy. Whether a reaction is feasible depends on ΔG = ΔH − TΔS: a negative ΔG means the reaction is feasible, although it may still be too slow to observe.
Must know
- q = mcΔT, then ΔH = −q ÷ n
- From formation data: ΔHr = ΣΔHf(products) − ΣΔHf(reactants)
- From combustion data: ΔHr = ΣΔHc(reactants) − ΣΔHc(products)
- ΔG = ΔH − TΔS with ΔS in kJ K⁻¹ mol⁻¹, feasible if ΔG is negative
What the syllabus asks you to do
- Explain and use the terms enthalpy change of formation, combustion, hydration, solution, neutralisation and atomisation, bond energy, lattice energy and standard conditions
- Calculate enthalpy changes from experimental results using heat change = mcΔT
- Explain qualitatively the effect of ionic charge and ionic radius on the magnitude of lattice energy
- Apply Hess's law to energy cycles, including Born-Haber cycles, enthalpy changes of formation from combustion data and average bond energies
- Explain and use the term entropy, and predict whether the entropy change for a process is positive or negative
- State and use ΔG = ΔH − TΔS to decide spontaneity, including its limitations and the effect of temperature
Key enthalpy changes and their usual signs
| Enthalpy change | Defined for | Usual sign |
|---|---|---|
| Formation, ΔHf | One mole of a compound from its elements in their standard states | Usually negative |
| Combustion, ΔHc | One mole of a substance completely burnt in oxygen | Negative |
| Neutralisation | One mole of water formed from an acid and a base | Negative |
| Atomisation | One mole of gaseous atoms formed from the element | Positive |
| Lattice energy | One mole of ionic solid formed from its gaseous ions | Negative |
| Hydration | One mole of gaseous ions dissolved in water | Negative |
| Bond energy | One mole of a bond broken in the gas phase | Positive |
Na(s) + ½Cl₂(g)
The elements in their standard states, the starting point
Atomise sodium
Na(s) gives Na(g), ΔHat = +107 kJ mol⁻¹
Ionise sodium
Na(g) gives Na⁺(g) + e⁻, IE1 = +496 kJ mol⁻¹
Atomise chlorine
½Cl₂(g) gives Cl(g), ΔHat = +122 kJ mol⁻¹
Add an electron to chlorine
Cl(g) + e⁻ gives Cl⁻(g), EA1 = −349 kJ mol⁻¹
Form the lattice
Na⁺(g) + Cl⁻(g) gives NaCl(s), LE = −787 kJ mol⁻¹
Compare with formation
The steps sum to ΔHf(NaCl) = −411 kJ mol⁻¹
Worked exam question
Use the data to calculate the lattice energy of sodium chloride. ΔHf(NaCl) = −411 kJ mol⁻¹, ΔHat(Na) = +107 kJ mol⁻¹, first ionisation energy of Na = +496 kJ mol⁻¹, ΔHat(Cl) = +122 kJ mol⁻¹, first electron affinity of Cl = −349 kJ mol⁻¹.
- Write the cycle: Na(s) + ½Cl₂(g) gives Na(g) + Cl(g), then Na⁺(g) + Cl⁻(g), then NaCl(s), and this route must equal ΔHf.
- By Hess's law, ΔHf = ΔHat(Na) + IE1(Na) + ΔHat(Cl) + EA1(Cl) + LE.
- Sum of the known steps = 107 + 496 + 122 + (−349) = +376 kJ mol⁻¹.
- So −411 = 376 + LE, giving LE = −411 − 376 = −787 kJ mol⁻¹.
- The value is negative, as expected for energy released when gaseous ions form a lattice.
Answer: The lattice energy of sodium chloride is −787 kJ mol⁻¹.
Common mistakes
- Adding the mass of a solid reactant to the mass of water in q = mcΔT.Use the mass of the solution being heated, usually taken from its volume at 1.00 g cm⁻³.
- Reporting ΔH as +56.8 kJ mol⁻¹ for a neutralisation in which the temperature rose.Use ΔH = −q ÷ n, so a temperature rise always gives a negative ΔH.
- Calculating ΔG as 178 − 298 × 161.Divide ΔS by 1000 first: 178 − 298 × 0.161 = +130 kJ mol⁻¹.
- Using the full Cl-Cl bond energy for the atomisation of chlorine in a Born-Haber cycle.Use half the Cl-Cl bond energy for ΔHat(Cl).
- Using bond energies for a reaction that forms H₂O(l).Use bond energies only when every species is a gas, otherwise use formation or combustion data.
- Saying diamond must burn quickly in air at room temperature because ΔG is negative.Remember ΔG decides whether a reaction can happen, while the activation energy decides how fast.
Exam tips
- Write the definition of each enthalpy change in full when asked, including 'one mole', the states and standard conditions. Definitions are a common source of easy marks lost.
- In Hess's law and Born-Haber questions, draw the cycle before writing any numbers, and write the multiplier beside each arrow.
- When asked to explain feasibility, quote the signs of ΔH and ΔS and say which term, ΔH or TΔS, dominates at the given temperature.
Notes 2 of 5
Reaction Kinetics
Key idea. Reaction kinetics measures how rate depends on concentration. The rate equation, rate = k[A]^(m)[B]^(n), has orders m and n that are found only by experiment, often by the initial rates method. A first-order reaction has a constant half-life, t½ = ln 2 ÷ k. The rate equation reflects the slow, rate-determining step of the mechanism.
Must know
- rate = k[A]^(m)[B]^(n), with orders from experiment, not from the equation
- First order: t½ = ln 2 ÷ k, the same whatever the starting concentration
- Units of k: s⁻¹ for first order, mol⁻¹ dm³ s⁻¹ for second order
- A catalyst lowers Ea, it does not move the Boltzmann curve
What the syllabus asks you to do
- Explain and use the terms rate of reaction, rate equation, order of reaction, rate constant, half-life, rate-determining step, activation energy and catalysis
- Deduce orders by the initial rates method, and justify zero and first order from concentration-time graphs
- Calculate a rate constant using the initial rates method
- Show that the half-life of a first-order reaction is independent of concentration, and use it in calculations
- Verify that a suggested mechanism is consistent with the observed kinetics, and predict the order from a mechanism
- Explain the effects of temperature and catalysts on the rate constant using the Boltzmann distribution and collision frequency
- Outline homogeneous and heterogeneous catalysis, including the Haber process and Fe²⁺ in the I⁻/S₂O₈²⁻ reaction
Justifying zero and first order from a concentration against time graph
| Order | Shape of the graph | Half-life |
|---|---|---|
| Zero | Straight line with a constant negative gradient, equal to −k | Gets shorter as the concentration falls |
| First | Curve whose gradient falls as the concentration falls | Constant, t½ = ln 2 ÷ k |
A curve falling from 0.80 mol dm⁻³ at 0 s. It halves to 0.40 at 30 s, to 0.20 at 60 s, to 0.10 at 90 s and to 0.05 at 120 s. Each halving takes the same 30 s, which is the sign of a first-order reaction.
Worked exam question
For the reaction A + B gives products, the initial rates were: Experiment 1, [A] = 0.10 and [B] = 0.10 mol dm⁻³, rate = 2.0 × 10⁻⁴ mol dm⁻³ s⁻¹. Experiment 2, [A] = 0.20 and [B] = 0.10, rate = 4.0 × 10⁻⁴. Experiment 3, [A] = 0.20 and [B] = 0.30, rate = 3.6 × 10⁻³. Deduce the rate equation and calculate k with units.
- Experiments 1 and 2: [B] is constant and [A] doubles, and the rate doubles from 2.0 × 10⁻⁴ to 4.0 × 10⁻⁴, so the reaction is first order with respect to A.
- Experiments 2 and 3: [A] is constant and [B] triples, and the rate rises from 4.0 × 10⁻⁴ to 3.6 × 10⁻³, which is 9 times, or 3², so the reaction is second order with respect to B.
- Rate equation: rate = k[A][B]², which is third order overall.
- Using experiment 1: k = rate ÷ ([A][B]²) = 2.0 × 10⁻⁴ ÷ (0.10 × 0.10²) = 2.0 × 10⁻⁴ ÷ 1.0 × 10⁻³ = 0.20.
- Units: (mol dm⁻³ s⁻¹) ÷ (mol dm⁻³)³ = mol⁻² dm⁶ s⁻¹.
- Check with experiment 3: 0.20 × 0.20 × 0.30² = 0.20 × 0.20 × 0.090 = 3.6 × 10⁻³, which matches.
Answer: rate = k[A][B]², with k = 0.20 mol⁻² dm⁶ s⁻¹.
Common mistakes
- Writing rate = k[NO₂]²[F₂] for 2NO₂ + F₂ gives 2NO₂F.Find orders only from experimental data or from the rate-determining step of a known mechanism.
- Giving s⁻¹ as the unit of k for rate = k[A][B].Divide the rate units by concentration raised to the overall order, giving mol⁻¹ dm³ s⁻¹ here.
- Explaining a temperature rise only by more frequent collisions.Say many more particles have energy at least equal to Ea, which is the main reason k increases.
- Drawing the catalysed Boltzmann curve shifted to the right.Keep the same curve and mark a lower Ea, with a larger shaded area beyond it.
- Calling F a catalyst in the mechanism for 2NO₂ + F₂ gives 2NO₂F.An intermediate is formed then consumed. A catalyst is consumed then formed again.
- Using t½ = ln 2 ÷ k for a zero-order reaction.Use a constant half-life only for first order, and check the graph for equal successive half-lives first.
Exam tips
- In initial rates questions, quote the pair of experiments you are comparing and state the factor by which each concentration and the rate change. That reasoning earns the marks, not just the order.
- Check your final rate equation and k against an experiment you did not use to calculate them.
- When drawing a Boltzmann distribution, start the curve at the origin, never let it touch the energy axis again, and keep the area the same at a higher temperature.
Notes 3 of 5
Chemical Equilibria
Key idea. A reversible reaction in a closed system reaches dynamic equilibrium when the forward and backward rates are equal and concentrations stay constant. The equilibrium constant, Kc or Kp, is products over reactants, each raised to its coefficient. Le Chatelier's principle predicts how the position shifts when conditions change, and only a change in temperature changes the value of K.
Must know
- Kc = [C]^(c)[D]^(d) ÷ ([A]^(a)[B]^(b)), leaving out pure solids and liquids
- Partial pressure = mole fraction × total pressure
- Raising temperature favours the endothermic direction and is the only change that alters K
- A catalyst changes neither the position of equilibrium nor K
What the syllabus asks you to do
- Explain, in terms of the rates of the forward and reverse reactions, what is meant by a reversible reaction and dynamic equilibrium
- State Le Chatelier's principle and apply it to deduce the effects of changes in concentration, pressure or temperature on a system at equilibrium
- Deduce whether changes in concentration, pressure or temperature or the presence of a catalyst affect the value of the equilibrium constant
- Deduce expressions for equilibrium constants in terms of concentrations, Kc, and partial pressures, Kp
- Calculate equilibrium constants and the quantities present at equilibrium from appropriate data
- Describe and explain the conditions used in the Haber process
Effect of changes on a system at equilibrium
| Change | Position of equilibrium | Value of K |
|---|---|---|
| Add a reactant | Shifts right | Unchanged |
| Increase total pressure | Shifts to the side with fewer moles of gas | Unchanged |
| Add an inert gas at constant volume | No change | Unchanged |
| Raise temperature, forward reaction exothermic | Shifts left | Decreases |
| Raise temperature, forward reaction endothermic | Shifts right | Increases |
| Add a catalyst | No change, equilibrium reached faster | Unchanged |
Two curves against time. The concentration of H₂, which equals that of I₂, falls from 1.00 mol dm⁻³ and levels off at 0.20 mol dm⁻³. The concentration of HI rises from zero and levels off at 1.60 mol dm⁻³. Once both curves are flat, equilibrium has been reached and the concentrations stay constant, though not equal.
Worked exam question
1.00 mol of hydrogen and 1.00 mol of iodine are sealed in a 1.00 dm³ flask and heated to a temperature at which H₂(g) + I₂(g) ⇌ 2HI(g) has Kc = 64. Calculate the equilibrium concentrations of all three gases.
- Initial concentrations: [H₂] = 1.00, [I₂] = 1.00 and [HI] = 0 mol dm⁻³.
- Let x mol dm⁻³ of H₂ react. Then I₂ also falls by x and HI rises by 2x, from the 1 : 1 : 2 ratio.
- Equilibrium: [H₂] = 1.00 − x, [I₂] = 1.00 − x, [HI] = 2x.
- Kc = (2x)² ÷ (1.00 − x)² = 64. Take the square root of both sides: 2x ÷ (1.00 − x) = 8.
- So 2x = 8 − 8x, giving 10x = 8 and x = 0.80.
- Check: [HI]² ÷ ([H₂][I₂]) = 1.60² ÷ (0.20 × 0.20) = 2.56 ÷ 0.0400 = 64, which matches Kc.
Answer: [H₂] = 0.20 mol dm⁻³, [I₂] = 0.20 mol dm⁻³ and [HI] = 1.60 mol dm⁻³. Kc has no units here because the moles of gas are equal on both sides.
Common mistakes
- Writing Kc for CaCO₃(s) ⇌ CaO(s) + CO₂(g) as [CaO][CO₂] ÷ [CaCO₃].Leave out pure solids and pure liquids, so Kc = [CO₂].
- Saying that increasing the pressure increases Kp for 2SO₂ + O₂ ⇌ 2SO₃.Only temperature changes K. Pressure changes Q, and the system shifts back to the same K.
- Writing that adding V₂O₅ increases the equilibrium yield of SO₃.Say a catalyst speeds up both reactions equally, so equilibrium is reached sooner with the same yield.
- Putting 0.40 mol straight into Kc when the flask is 2.00 dm³.Divide every amount by the volume first, so 0.40 mol in 2.00 dm³ is 0.20 mol dm⁻³.
- Stating that at equilibrium the reactant and product concentrations are equal.Say the forward and backward rates are equal, so the concentrations are constant, not equal.
- Predicting a shift when argon is added at constant volume.Check the partial pressures. At constant volume they do not change, so there is no shift.
Exam tips
- Set out every equilibrium calculation as an ICE table with initial, change and equilibrium rows, and state the volume used to convert moles to concentrations.
- In Le Chatelier answers, name the change, state the direction of the shift and give the reason in terms of reducing that change, such as 'to the side with fewer moles of gas'.
- Always check whether a question asks about the position of equilibrium, the value of K or the rate. These are three separate things and markers expect you to distinguish them.
Notes 4 of 5
Acid and Base Equilibria
Key idea. Acid and base equilibria describe proton transfer in water, with pH = −lg[H⁺]. Strong acids ionise fully, so [H⁺] equals the acid concentration, while for a weak acid [H⁺] ≈ √(Ka × c). Buffers, made of a weak acid and its conjugate base or a weak base and its conjugate acid, resist pH change, with pH = pKa + lg([A⁻] ÷ [HA]).
Must know
- pH = −lg[H⁺], and pH + pOH = 14.00 at 25 °C
- Weak acid on its own: [H⁺] = √(Ka × c)
- Buffer: pH = pKa + lg([A⁻] ÷ [HA])
- Conjugate pair: Ka × Kb = Kw = 1.0 × 10⁻¹⁴ mol² dm⁻⁶ at 25 °C
What the syllabus asks you to do
- Apply the Brønsted-Lowry theory of acids and bases, including conjugate acids and conjugate bases
- Explain qualitatively the differences between strong and weak acids and bases in terms of the extent of dissociation
- Explain the terms pH, Ka, pKa, Kb, pKb and Kw and apply them in calculations, including Kw = Ka × Kb
- Calculate [H⁺] and pH for strong acids, weak monobasic acids, strong bases and weak monoacidic bases
- Describe and explain the changes in pH during acid-base titrations, and explain the choice of suitable indicators
- Explain how buffer solutions control pH, including the role of CO₃²⁻/HCO₃⁻ in the oceans, and calculate the pH of buffer solutions
Finding the pH of each type of solution at 25 °C
| Solution | Method | Example |
|---|---|---|
| Strong monoprotic acid | [H⁺] = acid concentration | 0.010 mol dm⁻³ HCl, pH 2.00 |
| Strong base | Find [OH⁻], then pOH = −lg[OH⁻] and pH = 14.00 − pOH | 0.010 mol dm⁻³ NaOH, pH 12.00 |
| Weak acid | [H⁺] = √(Ka × c) | 0.10 mol dm⁻³ ethanoic acid, Ka = 1.8 × 10⁻⁵, pH 2.87 |
| Weak base | [OH⁻] = √(Kb × c), then pH from pOH | 0.20 mol dm⁻³ NH₃, Kb = 1.8 × 10⁻⁵, pH 11.28 |
| Buffer | pH = pKa + lg([A⁻] ÷ [HA]) | Equal concentrations of acid and salt give pH = pKa |
A line graph of pH against volume of NaOH from 0 to 50 cm³. The pH starts at 2.88, rises gently through the buffer region and passes 4.76 at 12.5 cm³, the half-equivalence point where pH equals pKa. It rises steeply at 25.0 cm³ through the equivalence point at pH 8.73, then levels off near pH 12.5.
Worked exam question
25.0 cm³ of 0.100 mol dm⁻³ ethanoic acid (pKa = 4.76) is titrated with 0.100 mol dm⁻³ sodium hydroxide at 25 °C. Calculate the pH (a) at the start, (b) after 12.5 cm³ of NaOH, (c) at the equivalence point, and suggest a suitable indicator.
- Ka = 10^(−4.76) = 1.74 × 10⁻⁵ mol dm⁻³.
- (a) [H⁺] = √(Ka × c) = √(1.74 × 10⁻⁵ × 0.100) = √(1.74 × 10⁻⁶) = 1.32 × 10⁻³ mol dm⁻³, so pH = 2.88.
- (b) Moles of acid = 0.00250 and moles of NaOH = 0.00125, so half the acid has become CH₃COO⁻. [HA] = [A⁻], so pH = pKa = 4.76.
- (c) At equivalence, 0.00250 mol of CH₃COO⁻ is in 50.0 cm³, so [CH₃COO⁻] = 0.00250 ÷ 0.0500 = 0.0500 mol dm⁻³.
- Kb of CH₃COO⁻ = Kw ÷ Ka = 1.0 × 10⁻¹⁴ ÷ 1.74 × 10⁻⁵ = 5.75 × 10⁻¹⁰ mol dm⁻³.
- [OH⁻] = √(5.75 × 10⁻¹⁰ × 0.0500) = √(2.88 × 10⁻¹¹) = 5.36 × 10⁻⁶ mol dm⁻³, so pOH = 5.27 and pH = 14.00 − 5.27 = 8.73.
- The equivalence point is at pH 8.73, so choose phenolphthalein, whose range of about 8.3 to 10 lies in the steep section.
Answer: (a) pH 2.88. (b) pH 4.76. (c) pH 8.73. Phenolphthalein is a suitable indicator.
Common mistakes
- Using [H⁺] = √(Ka × c) for a mixture of ethanoic acid and sodium ethanoate.For a buffer, use pH = pKa + lg([A⁻] ÷ [HA]).
- Giving pH 2.00 for 0.010 mol dm⁻³ NaOH.Find pOH = −lg[OH⁻] = 2.00 first, then pH = 14.00 − 2.00 = 12.00 at 25 °C.
- Using 0.100 mol dm⁻³ for ethanoate at equivalence after mixing 25.0 cm³ each of 0.100 mol dm⁻³ acid and alkali.Divide moles by the total volume: 0.00250 ÷ 0.0500 = 0.0500 mol dm⁻³.
- Stating that every titration has its equivalence point at pH 7.Ask which salt forms. A weak acid's salt is alkaline, and a weak base's salt is acidic.
- Explaining buffer action as 'the buffer neutralises the acid', with no equation.Write CH₃COO⁻ + H⁺ gives CH₃COOH for added acid, and CH₃COOH + OH⁻ gives CH₃COO⁻ + H₂O for added alkali.
- Calling pure water at 50 °C acidic because its pH is below 7.Neutral means [H⁺] = [OH⁻]. Kw rises with temperature, so neutral pH falls below 7.
Exam tips
- Before any pH calculation, classify the solution first: strong acid, strong base, weak acid, weak base, buffer or salt. Each has its own method, and naming it prevents using the wrong formula.
- For a mixture of a weak acid and a strong base, work in moles first to see what is left after reaction, then decide whether you have a buffer, a salt solution or excess base.
- When explaining an indicator choice, quote both the pH at equivalence and the indicator's working range.
Notes 5 of 5
Electrochemistry
Key idea. Electrochemistry links redox reactions to electricity. Standard electrode potentials, measured against the standard hydrogen electrode, give E°cell = E°(reduction) − E°(oxidation). A positive E°cell means the reaction is feasible, since ΔG° = −nFE°cell. In electrolysis, the amount of product follows from the charge passed: Q = It, and moles of electrons = Q ÷ F.
Must know
- E°cell = E°(reduction) − E°(oxidation), feasible if positive
- ΔG° = −nFE°cell, with F = 96 500 C mol⁻¹
- Q = It with t in seconds, and moles of electrons = Q ÷ F
- Never multiply E° values when balancing half-equations
What the syllabus asks you to do
- Define standard electrode potential and standard cell potential, and describe the standard hydrogen electrode
- Calculate a standard cell potential by combining two standard electrode potentials
- Use standard cell potentials to deduce the direction of electron flow and predict the spontaneity of a reaction, and understand their limitations
- State and apply ΔG = −nFE to electrochemical cells
- Predict qualitatively how an electrode potential varies with the concentration of the aqueous ion
- Predict the substance liberated in electrolysis, and calculate the charge passed and the mass or volume liberated, using F = Le
Selected standard electrode potentials at 298 K
| Half-equation | E° / V |
|---|---|
| Mg²⁺ + 2e⁻ ⇌ Mg | −2.38 |
| Zn²⁺ + 2e⁻ ⇌ Zn | −0.76 |
| Fe²⁺ + 2e⁻ ⇌ Fe | −0.44 |
| 2H⁺ + 2e⁻ ⇌ H₂ | 0.00 |
| Cu²⁺ + 2e⁻ ⇌ Cu | +0.34 |
| I₂ + 2e⁻ ⇌ 2I⁻ | +0.54 |
| Fe³⁺ + e⁻ ⇌ Fe²⁺ | +0.77 |
| Ag⁺ + e⁻ ⇌ Ag | +0.80 |
| Br₂ + 2e⁻ ⇌ 2Br⁻ | +1.07 |
| Cl₂ + 2e⁻ ⇌ 2Cl⁻ | +1.36 |
Convert time to seconds
32 min 10 s = 1930 s
Find the charge passed
Q = It = 0.500 × 1930 = 965 C
Find moles of electrons
965 ÷ 96 500 = 0.0100 mol
Use the half-equation
Ag⁺ + e⁻ gives Ag, so 0.0100 mol of silver forms
Find the mass of product
0.0100 × 108 = 1.08 g of silver
Worked exam question
A current of 0.500 A is passed through aqueous silver nitrate for 32 minutes 10 seconds using inert electrodes. Calculate the mass of silver deposited at the cathode and the volume of oxygen produced at the anode at room conditions. (F = 96 500 C mol⁻¹, Ar of Ag = 108, molar volume of gas = 24.0 dm³ mol⁻¹ at r.t.p.)
- Time in seconds: t = 32 × 60 + 10 = 1930 s.
- Charge: Q = It = 0.500 × 1930 = 965 C.
- Moles of electrons = Q ÷ F = 965 ÷ 96 500 = 0.0100 mol.
- Cathode: Ag⁺ + e⁻ gives Ag, so 0.0100 mol of Ag forms, and mass = 0.0100 × 108 = 1.08 g.
- Anode: 2H₂O gives O₂ + 4H⁺ + 4e⁻, so moles of O₂ = 0.0100 ÷ 4 = 0.00250 mol.
- Volume of O₂ = 0.00250 × 24.0 = 0.0600 dm³ = 60.0 cm³.
Answer: 1.08 g of silver is deposited and 60.0 cm³ of oxygen is produced at r.t.p.
Common mistakes
- Doubling 0.77 V to 1.54 V after doubling the Fe³⁺/Fe²⁺ half-equation.Multiply half-equations to balance electrons, but keep every E° value unchanged.
- Calculating E°cell for the zinc and copper cell as −0.76 − (+0.34) = −1.10 V.Use E°cell = E°(reduction) − E°(oxidation), with the more positive half-cell as the reduction.
- Using t = 32 for 32 minutes in Q = It.Convert time to seconds first, so 32 minutes is 1920 s.
- Taking moles of copper deposited as equal to moles of electrons.Read the ratio from Cu²⁺ + 2e⁻ gives Cu, so moles of Cu = moles of electrons ÷ 2.
- Saying electrons flow through the salt bridge.Electrons flow in the external wire. The salt bridge carries ions to keep each solution neutral.
- Labelling the anode positive in a zinc and copper cell.Define the anode by oxidation. It is negative in an electrochemical cell and positive in electrolysis.
Exam tips
- Write both half-equations with their E° values before calculating anything, then mark which one is reversed for oxidation. This keeps the sign of E°cell straight.
- In Faraday questions, set out the chain charge, moles of electrons, moles of product, then mass or volume, with the half-equation written beside the mole ratio step.
- When asked to explain feasibility, say both that E°cell is positive and that this makes ΔG° negative, and add a note on rate or non-standard conditions if the question hints that the reaction is not observed.
Syllabus checklist for the rest
What the syllabus asks you to do in the chapters, and parts of chapters, that do not have full notes yet. Each point is the syllabus learning outcome in plain words. Tick each one you can do without looking.
Chapter 1. Atomic Structure
3 points to check
- Describe protons, neutrons and electrons by relative charge and mass, and deduce how beams of each behave in an electric field.
- Describe the number, relative energies and shapes of s, p and d orbitals, and state the electronic configuration of atoms and ions.
- Explain what affects ionisation energies, and use successive ionisation energies to deduce configuration and position in the Periodic Table.
Chapter 2. Chemical Bonding
1 point to check
- Cover ionic, metallic, covalent and co-ordinate (dative) bonding, shapes and bond angles of simple molecules, bond and molecular polarity, intermolecular forces including hydrogen bonding, bond energies and lengths, lattice structures, and how bonding explains physical properties.
Chapter 3. The Gaseous State
1 point to check
- State the assumptions of the kinetic theory for an ideal gas, explain when real gases behave ideally, use pV = nRT including to find Mr, and use Dalton's Law for partial pressures.
Chapter 4. Theories of Acids and Bases
1 point to check
- Understand and apply the Arrhenius, Brønsted-Lowry and Lewis theories of acids and bases, including non-aqueous systems for the Lewis theory.
Chapter 5. The Periodic Table
3 points to check
- Explain trends in electronic configuration, atomic and ionic radius, ionisation energy, electronegativity, melting point and electrical conductivity across Period 3 and down Groups 2 and 17.
- Describe the oxides and chlorides of Period 3: oxidation numbers, bonding, reactions with water and acid or base behaviour.
- Describe Group 2 elements as reducing agents and Group 17 elements as oxidising agents, and the thermal stability of Group 2 carbonates and Group 17 hydrides.
Chapter 6. The Mole Concept and Stoichiometry
1 point to check
- Define relative atomic, isotopic, molecular and formula mass and the mole, find relative atomic mass from isotope abundances, find empirical and molecular formulae, balance equations, and calculate reacting masses and volumes of solutions and gases.
Chapter 11. Solubility Equilibria
1 point to check
- Use the solubility product Ksp, calculate Ksp from concentrations and back, and discuss how the common ion effect and complex ion formation change the solubility of ionic salts.
Chapter 12. Organic Chemistry
2 points to check
- Use empirical, molecular, structural, displayed, skeletal and stereochemical formulae, functional groups and naming, and curly arrows for mechanisms.
- The syllabus sections are: introduction, isomerism, organic reactions and mechanisms, hydrocarbons, halogen derivatives, hydroxy compounds, carbonyl compounds, carboxylic acids and derivatives, nitrogen compounds and polymers.
Chapter 14. Transition Elements
3 points to check
- Define a transition element, give the electronic configurations of first row elements and their ions, and explain their fairly constant atomic radii and first ionisation energies.
- Describe variable oxidation states and the redox systems Fe³⁺/Fe²⁺, MnO₄⁻/Mn²⁺ and Cr₂O₇²⁻/Cr³⁺, using standard electrode potentials to predict whether redox reactions happen.
- Define ligand and complex, explain ligand exchange and colour from d orbital splitting, and explain catalysis by transition elements.
If your child is struggling
- Find the exact chapter and point. The checklists and common mistakes on this page show where an answer breaks down.
- Go back to that chapter's worked question and do it again without the steps.
- Ask the school teacher which chapters the class has covered and which are tested next.
- If several chapters are weak, work on one at a time, with short timed questions, rather than rereading everything.