Electrochemical cells and cell potentialsEdexcel International A Level Chemistry: Revision notes
Section 1
Building an electrochemical cell
An electrochemical cell joins two half-cells so that a redox reaction drives electrons round an external circuit. Each half-cell contains a species and its ion (or two ions of one element with an inert platinum electrode).
- The more negative half-cell loses electrons: oxidation occurs at this negative electrode.
- Electrons flow through the wire to the more positive half-cell, where reduction occurs.
- A salt bridge (filter paper soaked in potassium nitrate solution) completes the circuit by letting ions move.
- A high-resistance voltmeter measures the cell voltage without drawing current.
For Zn and Cu: Zn(s) → Zn²⁺(aq) + 2e⁻ at the negative electrode, and Cu²⁺(aq) + 2e⁻ → Cu(s) at the positive electrode.
Section 2
Calculating the standard cell potential
The standard cell potential, E°cell, is the emf of a cell under standard conditions. It is found by combining two standard electrode potentials:
E°cell = E°(positive electrode) − E°(negative electrode), that is, E°(right) − E°(left) in the cell diagram.
Worked example, Zn|Zn²⁺ (−0.76 V) and Cu²⁺|Cu (+0.34 V): E°cell = (+0.34) − (−0.76) = +1.10 V.
A positive E°cell means the cell reaction as written is feasible. Do not multiply E° values by the stoichiometric numbers: for 2Ag⁺ + Cu → 2Ag + Cu²⁺, E°cell = (+0.80) − (+0.34) = +0.46 V, not 1.26 V.
Never double an E° value because you doubled a half-equation. E° is not dependent on the amount of substance.
Section 3
Writing cell diagrams
Cell diagrams use conventional notation. A single line | is a phase boundary, and a double line ‖ is the salt bridge.
- Write the oxidation (negative) half-cell on the left and the reduction (positive) half-cell on the right.
- Put the electrode at the outside, then the species in the order they react across the cell.
- For two ions of one element, use a comma and an inert Pt electrode.
Examples:
Zn(s) | Zn²⁺(aq) ‖ Cu²⁺(aq) | Cu(s)
Pt(s) | Fe²⁺(aq), Fe³⁺(aq) ‖ Ag⁺(aq) | Ag(s)
With this arrangement E°cell = E°(right) − E°(left) is positive for a feasible reaction.
Left = oxidation = negative electrode. If you reverse the diagram, the sign of E°cell reverses.
Section 4
Why the conditions matter
Each electrode reaction is an equilibrium, so its potential depends on temperature, concentration and pressure. E° values are only quoted at 298 K, 100 kPa and 1.00 mol dm⁻³, so that every half-cell can be compared fairly.
If conditions change, the equilibrium shifts (Le Chatelier). For Ag⁺ + e⁻ ⇌ Ag, a lower [Ag⁺] shifts the equilibrium left, making the electrode less positive. The measured emf then differs from E°cell.
Practical points: use a high-resistance voltmeter so no current flows, use fresh solutions, clean the electrodes, and use a KNO₃ salt bridge rather than KCl when silver ions are present (AgCl would precipitate).
Section 5
Standard reduction potentials and the electrochemical series
Standard electrode potentials are also called standard reduction potentials, because every half-equation is written as a reduction: oxidised form + ne⁻ ⇌ reduced form.
Listing half-cells in order of E° gives the electrochemical series:
- Most negative E°: the reduced form (e.g. Mg, Li) is a strong reducing agent.
- Most positive E°: the oxidised form (e.g. Cl₂, MnO₄⁻) is a strong oxidising agent.
A species can reduce any oxidised form that lies below it (more positive). Example: Mg (−2.37 V) reduces Ni²⁺ (−0.25 V), and the cell has E°cell = +2.12 V.
Must Know
- E°cell = E°(right) − E°(left); a positive value means a feasible reaction
- Never multiply E° by the number of electrons or moles
- Cell diagram: oxidation on the left, ‖ for the salt bridge, Pt for ions of one element
- E° is only defined at 298 K, 100 kPa, 1.00 mol dm⁻³ because the electrode reactions are equilibria
- Standard reduction potentials ordered give the electrochemical series
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Electrochemical cells and cell potentials
- A cell is built from a zinc half-cell and a copper half-cell under standard conditions. Standard electrode potentials: Zn²⁺(aq)|Zn(s) −0.76 V; Cu²⁺(aq)|Cu(s) +0.34 V.Write the overall ionic equation for the cell reaction, including state symbols, and state the direction of electron flow in the external circuit.2 marks
- A student builds a cell from a silver half-cell and an iron(III)/iron(II) half-cell, both under standard conditions. Standard electrode potentials: Ag⁺(aq)|Ag(s) +0.80 V; Fe³⁺(aq)|Fe²⁺(aq) +0.77 V.The student repeats the experiment using 0.50 mol dm⁻³ silver nitrate solution in the silver half-cell. Explain why the cell emf would no longer be the standard value of E°cell.2 marks
- A data table gives these standard electrode potentials: Mg²⁺(aq)|Mg(s) −2.37 V; Ni²⁺(aq)|Ni(s) −0.25 V; Ag⁺(aq)|Ag(s) +0.80 V; Cl₂(g)|Cl⁻(aq) +1.36 V.A cell is made from the Mg²⁺|Mg and Ni²⁺|Ni half-cells under standard conditions. Write the cell diagram and calculate E°cell.3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).