Key facts and data: learn it, or look it up?

AQA: AS and A level · OCR A: AS and A level

protons = Z; neutrons = A − Z; electrons = Z − q

For atomic number Z, mass number A and signed ion charge number q, protons = Z, neutrons = A − Z and electrons = Z − q. A positive ion has lost electrons; changing charge does not change its proton number.

Z

atomic number

A

mass number

q

signed ion charge number

AQA: AS and A level · OCR A: AS and A level

Isotopes of one element have the same proton number and different neutron numbers. Their mass numbers differ. Isotope identity and ion charge are separate: removing an electron changes charge without changing the nucleus.

AQA: AS and A level · OCR A: AS and A level

Aᵣ = Σ(fᵢmᵢ) / Σfᵢ

Relative atomic mass compares the abundance-weighted mean atomic mass with one twelfth of the mass of a carbon-12 atom. Multiply each supplied relative isotopic mass by its abundance and divide the sum by total abundance; use one abundance scale throughout. For isotope peaks from 1+ ions, use the supplied m / z values. If only mass numbers are given, use them as the usual relative-isotopic-mass approximations. Relative atomic mass has no unit.

fᵢ

relative abundance of isotope i

mᵢ

relative isotopic mass of isotope i

AQA: AS and A level · OCR A: AS and A level

One orbital holds at most two electrons. The s, p and d subshells contain one, three and five orbitals, so their maximum occupancies are 2, 6 and 10 electrons. Fill subshells in the usual energy order, including 4s before 3d. When forming s- and p-block positive ions, remove electrons from the highest occupied shell first, taking p electrons before s electrons within that shell.

AQA: Learn the Cr ([Ar] 3d⁵ 4s¹) and Cu ([Ar] 3d¹⁰ 4s¹) exceptions. For d-block cations up to Z = 36, remove 4s electrons before any 3d electrons; Fe²⁺ is [Ar] 3d⁶. Use any spin rules stated in the question.

OCR A: Two electrons in the same orbital have opposite spins. Fill equal-energy orbitals singly, with parallel spins, before pairing. Cr and Cu are exceptions to the simple filling order; this app supplies their configurations in OCR AS items when needed. For OCR A Level transition elements, write Sc–Zn atom and ion configurations: Cr is [Ar] 3d⁵ 4s¹ and Cu [Ar] 3d¹⁰ 4s¹. Remove 4s electrons before 3d when forming these cations.

AQA: AS and A level · OCR A: AS and A level

First ionisation energy is the energy needed to remove one electron from each atom in one mole of gaseous atoms, forming one mole of gaseous 1+ ions: X(g) → X⁺(g) + e⁻. It is positive and is measured in kJ mol⁻¹. State gaseous species.

AQA: AS and A level · OCR A: AS and A level

When a supplied main-group sequence starts at the first ionisation energy and includes an outer-shell-to-inner-shell jump, the number of electrons removed before that jump is the outer-shell population and helps identify the group. The next electron is closer to the nucleus and less shielded. Use the position of the jump, not the size of a single value.

AQA: AS and A level · OCR A: AS and A level

n = m / M

Divide sample mass by the corresponding molar mass. Match the mass units: using mass in grams with molar mass in g mol⁻¹ gives amount in moles. Relative formula mass is unitless; molar mass carries units.

n

amount (mol)

m

mass (g)

M

molar mass (g mol⁻¹)

AQA: AS and A level · OCR A: AS and A level

N = nNₐ

Multiply the amount by the Avogadro constant to find the number of specified entities. Identify whether the entities are atoms, molecules, ions or formula units; atoms within a molecule require its atom count as well.

OCR A: The common data sheet p2 prints Nₐ = 6.02 × 10²³ mol⁻¹. The relationship N = nNₐ is not printed there.

N

number of specified entities

n

amount of those entities (mol)

Nₐ

Avogadro constant (mol⁻¹)

AQA: AS and A level · OCR A: AS and A level

c = n / V

Use the amount of solute divided by the final solution volume. With volume in dm³, concentration is in mol dm⁻³. Convert cm³ to dm³ by dividing by 1000. A solvent volume before mixing is not automatically the final solution volume.

c

amount concentration (mol dm⁻³)

n

solute amount (mol)

V

solution volume (dm³)

AQA: AS and A level · OCR A: AS and A level

pV = nRT

For the ideal-gas model, use absolute temperature in kelvin and consistent units. With R = 8.314 J mol⁻¹ K⁻¹, use pressure in Pa and volume in m³. Convert kPa to Pa (× 1000), dm³ to m³ (÷ 1000) and cm³ to m³ (÷ 10⁶) before substitution.

OCR A: R = 8.314 J mol⁻¹ K⁻¹ is printed on common data sheet p2. The ideal gas equation itself is not printed there.

p

gas pressure (Pa)

V

gas volume (m³)

n

gas amount (mol)

R

gas constant (J mol⁻¹ K⁻¹)

T

absolute temperature (K)

AQA: AS and A level · OCR A: AS and A level

n(A) / a = n(B) / b

Use the balanced coefficients as mole ratios, not mass ratios. For aA + bB → products, complete stoichiometric reaction requires n(A) / a = n(B) / b. For limiting-reagent work, divide each available amount by its coefficient. The reactant with the smallest quotient is limiting and sets the maximum product amount.

a, b

balanced stoichiometric coefficients

n(A), n(B)

reacting amounts (mol)

AQA: AS and A level · OCR A: AS and A level

yield / % = 100 × actual product / theoretical product

Compare actual isolated amount with the theoretical amount calculated from the limiting reagent. Both quantities must describe the same product on the same amount or mass basis. A high yield does not by itself establish high purity.

AQA: AS and A level · OCR A: AS and A level

atom economy / % = 100 × ν(desired)M(desired) / Σᵢ(νᵢMᵢ)

Use the balanced equation to compare the desired product mass contribution with the total reactant mass contribution. Include coefficients. Atom economy is a reaction-design quantity; it does not measure losses during an experimental work-up.

OCR A: For a balanced equation, the coefficient-weighted molar-mass totals of reactants and all products are equal. Using the total for all products as the denominator therefore gives the same atom economy.

ν

balanced coefficient (νᵢ for reactant i)

M

molar mass (Mᵢ for reactant i) (g mol⁻¹)

i

index over all reactants

AQA: AS and A level · OCR A: AS and A level

With percentage-by-mass data, treat each percentage as the mass in grams in a 100 g sample. For each element, divide its mass by its atomic molar mass in g mol⁻¹ to obtain moles of atoms. Divide all amounts by the smallest and find a consistent small whole-number ratio. Use atomic, not diatomic, molar masses and do not round a fractional ratio prematurely.

AQA: AS and A level · OCR A: AS and A level

r = Mᵣ / empirical relative formula mass

The molecular formula is an integer multiple of the empirical formula. Divide the supplied relative molecular mass by the empirical relative formula mass, then multiply every empirical subscript by that same integer. Use a consistent mass convention.

r

common integer subscript multiplier

AQA: AS and A level · OCR A: AS and A level

An ionic lattice is held by electrostatic attraction between oppositely charged ions. Electron transfer can explain formation of ions, but the ionic bond is the attraction in the resulting structure. Ions need mobility for bulk electrical conduction.

AQA: AS and A level · OCR A: AS and A level

A single covalent bond is the strong electrostatic attraction between a shared electron pair and the nuclei of the bonded atoms. Double and triple bonds contain two and three shared pairs. In forming a coordinate (dative covalent) bond, both electrons in the shared pair come from one donor. Count the resulting bonding pair in the structure; the donor origin does not create a second pair.

AQA: AS and A level · OCR A: AS and A level

Metallic bonding is the electrostatic attraction between positive metal ions and delocalised electrons within a giant lattice. The electrons can move, so metals conduct electricity when solid or molten. Layers of ions can slide while remaining attracted to the delocalised electrons, so a typical metal is malleable and ductile.

AQA: AS and A level · OCR A: AS and A level

Arrange electron-density groups around a central atom to minimise repulsion. A multiple bond counts as one group. Repulsion is strongest between two lone pairs, then a lone pair and a bonding pair, then two bonding pairs. In a tetrahedral electron-group arrangement, lone pairs compress the bond angles: CH₄ 109.5°, NH₃ 107°, H₂O 104.5°. Distinguish the electron-group arrangement from the molecular shape.

AQA: AS and A level · OCR A: AS and A level

Electronegativity differences can make bonds polar. Molecular polarity also depends on geometry: bond dipoles may cancel by symmetry. The presence of a polar bond alone therefore does not establish a permanent molecular dipole.

AQA: AS and A level · OCR A: AS and A level

Fluctuating electron distributions create instantaneous and induced dipoles. London forces act between all atoms and molecules. For comparable molecules, more electrons generally mean greater polarisability and stronger London forces. Greater accessible contact area, such as in less-branched isomers, also strengthens these attractions. Molar mass alone is not a reliable boiling-point predictor.

AQA: AQA calls these induced dipole–dipole forces, also termed van der Waals forces. These are London (dispersion) forces.

AQA: AS and A level · OCR A: AS and A level

In the usual A Level model, an H atom bonded to N, O or F can interact with a lone pair on a suitable N, O or F atom. Distinguish that intermolecular interaction from the covalent bond attaching H to its own molecule.

AQA: AS and A level · OCR A: AS and A level

q = mcΔT

For the stated constant-specific-heat model, heat gained by the sample is its mass times specific heat capacity times temperature rise. Here m is the mass of the water or solution whose temperature changes, not the mass of fuel burned or of solid reactant alone. Use the heated material’s specific heat capacity. A temperature difference of 1 K equals a difference of 1 °C. Include any separately specified vessel contribution.

OCR A: The common data sheet p2 prints water’s c = 4.18 J g⁻¹ K⁻¹. The heat equation is Learn this; that value is not a default for every sample.

q

heat gained by the sample (J)

m

mass of the substance heated (e.g. water or solution) (g)

c

specific heat capacity (J g⁻¹ K⁻¹)

ΔT

final minus initial temperature (K)

AQA: AS and A level · OCR A: AS and A level

ΔH = −qₛ / n

At constant pressure in the stated calorimetry model, reaction heat is the negative of the accounted surroundings’ heat gain. Divide by the amount corresponding to the reaction as written. Convert J to kJ when reporting kJ mol⁻¹.

ΔH

molar reaction enthalpy (kJ mol⁻¹)

qₛ

accounted surroundings heat gain, converted to kJ (kJ)

n

stoichiometric reaction amount (mol)

AQA: AS and A level · OCR A: AS and A level

For fixed initial and final states and conditions, reaction enthalpy is independent of the route taken. Reverse an equation and reverse its enthalpy sign; multiply an equation and multiply its enthalpy by the same factor. Add only steps that cancel to the required overall equation.

AQA: AS and A level · OCR A: AS and A level

Standard enthalpy of formation is the enthalpy change when one mole of a substance forms from its constituent elements, with all reactants and products in their standard states at 100 kPa and a stated temperature, often 298 K. An element in its standard state has zero standard formation enthalpy by definition. State the physical forms.

AQA: AS and A level · OCR A: AS and A level

ΔH ≈ ΣE(bonds broken) − ΣE(bonds formed)

Estimate a gaseous reaction enthalpy by adding energies for bonds broken and subtracting energies released for bonds formed. Count bonds using the balanced equation. Mean values are averages, so the estimate is not an exact value for every molecule or condensed-phase reaction.

E

positive molar bond enthalpy (kJ mol⁻¹)

AQA: A level · OCR A: A level

ΔS° = ΣνS°(products) − ΣνS°(reactants)

Multiply each supplied standard molar entropy by its balanced coefficient. Subtract the reactant total from the product total. Include elements: the zero convention for standard formation enthalpy does not apply to standard molar entropy. Entropy units are normally J mol⁻¹ K⁻¹; coefficient-weighted values refer to the reaction as written.

S°

standard molar entropy (J mol⁻¹ K⁻¹)

ν

balanced coefficient

AQA: A level · OCR A: A level

ΔG = ΔH − TΔS

Use a consistent energy unit for ΔH and TΔS, with absolute temperature in kelvin. If ΔH is in kJ mol⁻¹ and ΔS in J mol⁻¹ K⁻¹, divide the entropy value by 1000 first. A temperature extrapolation assumes the stated ΔH and ΔS model remains applicable.

ΔG, ΔH

molar energy changes, on one unit basis (kJ mol⁻¹)

T

absolute temperature (K)

ΔS

entropy change, converted to matching energy units (kJ mol⁻¹ K⁻¹)

AQA: A level · OCR A: A level

For ΔG = ΔH − TΔS with matched units at a stated temperature, a negative result supports thermodynamic feasibility in the stated exam model. It does not predict reaction speed: an activation barrier can make a feasible reaction slow. Zero marks the model’s feasibility boundary, with no net thermodynamic drive. A standard ΔG° value alone does not establish that a given mixture is at equilibrium.

AQA: AQA classifies ΔG ≤ 0 as feasible. ΔG = 0 is the threshold, often described as where a reaction just becomes feasible; zero gives no net thermodynamic drive.

OCR A: OCR uses ΔG < 0 for thermodynamic feasibility. Treat zero as the boundary, with no net thermodynamic drive.

AQA: AS and A level · OCR A: AS and A level

average disappearance rate = −Δ[A] / Δt

At constant volume, a reactant disappearance rate is the negative concentration gradient. A secant gives an interval average and a tangent gives an instantaneous value. State which species and quantity the rate refers to, with matching units.

[A]

reactant concentration (mol dm⁻³)

t

time (s)

AQA: A level · OCR A: A level

rate₂ / rate₁ = ([A]₂ / [A]₁)ᵐ

At fixed temperature and with other relevant concentrations matched, multiplying [A] by a factor f multiplies the rate by fᵐ for order m in A. Deduce orders from experimental evidence; overall-equation coefficients do not generally supply them.

AQA: A level · OCR A: A level

rate = k[A]ᵐ[B]ⁿ

The exponents are the experimentally determined orders in the stated model. Overall order is their sum. A species can have order zero while still appearing in the overall chemical equation. The rate constant applies to the specified reaction and conditions.

m, n

orders in A and B

k

rate constant; units depend on overall order

AQA: A level · OCR A: A level

units(k) = (mol dm⁻³)¹⁻ᵖ s⁻¹

Derive units by dividing the stated rate unit by the concentration units raised to the total order. With rate in mol dm⁻³ s⁻¹ and overall order p, the concentration power is 1 − p. Keep the time unit consistent.

p

overall reaction order

OCR A: A level

k = ln(2) / t½

For a first-order process with constant k, successive half-lives are equal. The rate constant is ln(2) divided by the half-life. Use the same time unit for half-life and the reciprocal-time unit of k. This specific relationship is OCR A Level scope.

k

first-order rate constant (s⁻¹)

t½

half-life when time is measured in seconds (s)

AQA: A level · OCR A: A level

k = A exp(−Eₐ/(RT)); ln k = ln A − Eₐ/(RT)

In the stated Arrhenius model, use T in kelvin and match Eₐ to the units of R. A plot of ln k against 1 / T has slope −Eₐ/R and intercept ln A. A has the same units as k.

OCR A: Both exponential and linear forms are printed on the common H032 / H432 data sheet p2. This card remains A Level only by course scope.

AQA: Not printed on the AQA 2025 insert. This app supplies the equation and R in its AQA items. Use the supplied exponential form, rearrange it to ln k = ln A − Eₐ/(RT), and plot ln k against 1 / T with gradient −Eₐ/R.

k

rate constant; units depend on overall order

Eₐ

activation energy; convert kJ to J for this R (J mol⁻¹)

R

gas constant; use the supplied value (usually 8.314) (J mol⁻¹ K⁻¹)

T

absolute temperature (K)

A

pre-exponential factor (same units as k)

AQA: AS and A level · OCR A: AS and A level

Activation energy is the minimum collision energy needed for particles to react by the stated pathway. On an enthalpy profile it is the height from the reactants to the transition-state maximum. Reaction enthalpy compares products with reactants; changing the activation barrier does not by itself change the overall enthalpy.

AQA: AS and A level · OCR A: AS and A level

Raising temperature changes the molecular-energy distribution. In the usual collision model, the main reason for the rate increase is the larger fraction of collisions with energy at or above the activation energy. Collision frequency also rises slightly; this is a secondary contribution.

AQA: AS and A level · OCR A: AS and A level

A catalyst provides an alternative reaction pathway with a lower activation barrier and is regenerated overall. It does not change the reaction’s initial and final states or its overall enthalpy change. For a reversible system it accelerates approach to equilibrium in both directions.

OCR A: In OCR AS rates, a homogeneous catalyst shares the reactants’ phase; a heterogeneous catalyst is in a different phase, such as a solid catalyst with gaseous reactants.

AQA: AS and A level · OCR A: AS and A level

In a closed system at equilibrium, forward and reverse reactions continue at equal rates, so macroscopic composition remains constant under fixed conditions. Equal rates do not require equal concentrations of reactants and products.

AQA: AS and A level · OCR A: AS and A level

Kc = [C]ᶜ[D]ᵈ / ([A]ᵃ[B]ᵇ)

For the stated homogeneous equilibrium aA + bB ⇌ cC + dD, use equilibrium concentrations and raise each to its balanced coefficient. Keep one concentration-unit convention. Initial concentrations are not interchangeable with equilibrium concentrations.

[X]

equilibrium concentration of species X (mol dm⁻³)

AQA: AS and A level · OCR A: A level

units(Kc) = (mol dm⁻³) raised to Δν

For a homogeneous concentration expression, subtract the sum of reactant coefficients from the sum of product coefficients. This difference, Δν, gives the power of the chosen concentration unit in the exam Kc convention. If Δν = 0, Kc has no units.

Δν

sum of product coefficients minus sum of reactant coefficients

AQA: A level · OCR A: A level

xᵢ = nᵢ / n(total gas); pᵢ = xᵢP

For an ideal gas mixture, a component’s partial pressure equals its mole fraction times total pressure. Include all gas species present when calculating total gas amount. Mole fraction is dimensionless; partial and total pressure use the same pressure unit.

xᵢ

gas mole fraction

pᵢ

partial pressure of gas i (kPa)

P

total pressure (kPa)

AQA: A level · OCR A: A level

Kp = p(C)ᶜp(D)ᵈ / (p(A)ᵃp(B)ᵇ)

For the stated homogeneous gas equilibrium aA + bB ⇌ cC + dD, use equilibrium partial pressures raised to their coefficients. Use one pressure unit throughout, matching the quoted Kp. In this exam convention, Kp has units (pressure unit) raised to Δν, where Δν is the sum of gaseous product coefficients minus the sum of gaseous reactant coefficients. If Δν = 0, Kp has no units.

OCR A: OCR A Level also uses Kc and Kp for heterogeneous equilibria. Omit pure solids and pure liquids; Kp contains only gaseous partial-pressure terms.

p(X)

equilibrium partial pressure of gas X (chosen pressure unit, e.g. kPa)

Δν

sum of gas product coefficients minus sum of gas reactant coefficients

AQA: AS and A level · OCR A: A level

For a fixed reaction definition, the equilibrium constant (Kc, or Kp where in scope) depends on temperature. Increasing temperature increases the equilibrium constant for an endothermic forward reaction and decreases it for an exothermic one. A changed equilibrium composition at fixed temperature does not by itself mean that this constant changed.

AQA: AS and A level · OCR A: AS and A level

For the usual ideal-gas equilibrium model at fixed temperature, decreasing volume favours the side with fewer gas moles. Count balanced gaseous coefficients. Equal gas totals give no such composition shift. Adding an inert gas at fixed volume raises total pressure but leaves the reacting gases’ partial pressures unchanged, so the ideal-gas equilibrium composition does not shift.

AQA: AS and A level · OCR A: AS and A level

At a fixed temperature, adding a catalyst changes how quickly equilibrium is approached, without changing the equilibrium composition (position) for the same closed system and species model. A faster approach is distinct from a larger equilibrium yield.

AQA: A level · OCR A: A level

pH = −log₁₀([H⁺])

In the dilute concentration model, use the numerical hydrogen-ion concentration in mol dm⁻³ inside the logarithm. A tenfold increase lowers pH by one. For the inverse, raise 10 to the power (−pH) to obtain the numerical hydrogen-ion concentration in mol dm⁻³. pH has no concentration unit.

AQA: A level · OCR A: A level

[H⁺] ≈ c(acid)

For a fully dissociated monoprotic acid, hydrogen-ion concentration is approximately the analytical acid concentration when water autoionisation is negligible. Strong describes dissociation, while concentrated describes amount per volume. Dilution can make the water contribution important.

c(acid)

analytical monoprotic-acid concentration (mol dm⁻³)

AQA: A level · OCR A: A level

Kw = [H⁺][OH⁻]; pKw = −log₁₀ Kw

At a fixed temperature, the water ionic product relates hydrogen- and hydroxide-ion concentrations in the stated dilute model. Neutrality means equal concentrations, so neutral pH is half pKw. pH 7 is the usual neutral value at 298 K, not at every temperature. Use the numerical Kw value in the stated concentration convention.

OCR A: The common data sheet p2 prints Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K. The concentration-product relationship on this card is Learn this.

Kw

water ionic product in the concentration convention (mol² dm⁻⁶)

AQA: A level · OCR A: A level

[H⁺] = Kw / [OH⁻]; pH = −log₁₀([H⁺])

Find hydroxide-ion concentration from dissociation and stoichiometry, then use the water ionic product at the stated temperature. For ordinary dilute NaOH with negligible water contribution, [OH⁻] ≈ c(NaOH). A base supplying two hydroxide ions per formula unit needs that factor.

AQA: A level · OCR A: A level

Ka = [H⁺][A⁻] / [HA]

For HA ⇌ H⁺ + A⁻, use equilibrium concentrations in the stated dilute aqueous convention. Undissociated HA belongs in the denominator. If the equilibrium is instead written using H₂O and H₃O⁺, omit [H₂O] from Ka because water is the solvent in large excess.

Ka

acid dissociation constant in this concentration convention (mol dm⁻³)

AQA: A level · OCR A: A level

[H⁺] ≈ √(Ka c)

For a pure monoprotic weak acid with negligible water contribution and small fractional dissociation, take [A⁻] ≈ [H⁺] and [HA] ≈ c. Then hydrogen-ion concentration is approximately the square root of Ka times c. Check that the resulting dissociation is small.

c

analytical acid concentration (mol dm⁻³)

AQA: A level · OCR A: A level

pKa = −log₁₀ Ka

Within the stated dilute concentration convention at a fixed temperature, a larger Ka means greater dissociation and a smaller pKa. Acid strength differs from the concentration of a prepared solution. Use the numerical Ka value in this convention. For the inverse, raise 10 to the power (−pKa) to recover that numerical Ka in mol dm⁻³.

Ka

acid dissociation constant in this concentration convention (mol dm⁻³)

pKa

negative base-10 logarithm of the specified numerical Ka

AQA: A level · OCR A: A level

[H⁺] ≈ Ka × n(HA) / n(A⁻)

For the stated weak-acid / conjugate-base buffer model, obtain the acid and conjugate-base amounts after any neutralisation first. With both present in the same solution volume, their concentration ratio equals their amount ratio. The approximation needs appreciable amounts of both partners.

n(HA), n(A⁻)

amounts remaining after reaction (mol)

AQA: A level · OCR A: A level

The conjugate base consumes small added amounts of H⁺, while the weak acid consumes small added amounts of OH⁻. This limits the change in their ratio and hence in pH. Buffer capacity is finite; a buffer does not hold one exact pH after unlimited additions.

AQA: A level · OCR A: A level

at half-neutralisation: pH ≈ pKa

In a monoprotic weak-acid / strong-base titration, the usual buffer approximation gives equal remaining acid and formed conjugate-base amounts at half the stoichiometric equivalence volume. Their ratio is one, so pH ≈ pKa. This requires the stated valid buffer model.

AQA: A level · OCR A: A level

Choose a stated indicator whose transition interval lies within the steep pH change around the titration’s equivalence region. Use the colour change expected for the actual titration direction (which solution is in the burette). The end point is the observed colour change and should fall within the equivalence region. Equivalence need not be pH 7, and a shallow weak-acid / weak-base curve lacks a suitable sharp colour endpoint in the usual model.

AQA: AS and A level · OCR A: AS and A level

Σ(number of atoms × oxidation number) = species charge

The sum of oxidation numbers equals the species’ overall charge. For a neutral formula the sum is zero; for an ion use its signed charge. Apply the relevant element rules and exceptions to the actual species before solving for an unknown oxidation number.

AQA: AS and A level · OCR A: AS and A level

Oxidation loses electrons and increases oxidation number. Reduction gains electrons and decreases oxidation number. Identify which species changes rather than relying solely on oxygen gain or loss; electron transfer also occurs in reactions without oxygen.

AQA: AS and A level · OCR A: A level

An oxidising agent accepts electrons and is itself reduced. A reducing agent donates electrons and is itself oxidised. The agent’s name describes the change it causes in its partner, not the direction of its own oxidation-number change.

OCR A: Formal oxidising-agent and reducing-agent terminology is OCR A Level 5.2.3 content. This app supplies definitions of these terms in its OCR AS items when they are used.

AQA: AS and A level · OCR A: A level

Balance the atoms and the net charge with electrons. In the stated acidic aqueous method, balance oxygen using H₂O and hydrogen using H⁺. Multiply half-equations to cancel equal electron amounts before adding them; do not leave net electrons in the final overall equation.

AQA: A level · OCR A: A level

E°cell = E°(reduction half-cell) − E°(oxidation half-cell)

For the stated forward cell direction, subtract the oxidation-side standard reduction potential from the reduction-side value. Use tabulated values in their reduction convention. Multiplying half-equations to balance electrons does not multiply these intensive potentials.

AQA: In conventional cell representation, E°cell = E°(right) − E°(left).

OCR A: For a galvanic cell under standard conditions, E°cell = E°(positive electrode) − E°(negative electrode). The positive electrode is the more positive half-cell, where reduction occurs. Measure cell potential at negligible current.

E°

standard reduction potential on the same reference scale (V)

AQA: A level · OCR A: A level

The standard hydrogen electrode is assigned E° = 0.00 V for 2H⁺(aq) + 2e⁻ ⇌ H₂(g). The A Level model uses an inert platinum contact, H₂ at 100 kPa, H⁺ at 1.00 mol dm⁻³ and a specified temperature, here 298 K. Match these conditions when comparing standard values.

AQA: A level · OCR A: A level

A positive E°cell supports thermodynamic feasibility for the stated forward reaction under the standard model. It does not guarantee a rapid observable reaction. Concentration changes can alter actual electrode potentials, so a standard prediction does not establish the direction for every nonstandard mixture.