Electrode potentials and electrochemical cells (A-level only)

Overview and exam scope

Electrode potentials and electrochemical cells are A-level-only content, not a GCSE Chemistry topic. They build on GCSE oxidation and reduction but extend this into quantitative prediction, cell notation, standard conditions and fuel-cell evaluation. This guide is aligned to AQA A-level Chemistry 3.1.11; if you study another board, use the chemistry here but check its precise wording and data-booklet conventions. AQA expects candidates to understand standard electrode potentials, use them to predict feasibility, represent cells conventionally and know commercial cells including lithium cells and alkaline hydrogen–oxygen fuel cells.1

This is a high-yield topic because one question can assess definitions, practical method, equation writing, calculation, data interpretation and evaluation. In a typical 5-mark response, candidates should plan for about 6 minutes: identify the command word, state the relevant redox direction, show the calculation or equation, then provide a precise conclusion. Marks are awarded for chemical reasoning, not just a correct final number.

Core idea: turning redox into electricity

An electrochemical cell converts chemical energy into electrical energy. Two half-cells are linked by an external wire and a salt bridge. Oxidation occurs at the more negative electrode, producing electrons; reduction occurs at the more positive electrode, using those electrons. Electrons travel in the external wire from the negative electrode to the positive electrode. Ions—not electrons—travel through the salt bridge to maintain electrical neutrality.

Memory hook: LEO says GER: Lose Electrons = Oxidation; Gain Electrons = Reduction.

Consider a zinc–copper cell. Zinc atoms oxidise because zinc releases electrons more readily:

[Zn_{(s)} \rightarrow Zn^{2+}_{(aq)} + 2e^-]

Copper(II) ions are reduced:

[Cu^{2+}{(aq)} + 2e^- \rightarrow Cu{(s)}]

Adding the half-equations gives the overall reaction:

[Zn_{(s)} + Cu^{2+}{(aq)} \rightarrow Zn^{2+}{(aq)} + Cu_{(s)}]

A candidate who says “electrons travel through the salt bridge” cannot receive credit for the mechanism. State instead: “The salt bridge permits ion migration, preventing charge accumulation in either half-cell.”

The standard hydrogen electrode and standard conditions

$A single half-cell has no measurable absolute potential; it must be compared with a reference. The standard hydrogen electrode (SHE) has a defined standard electrode potential, (E^\theta), of 0.00 V. It consists of hydrogen gas passed over an inert platinum electrode in aqueous hydrogen ions. Platinum conducts but does not take part in the redox reaction.$

A standard electrode potential is the potential difference when a half-cell is connected to the SHE under standard conditions. For full credit, state all three values and units:

Memory hook: “Two-nine-eight, one hundred, one point zero zero.” Say the units as well.

The apparatus needs a high-resistance voltmeter so that negligible current flows. If a current flowed significantly, reactant concentrations would change and the measured potential difference would fall from its maximum value. The salt bridge is usually filter paper soaked in an unreactive electrolyte such as potassium nitrate; it completes the circuit by allowing ions to migrate.1

Standard electrode potentials and feasibility

$The electrochemical series lists standard electrode potentials as reduction half-equations. A more positive (E^\theta) value means that reduction is more favourable. When two half-cells are connected, the more positive half-equation proceeds as written; the less positive half-equation is reversed and becomes oxidation.$

Memory hook: “Right is Reduction; Positive is Preferred.” Put the more positive half-cell on the right of the cell diagram, where it is reduced.

A positive calculated cell EMF shows that the redox reaction is thermodynamically feasible under standard conditions. Do not overclaim: a positive EMF does not guarantee that the reaction will be fast; activation energy and kinetics can still prevent an observable reaction.

Cell notation: presentation marks

Cell notation is a compact way to represent a cell. The more negative, oxidised half-cell goes on the left; the more positive, reduced half-cell goes on the right. A single vertical line (|) is a phase boundary, and a double line (||) represents the salt bridge. Place the species with the higher oxidation state nearest the salt bridge.

For zinc–copper:

[Zn_{(s)} | Zn^{2+}{(aq)} || Cu^{2+}{(aq)} | Cu_{(s)}]

For a half-cell containing only aqueous ions, such as (Fe^{3+}/Fe^{2+}), include an inert platinum electrode. A correct representation with the iron half-cell on the right is:

[Pt_{(s)} | H_{2(g)} | H^+{(aq)} || Fe^{3+}{(aq)}, Fe^{2+}{(aq)} | Pt{(s)}]

Candidates lose marks by putting the cell in the wrong order, omitting platinum when no conducting metal is present, or using a single line instead of a double line for the salt bridge.

Formula, equations and unit discipline

The AQA data booklet is supplied with the paper, but candidates should not assume that the EMF equation will be printed; learn it.2 Standard electrode potential tables are normally supplied in the question or insert when required.

There are no Foundation/Higher tiers at A-level. Treat all material in this guide as A-level content. For data skills, first identify which value is more positive, then label the reduction and oxidation directions before calculating. This topic rarely requires a plotted graph, but candidates must interpret tables accurately, retain signs, quote units, and distinguish a data-table prediction from experimental proof.

Worked calculation: examiner method

Question (2 marks): A silver half-cell has (E^\theta=+0.80,V) and a zinc half-cell has (E^\theta=-0.76,V). Calculate the standard cell EMF.

Step 1: Silver is more positive, so it is reduced and placed on the right. Zinc is on the left.

Step 2: Write the formula: (E^\theta_{cell}=E^\theta_{right}-E^\theta_{left}).

Step 3: Substitute with signs: ((+0.80)-(-0.76)).

Answer: (E^\theta_{cell}=+1.56,V).

Examiner commentary: One mark is awarded for correct substitution; one for +1.56 V. A candidate who writes (-0.76-(+0.80)) has reversed the order. The common error is not arithmetic; it is failing to decide which half-cell is reduced before calculating.

Commercial cells and evaluation

Rechargeable lithium cells are lightweight and have high energy density. During discharge, lithium is oxidised at the negative electrode, releasing electrons; an external power supply reverses the reactions during charging. Hydrogen–oxygen fuel cells produce electricity continuously while reactants are supplied, rather than needing a recharge cycle.

For an alkaline hydrogen–oxygen fuel cell, the product at the point of use is water. This gains credit as a local-emissions advantage. To reach the highest level in an evaluation, add limitations: hydrogen is difficult to store, is highly flammable, and may be produced using fossil fuels. A balanced conclusion is stronger than “fuel cells are green”: their environmental value depends on how the hydrogen and electricity are produced.

Practical method: measuring the EMF of a cell

Although boards label practical work differently, measuring the EMF of an electrochemical cell is an essential practical skill. Use two beakers, metal strips such as zinc and copper, 1.00 mol dm⁻³ solutions of their ions, a high-resistance voltmeter, leads, filter paper and potassium nitrate solution. First clean each metal with emery paper to remove oxide. Place each metal in its matching ion solution. Soak filter paper in potassium nitrate and use it as a salt bridge. Connect the metals to the voltmeter and record the maximum steady reading.

The expected result for a zinc–copper cell is a positive EMF of about +1.10 V under standard conditions, with zinc acting as the negative electrode and copper as the positive electrode. Marks are awarded for linking a procedural choice to its purpose: clean electrodes improve contact; a salt bridge maintains charge balance; a high-resistance voltmeter prevents significant current flow.

Common errors include dirty oxide-coated electrodes, solutions at the wrong concentration, an unsuitable salt bridge that reacts with ions, or reversed leads producing a negative displayed reading. In an exam, distinguish accuracy from reliability: polishing electrodes improves accuracy of the potential measurement; repeating the reading and calculating a mean improves reliability.

Synoptic links and final checklist

This topic connects to redox chemistry, because oxidation states and half-equations identify electron loss and gain. It connects to equilibria, because changing concentration or temperature shifts a half-cell equilibrium and changes the electrode potential. It connects to energetics and kinetics, because a positive EMF indicates thermodynamic feasibility, but rate depends on activation energy. It also links to transition metals, where pairs such as (Fe^{3+}/Fe^{2+}) require an inert platinum electrode.

Before an exam, say this final checklist aloud: standard conditions are 298 K, 100 kPa and 1.00 mol dm⁻³; more positive means reduction; EMF equals right minus left; electrons use the wire; ions use the salt bridge.

References