Electrochemistry: Electrolysis and Standard Electrode Potentials

This guide covers subtopics 24.1, Electrolysis, and 24.2, Standard electrode potentials, standard cell potentials and the Nernst equation, from Topic 24, Electrochemistry, of Cambridge International AS & A Level Chemistry 9701, 2025–2027 series. This is A Level content.

Before studying this

This resource assumes oxidation numbers and half-equation construction from Redox Processes: Oxidation Numbers and Electron Transfer — electrode potentials describe the same electron-transfer chemistry, now quantified as a voltage. It also assumes the mole concept from Atoms, Molecules and Stoichiometry.

Syllabus coverage

CAMBRIDGE INTERNATIONAL AS & A LEVEL CHEMISTRY 9701 — A Level, Topic 24

24.1 Electrolysis — predicting the substances liberated during electrolysis from the electrolyte’s state, its position in the redox series, and concentration; stating and applying F = Le; calculating charge passed (Q = It) and the mass or volume of substance liberated; describing the determination of the Avogadro constant by an electrolytic method.

24.2 Standard electrode potentials, standard cell potentials and the Nernst equation — defining standard electrode and cell potential; describing the standard hydrogen electrode and methods of measuring standard electrode potentials; calculating a standard cell potential from two electrode potentials; using standard cell potentials to deduce electrode polarity, electron flow direction and reaction feasibility; deducing relative reactivity of oxidising/reducing agents from E° values; constructing redox equations from half-equations; predicting qualitatively and (via the Nernst equation) quantitatively how electrode potential varies with ion concentration; using ΔG = −nE°_cellF.

Electrolysis

Electrolysis uses electrical energy to drive a non-spontaneous redox reaction, decomposing an electrolyte at two electrodes: reduction at the cathode (negative electrode, attracting cations), oxidation at the anode (positive electrode, attracting anions).

Predicting what’s liberated depends on three things together: whether the electrolyte is molten (only the compound’s own ions are present, so those are what’s discharged) or aqueous (H⁺/OH⁻ from water compete with the electrolyte’s ions); each species’ position in the redox series (the species that is more easily reduced/oxidised — closer to the top of a table of E° values in the appropriate direction — is discharged preferentially); and concentration (chloride, whose E⦵ of +1.36 V is only a little more positive than water’s +1.23 V, is discharged in preference to oxygen when concentrated — concentrated NaCl(aq) gives Cl₂ at the anode, very dilute NaCl(aq) gives O₂; iodide and bromide, with less positive E⦵ than water, are discharged even when dilute).

Quantitative electrolysis links the charge passed to the amount of substance produced. The Faraday constant, F, is the charge on one mole of electrons:

F = Le

where L is the Avogadro constant and e is the charge on a single electron. Charge passed is:

Q = It

Worked example. A current of 2.00 A is passed through aqueous CuSO₄ electrolyte with copper electrodes for 3860 s. Calculate the mass of copper deposited. (F = 96 500 C mol⁻¹, A_r(Cu) = 63.5)

Q = It = 2.00 × 3860 = 7720 C

moles of electrons = Q / F = 7720 / 96 500 = 0.0800 mol

Cu²⁺ + 2e⁻ → Cu, so moles of Cu = 0.0800 / 2 = 0.0400 mol

mass of Cu = 0.0400 × 63.5 = 2.54 g

The Avogadro constant by electrolysis. Passing a measured, known charge through a cell and measuring the mass (or volume, for a gas) of product liberated lets you calculate the number of electrons needed per mole of product from the balanced half-equation, and hence a value for the Avogadro constant, using F = Le with an independently known value of e.

Standard electrode potentials

A standard electrode (reduction) potential, E°, is the EMF measured for a half-cell connected to a standard hydrogen electrode, under standard conditions (298 K, 101 kPa, 1 mol dm⁻³ solutions — the syllabus assumes 101 kPa), with the half-cell written as a reduction.

The standard hydrogen electrode is the reference against which all other electrode potentials are measured, arbitrarily assigned E° = 0.00 V. It consists of H₂ gas at 101 kPa bubbled over a platinum electrode (platinised for a large surface area and catalytic activity), immersed in 1 mol dm⁻³ H⁺(aq) at 298 K.

Measuring electrode potentials. For a metal/metal-ion half-cell (e.g. Zn²⁺(aq)/Zn(s)), the metal itself acts as the electrode, connected to the hydrogen electrode via a wire (through a voltmeter) and a salt bridge. For a half-cell involving ions of the same element in different oxidation states in solution (e.g. Fe³⁺(aq)/Fe²⁺(aq)), an inert platinum electrode is used to make electrical contact without itself taking part in the reaction.

Standard cell potential, E°_cell, is the EMF of a cell built from two half-cells under standard conditions, calculated as:

E°_cell = E°(reduction, cathode) − E°(reduction, anode)

or equivalently, the more positive E° minus the more negative E°.

Worked example. Calculate the standard cell potential for a cell combining Cu²⁺(aq)/Cu(s) (E° = +0.34 V) and Zn²⁺(aq)/Zn(s) (E° = −0.76 V), and deduce the direction of electron flow.

E°_cell = E°(more positive) − E°(more negative) = (+0.34) − (−0.76) = +1.10 V

Copper has the more positive (less negative) E°, so Cu²⁺ is preferentially reduced (Cu²⁺ + 2e⁻ → Cu) — copper is the cathode (positive electrode). Zinc, with the more negative E°, is oxidised (Zn → Zn²⁺ + 2e⁻) — zinc is the anode (negative electrode), and electrons flow through the external circuit from zinc to copper.

Predicting feasibility of a specified reaction uses a different construction from finding the EMF of a freely-combined cell. For a reaction written as given — with one species stated to be reduced and another stated to be oxidised — calculate:

E°_cell = E°(species reduced) − E°(species oxidised)

A positive result means the reaction is thermodynamically feasible as written (though, as with ΔG, this says nothing about rate); a negative result means it is not feasible as written — the reverse reaction is the feasible one instead.

Worked example (infeasible reaction). Does Fe³⁺(aq) oxidise Cl⁻(aq) to Cl₂? Fe³⁺/Fe²⁺ has E° = +0.77 V (species reduced: Fe³⁺ → Fe²⁺) and Cl₂/Cl⁻ has E° = +1.36 V (species oxidised: Cl⁻ → ½Cl₂).

E°_cell = E°(reduced) − E°(oxidised) = (+0.77) − (+1.36) = −0.59 V

The negative value shows this reaction is not feasible — iron(III) is not a strong enough oxidising agent to oxidise chloride ions. (The reverse reaction, chlorine oxidising Fe²⁺ to Fe³⁺, is feasible instead, with E°_cell = +1.36 − 0.77 = +0.59 V.)

Constructing redox equations. Write the two relevant half-equations, multiply each by a whole number if needed so the electrons lost equal the electrons gained, then add them together and cancel the electrons — the same method as balancing by oxidation-number change, applied via half-equations instead.

Deducing relative reactivity. A more negative E° indicates a stronger reducing agent (more readily oxidised — loses electrons more easily); a more positive E° indicates a stronger oxidising agent (more readily reduced — gains electrons more easily). Ranking half-equations by E° value gives an electrochemical series, from which the feasible direction of any redox reaction between two of the species can be read off.

The Nernst equation

Electrode potential is not fixed — it varies with the concentration of the aqueous ions involved, since concentration affects the equilibrium position of the electrode half-reaction.

Qualitatively: increasing the concentration of the oxidised species (the species being reduced, on the left of the half-equation as written for reduction) makes E more positive (more oxidising); increasing the concentration of the reduced species makes E less positive (more reducing) — consistent with Le Chatelier’s principle applied to the half-equilibrium.

Quantitatively, the Nernst equation:

E = E° + (0.059 / z) log([oxidised species] / [reduced species])

where z is the number of electrons transferred in the half-equation. For example, for Cu²⁺(aq) + 2e⁻ ⇌ Cu(s), z = 2 and the expression uses [Cu²⁺]; for Fe³⁺(aq) + e⁻ ⇌ Fe²⁺(aq), z = 1 and the expression uses [Fe³⁺]/[Fe²⁺] (Cu(s) does not appear, as a pure solid has no meaningful “concentration” in this expression).

Worked example. Calculate the electrode potential of a Cu²⁺(aq)/Cu(s) half-cell in which [Cu²⁺] has been diluted to 0.0100 mol dm⁻³ (E° = +0.34 V, z = 2).

E = E° + (0.059 / z) log[Cu²⁺] = 0.34 + (0.059 / 2) × log(0.0100)

E = 0.34 + (0.0295 × (−2)) = 0.34 − 0.059 = +0.281 V

Diluting the Cu²⁺(aq) shifts the half-equilibrium to favour the reverse (oxidation) direction by Le Chatelier’s principle, so the electrode potential becomes less positive than the standard value — consistent with the qualitative rule above.

Linking to Gibbs free energy:

ΔG = −nE°_cellF

where n is the number of electrons transferred in the overall cell reaction. This connects the electrochemical feasibility criterion (positive E°_cell) directly to the thermodynamic one (negative ΔG) — a positive E°_cell always corresponds to a negative ΔG, since F is always positive.

Common mistakes

Reversing the E°_cell subtraction. When finding the EMF of a cell freely built from two half-cells, it is always more positive E° minus more negative E°, regardless of which half-cell is “written first” — this always gives a positive value. But when testing the feasibility of a specified reaction, use E°(species reduced) − E°(species oxidised) as the reaction is actually written: this can come out negative, and a negative value correctly signals that the reaction is not feasible. Applying the “more positive minus more negative” shortcut to a feasibility question makes every reaction look feasible, which is wrong — see the iron(III)/chloride example above.

Forgetting that a pure solid or liquid doesn’t appear in the Nernst equation’s concentration term. Only aqueous ionic species (or gas pressures, where relevant) go in the log term.

Assuming a positive E°_cell guarantees a fast reaction. Electrode potentials, like ΔG, describe feasibility (whether a reaction can happen), not rate (how fast it does) — a reaction can be feasible and still proceed extremely slowly if it has a high activation energy.

Confusing anode and cathode polarity between electrolytic and voltaic (cell) setups. In an electrolytic cell the anode is positive and forces oxidation; in a voltaic (galvanic) cell like the Cu/Zn example above, oxidation still happens at the anode, but the anode is the negative electrode, since it’s the source of electrons the cell itself generates.

Quick revision checklist

• Q = It; moles of electrons = Q / F
• Standard hydrogen electrode: E° = 0.00 V by definition, 298 K, 101 kPa, 1 mol dm⁻³
• E°_cell of a freely-built cell = more positive E° − more negative E° (always positive)
• Feasibility of a specified reaction: E°_cell = E°(species reduced) − E°(species oxidised) — can be negative, meaning infeasible
• More negative E° = stronger reducing agent; more positive E° = stronger oxidising agent
• Positive E°_cell (by the reduced-minus-oxidised construction) = feasible reaction; negative = not feasible as written
• Nernst equation: E = E° + (0.059/z) log([ox]/[red])
• ΔG = −nE°_cellF links electrode potential to Gibbs free energy

Related resources

• Redox Processes: Oxidation Numbers and Electron Transfer — half-equations and oxidation numbers this topic builds on
• Transition Elements: Properties, Complexes and Redox Chemistry — applies E° values to predict feasibility of transition-metal redox reactions
• Chemical Energetics: Lattice Energy, Entropy and Gibbs Free Energy — the ΔG feasibility criterion this topic connects to
• Cambridge AS & A Level Chemistry hub

Written against Cambridge International AS & A Level Chemistry 9701, 2025–2027 series. Always check the current syllabus for your examination year.