24.2 A2 Level

Standard electrode potentials E

Cambridge A-Level Chemistry (9701)  · Unit 24: Electrochemistry  · 11 flashcards

Standard electrode potentials E is topic 24.2 in the Cambridge A-Level Chemistry (9701) syllabus , positioned in Unit 24 — Electrochemistry , alongside Electrolysis.  In one line: E⦵ is the potential difference of a half-cell connected to a standard hydrogen electrode (SHE) under standard conditions (298K, 1 atm/100kPa, 1 mol dm⁻³ solution of ions). It represents the tendency of a species to be reduced.

Marked as A2 Level: examined at A Level in Paper 4 (A Level Structured Questions) and Paper 5 (Planning, Analysis and Evaluation). It is not tested on the AS-only papers (Papers 1, 2 and 3).

The deck below contains 11 flashcards — 1 definition, 7 key concepts and 3 calculations — covering the precise wording mark schemes reward.  Use the definition card to lock down command-word answers (define, state), then move on to the concept and calculation cards to handle explain, describe, calculate and compare questions.

Key definition

Standard electrode (reduction) potential, E⦵

E⦵ is the potential difference of a half-cell connected to a standard hydrogen electrode (SHE) under standard conditions (298K, 1 atm/100kPa, 1 mol dm⁻³ solution of ions). It represents the tendency of a species to be reduced.

What the Cambridge 9701 syllabus says

Official 2025-2027 spec · A2 Level

These are the exact learning outcomes Cambridge sets for this topic. The candidate is expected to be able to do each of these on the relevant paper.

  1. define the terms: (a) standard electrode (reduction) potential (b) standard cell potential
  2. describe the standard hydrogen electrode
  3. describe methods used to measure the standard electrode potentials of: (a) metals or non-metals in contact with their ions in aqueous solution (b) ions of the same element in different oxidation states
  4. calculate a standard cell potential by combining two standard electrode potentials
  5. use standard cell potentials to: (a) deduce the polarity of each electrode and hence explain/deduce the direction of electron flow in the external circuit of a simple cell (b) predict the feasibility of a reaction
  6. deduce from E ⦵ values the relative reactivity of elements, compounds and ions as oxidising agents or as reducing agents
  7. construct redox equations using the relevant half-equations
  8. predict qualitatively how the value of an electrode potential, E, varies with the concentrations of the aqueous ions
  9. use the Nernst equation, e.g. E = E ⦵ + (0.059/z) log [oxidised species] [reduced species] ,
  10. understand and use the equation ΔG ⦵ = –nE ⦵ cell F

Cambridge syllabus keywords to use in your answers

These are the official Cambridge 9701 terms tagged to this section. Mark schemes credit responses that use the exact term — weave them into your answers verbatim rather than paraphrasing.

standard electrode potential standard cell potential standard hydrogen electrode Nernst equation feasibility of a reaction

Tips to avoid common mistakes in Standard electrode potentials E

Definition Flip

Define standard electrode (reduction) potential, E⦵.

Answer Flip

E⦵ is the potential difference of a half-cell connected to a standard hydrogen electrode (SHE) under standard conditions (298K, 1 atm/100kPa, 1 mol dm⁻³ solution of ions). It represents the tendency of a species to be reduced.

Key Concept Flip

Describe the key features of the standard hydrogen electrode (SHE).

Answer Flip

The SHE consists of a platinum electrode immersed in a 1 mol dm⁻³ solution of H⁺ ions, with hydrogen gas bubbled through at 1 atm/100kPa and a temperature of 298K. The platinum electrode acts as a surface for the equilibrium: 2H⁺(aq) + 2e⁻ ⇌ H₂(g). By definition, the E⦵ value is 0.00V.

Key Concept Flip

How can you measure the standard electrode potential of a metal in contact with its ions?

Answer Flip

Set up a half-cell with the metal electrode in a 1 mol dm⁻³ solution of its ions under standard conditions. Connect this half-cell to the SHE via a salt bridge. The voltmeter reading gives the standard electrode potential (E⦵) of the metal.

Key Concept Flip

How can you measure the standard electrode potential of ions of the same element in different oxidation states (e.g., Fe²⁺/Fe³⁺)?

Answer Flip

Construct a half-cell containing a platinum electrode in a solution containing both ions at 1 mol dm⁻³ concentration each, under standard conditions. Connect this half-cell to the SHE via a salt bridge. The voltmeter reading gives the standard electrode potential (E⦵) for the ion pair.

Calculation Flip

Calculate the standard cell potential (E⦵cell) for a cell composed of Zn²⁺/Zn (E⦵ = -0.76V) and Cu²⁺/Cu (E⦵ = +0.34V) half-cells.

Answer Flip

E⦵cell = E⦵(reduction) - E⦵(oxidation). In this case, copper is reduced and zinc is oxidized. E⦵cell = (+0.34V) - (-0.76V) = +1.10V.

Key Concept Flip

Given E⦵cell = +1.10V for a Zn/Cu cell, deduce the polarity of each electrode and the direction of electron flow.

Answer Flip

Since E⦵cell is positive, the reaction is spontaneous. Copper is the positive electrode (cathode, reduction) and zinc is the negative electrode (anode, oxidation). Electrons flow from the zinc electrode to the copper electrode in the external circuit.

Key Concept Flip

How can standard electrode potentials be used to predict the feasibility of a redox reaction?

Answer Flip

A reaction is feasible (spontaneous) if the calculated E⦵cell is positive. A positive E⦵cell indicates that the reaction will proceed as written under standard conditions.

Key Concept Flip

How does a more positive E⦵ value relate to the strength of an oxidizing agent?

Answer Flip

A more positive E⦵ value indicates a greater tendency for reduction, meaning the species is a stronger oxidizing agent. It has a greater ability to accept electrons.

Key Concept Flip

Write the balanced redox equation for the reaction between Fe²⁺(aq) and MnO₄⁻(aq) in acidic solution, given the half-equations.

Answer Flip

First, balance the half equations. Then, multiply each half-equation by a suitable factor so that the number of electrons is the same in both. Add the equations together and cancel the electrons. Result should be: 5Fe²⁺ + MnO₄⁻ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O

Calculation Flip

State the Nernst equation and explain how it predicts the change of electrode potential with changes in ion concentrations.

Answer Flip

The Nernst equation is: E = E⦵ + (0.059/z) log ([oxidised species]/[reduced species]). Increasing the concentration of the oxidised species increases the electrode potential (E) making it more positive and favouring reduction. The reverse is true if you increase the concentration of the reduced species.

Calculation Flip

State the equation that relates the change in Gibbs Free Energy to the Standard Cell Potential and define each of the terms.

Answer Flip

ΔG⦵ = –nFE⦵cell, where ΔG⦵ is the standard Gibbs free energy change, n is the number of moles of electrons transferred in the balanced equation, F is the Faraday constant (96500 C mol⁻¹), and E⦵cell is the standard cell potential.

More Chemistry flashcards

Browse every 9701 flashcard topic by syllabus area.

All Chemistry Flashcards
24.1 Electrolysis 25.1 Acids and bases

More topics in Unit 24 — Electrochemistry

Standard electrode potentials E sits alongside these A-Level Chemistry decks in the same syllabus unit. Each uses the same spaced-repetition system, so progress in one informs the next.

Key terms covered in this Standard electrode potentials E deck

Every term below is defined in the flashcards above. Use the list as a quick recall test before your exam — if you can't define one of these in your own words, flip back to that card.

Standard electrode (reduction) potential, E⦵

How to study this Standard electrode potentials E deck

Start in Study Mode, attempt each card before flipping, then rate Hard, Okay or Easy. Cards you rate Hard come back within a day; cards you rate Easy push out to weeks. Your progress is saved in your browser, so come back daily for 5–10 minute reviews until every card reads Mastered.