Explain how an electrode potential is measured and what standard conditions and the salt bridge contribute to a valid comparison.
An isolated half-cell has no directly measurable absolute potential
A half-cell combines a redox couple with an electrical conductor. A metal/ion half-cell uses the metal dipped into its aqueous ions. When both oxidation states are dissolved, such as Fe³⁺/Fe²⁺, an inert platinum electrode provides a conducting surface without appearing in the net reaction.
A voltmeter measures a potential difference between two electrodes, so a reference is needed. The standard hydrogen electrode is assigned 0.00 V. The standard electrode potential E° of a half-cell is its measured potential relative to that reference under standard conditions, using the reduction convention.
Standard electrode data are written with electrons on the left: oxidised form + electrons ⇌ reduced form. A more positive E° indicates a stronger tendency for the reduction relative to the reference under those conditions. The value is an electrical potential, not an amount of energy per mole that scales with coefficients.
Construct the standard hydrogen electrode
The hydrogen electrode uses hydrogen gas at 100 kPa in contact with a platinum surface and aqueous H⁺ of concentration 1.00 mol dm⁻³, at a specified temperature conventionally 298 K for tabulated values. Platinum conducts and provides a surface for the hydrogen equilibrium; it is not the source of H⁺.
Connect it to the comparison half-cell through wires and a high-resistance voltmeter, and join the solutions through a salt bridge containing a suitable inert electrolyte. A gas electrode needs effective contact among gas, electrode and electrolyte, not a gas label beside a disconnected wire.
Read the polarity as well as the magnitude. If the test electrode is the positive terminal against the hydrogen reference, its reduction potential is positive. If the connections are swapped the meter sign reverses; the chemistry has not changed.
Why the circuit needs both electron and ion movement
Electrons travel through the external wire from the electrode where oxidation occurs to the electrode where reduction occurs. The salt bridge completes the internal circuit through ion movement and limits charge build-up. It does not transfer electrons between solutions.
Use ions that do not react materially with the half-cell components. For example, chloride is unsuitable if it precipitates a metal ion in the chosen system. The bridge must make contact with both solutions; metal electrodes must contact their electrolytes. A high-resistance voltmeter draws very little current, limiting changes caused by the measurement.
For a zinc–copper cell, zinc dissolves as Zn²⁺ while Cu²⁺ is reduced and copper deposits. Anions from the bridge compensate for positive charge accumulating in the zinc half-cell; cations compensate for cations consumed in the copper half-cell. This explains the direction of ionic movement instead of treating the bridge as a decorative tube.
Worked example: use reduction potentials consistently
For an illustrative standard pair, Zn²⁺ + 2e⁻ ⇌ Zn has E° = −0.76 V and Cu²⁺ + 2e⁻ ⇌ Cu has E° = +0.34 V. Copper is the reduction electrode and zinc the oxidation electrode in the spontaneous cell. E°cell = E°reduction − E°oxidation = +0.34 −(−0.76) = +1.10 V.
The overall reaction is Zn + Cu²⁺ → Zn²⁺ + Cu. Balance electrons before adding equations, but do not multiply E° by the electron coefficient. If one half-equation needed doubling, its potential would still be the same: voltage is not doubled by rewriting a chemical equation.
A complete PAG8-style measurement records temperature, concentrations, electrode identities, electrolyte contact, bridge and voltmeter polarity. Poorly cleaned electrodes or reacting bridge ions can make a real result differ from a tabulated standard value.
Quick checks
Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official OCR A mark allocations.
Q1. Why is platinum used for Fe³⁺/Fe²⁺ rather than a piece of iron?Show answer
Both members of the intended couple are dissolved ions. Platinum provides an inert conductor. Iron metal would introduce an additional Fe/Fe²⁺ redox system and potentially react with Fe³⁺.
Q2. Explain why a salt bridge is not an electron bridge.Show answer
Its electrolyte ions move to maintain electroneutrality and complete the internal circuit. Electrons flow in the external conductor between electrodes, not through the salt solution as free electrons.
Q3. Calculate E°cell if the reduction electrode is +0.80 V and the oxidation couple is +0.34 V.Show answer
E°cell = +0.80 −(+0.34) = +0.46 V. Both quoted numbers are reduction potentials; subtract the oxidation electrode’s listed reduction value.
Q4. Must a +0.80 V half-cell have its potential doubled when its half-equation is multiplied by two?Show answer
No. The number of transferred electrons is scaled to balance the chemical equation; electrode potential is unchanged. Multiplying it would produce an incorrect cell voltage.
Q5. Name the three standard-condition labels needed for a hydrogen reference at the usual data temperature.Show answer
H₂ at 100 kPa, H⁺ concentration 1.00 mol dm⁻³ and temperature 298 K. Include a platinum electrode in contact with both gas and solution and connect the reference to the other half-cell.
Sources
Sources and examiner guidance (reviewed 6 October 2026)
- OCR A H432 specification — version 3.1 — 5.2.3, printed pp. 49–50; outcomes and additional guidance, with relevant Module 1 practical skills.
- Chemrevise — OCR A 5.2.3 — Pages 1–10; secondary coverage cross-check. Lesson explanations, data and questions are original Finesse material.
- OCR H432/01 mark scheme — June 2025 — Q20–21(d); printed pp. 26–30. Question-specific evidence, not universal marking rules.
- OCR H432/01 examiner report — June 2025 — Q20–21(d); printed pp. 42–52. Read with the corresponding question context.
- OCR H432/01 question paper — June 2025 — Q20–21(d); context for the assessment references, not reproduced questions.
Finesse Tuition is not endorsed by AQA or Chemrevise. All explanations and examples here are our own.
