# Redox Equilibria

> Edexcel International A-Level Chemistry · IAL Chemistry Unit 5
> Source: https://www.owlsprep.com/study/edexcel-ial-chemistry-u5-redox-equilibria/

This guide covers all Edexcel IAL Chemistry Unit 5 (WCH15) content for redox equilibria, including electrode potentials, cell calculations, redox titrations, feasibility predictions, and fuel cells, aligned to spec points 16.1–16.19.

**Prerequisites:** [Redox half-equations and oxidation number calculation (Unit 2)](https://www.owlsprep.com/study/edexcel-ial-chemistry-u2-redox-reactions/); [Enthalpy and entropy basics (Unit 4)](https://www.owlsprep.com/study/edexcel-ial-chemistry-u4-energetics/)

## Learning objectives

- Define standard electrode potential and describe its measurement using the standard hydrogen electrode
- Calculate standard cell emf and construct valid cell diagrams for electrochemical cells
- Use E° values to predict thermodynamic feasibility of redox reactions, including disproportionation
- Perform accurate calculations for redox titrations involving MnO₄⁻/Fe²⁺ and I₂/S₂O₃²⁻
- Write electrode equations for hydrogen-oxygen fuel cells in acidic and alkaline electrolyte conditions
- Explain limitations of E° feasibility predictions, including kinetic stability and non-standard conditions

## Standard Electrode Potentials and Electrochemical Cells

**Standard Electrode Potential** — The potential of a half-cell measured relative to the standard hydrogen electrode (SHE) under standard conditions.

*Example:* E° for Cu²⁺(aq) + 2e⁻ → Cu(s) is +0.34 V

Standard conditions are defined as 298 K (25°C), 100 kPa gas pressure, and 1.00 mol dm⁻³ concentration of all aqueous species. The SHE is the universal reference half-cell, consisting of a platinum electrode submerged in 1.00 mol dm⁻³ H⁺ solution with H₂ gas bubbled over the electrode at 100 kPa, and is assigned an E° value of 0 V by definition.

E° can be measured for three types of half-cells: metal/metal ion (metal strip dipped in its ion solution), non-metal/ion (platinum electrode with non-metal gas bubbled over, submerged in its ion solution), and half-cells with the same element in two oxidation states (platinum electrode in solution with equal concentrations of both ion species).

> **note**
>
> All E° values in the Edexcel data booklet are written as reduction half-equations, with the most negative values listed first.

**Worked example:** Describe the setup used to measure the standard electrode potential of the Zn²⁺/Zn half-cell (E° = -0.76 V).

1. 1. Set up the standard hydrogen electrode (SHE) as the reference half-cell: platinum electrode in 1.00 mol dm⁻³ H⁺ solution, H₂ gas bubbled at 100 kPa, temperature 298 K.
2. 2. Set up the zinc half-cell: clean zinc metal strip submerged in 1.00 mol dm⁻³ Zn²⁺ solution at 298 K.
3. 3. Connect the two half-cells with a salt bridge (e.g. KNO₃ soaked filter paper) to complete the ionic circuit and balance charge.
4. 4. Connect the zinc and platinum electrodes to a high-resistance voltmeter; the measured potential difference is the E° of the Zn²⁺/Zn half-cell: -0.76 V.

> **Exam tip:** You will not be asked to draw full cell setups, but you must be able to label key components of E° measurements for 1-2 mark questions.

## Cell Diagrams and Standard Cell Emf Calculations

**Standard Cell Emf** — The maximum potential difference between two connected half-cells under standard conditions, when no current flows through the circuit.

Cell diagrams follow a standard convention: the oxidation half-cell (more negative E°, stronger reducing agent) is written on the left, the reduction half-cell (more positive E°, stronger oxidising agent) on the right. A single vertical line | represents a phase boundary, and a double vertical line || represents the salt bridge. Platinum electrodes are included for half-cells with no solid metal reactant.

$$E^\circ_{\text{cell}} = E^\circ_{\text{reduction (right)}} - E^\circ_{\text{oxidation (left)}}$$

A positive E°cell value confirms the reaction is thermodynamically feasible under standard conditions.

**Worked example:** Given E°(Cu²⁺/Cu) = +0.34 V and E°(Zn²⁺/Zn) = -0.76 V, write the cell diagram and calculate the standard cell emf.

1. 1. Identify the half-cells: Zn has a more negative E°, so it is oxidised (left side of diagram), Cu is reduced (right side).
2. 2. Write the cell diagram: Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s)
3. $$E^\circ_{\text{cell}} = +0.34 - (-0.76) = +1.10 V$$
4. 4. The positive value confirms the reaction is feasible as written.

> **warning**
>
> Never flip the sign of E° values when calculating E°cell. Always use the reduction half-equation values directly from the data booklet in the formula.

**Check your understanding**

1. What is the correct cell diagram for a cell containing Fe³⁺/Fe²⁺ (E°=+0.77 V) and I₂/I⁻ (E°=+0.54 V) half-cells?

   - Pt(s) | I⁻(aq), I₂(aq) || Fe³⁺(aq), Fe²⁺(aq) | Pt(s)
   - Pt(s) | Fe³⁺(aq), Fe²⁺(aq) || I⁻(aq), I₂(aq) | Pt(s)
   - Fe(s) | Fe³⁺(aq), Fe²⁺(aq) || I⁻(aq), I₂(aq) | I(s)

   *Answer:* Pt(s) | I⁻(aq), I₂(aq) || Fe³⁺(aq), Fe²⁺(aq) | Pt(s)

   *Why:* The I₂/I⁻ half-cell has a lower E° so is oxidised (placed on the left). Both half-cells use platinum electrodes as there is no solid metal reactant.

## Feasibility of Redox Reactions and Limitations

A positive E°cell means a reaction is thermodynamically feasible under standard conditions. E°cell is proportional to total entropy change and ln K, so a larger positive E°cell corresponds to a more complete reaction at equilibrium.

To predict the feasibility of disproportionation, calculate E°cell for the reaction where the same element is both oxidised and reduced: if the value is positive, the reaction is feasible. For example, Cu⁺ disproportionates because E°cell = E°(Cu⁺/Cu) - E°(Cu²⁺/Cu⁺) = +0.52 V - +0.15 V = +0.37 V.

> **warning**
>
> Feasibility does not guarantee an observable reaction. Kinetic stability (high activation energy) or non-standard conditions can make reactions very slow or non-spontaneous even with a positive E°cell.

**Worked example:** Predict if the reaction between MnO₄⁻ (E°(MnO₄⁻/Mn²⁺) = +1.51 V) and Cl⁻ (E°(Cl₂/Cl⁻) = +1.36 V) is feasible under standard conditions, and state one limitation of this prediction.

1. 1. MnO₄⁻ has a higher E° so is reduced, Cl⁻ is oxidised.
2. $$E^\circ_{\text{cell}} = 1.51 - 1.36 = +0.15 V$$
3. 3. The positive E°cell means the reaction is thermodynamically feasible under standard conditions.
4. 4. Limitation: The reaction has a high activation energy, so it is very slow at room temperature unless heated or a catalyst is added, so no observable reaction may occur.

## Redox Titrations

**Redox Titration** — A volumetric analysis technique that uses a redox reaction between an analyte and titrant to determine the unknown concentration of the analyte.

Two standard redox titrations are specified for this unit: 1. Manganate(VII) (MnO₄⁻) titration with Fe²⁺: carried out in acidic conditions, self-indicating (purple MnO₄⁻ turns colourless when reduced to Mn²⁺, endpoint is a permanent pale pink), mole ratio MnO₄⁻:Fe²⁺ = 1:5. 2. Iodine-thiosulfate titration: I₂ is produced from the oxidising analyte, titrated with S₂O₃²⁻, starch indicator (blue-black to colourless at endpoint), mole ratio S₂O₃²⁻:I₂ = 2:1.

**Worked example:** 25.0 cm³ of Fe²⁺ solution is titrated with 0.0200 mol dm⁻³ KMnO₄ solution, requiring 22.4 cm³ to reach the endpoint. Calculate the concentration of Fe²⁺ in mol dm⁻³.

1. $$n(\text{MnO}_4^-) = c \times V = 0.0200 \times 22.4 / 1000 = 4.48 \times 10^{-4} \text{ mol}$$
2. 2. Mole ratio MnO₄⁻:Fe²⁺ = 1:5, so n(Fe²⁺) = 5 × 4.48 × 10⁻⁴ = 2.24 × 10⁻³ mol
3. $$c(\text{Fe}^{2+}) = n / V = 2.24 \times 10^{-3} / 0.0250 = 0.0896 \text{ mol dm}^{-3}$$

> **Exam tip:** Always state the mole ratio between reactants in your calculation steps to gain method marks, even if you use shortcuts to solve the problem.

## Fuel Cells

**Fuel Cell** — An electrochemical cell that converts the chemical energy of a fuel (e.g. H₂, methanol) and oxidising agent (O₂) directly into electrical energy continuously, as long as fuel is supplied.

Hydrogen-oxygen fuel cells are the most common type, with either acidic (e.g. H₂SO₄) or alkaline (e.g. KOH) electrolytes. They have higher efficiency than combustion engines, produce only water as waste for H₂ fuel, and are used in electric vehicles and space missions.

**Worked example:** Write the half-equations and overall equation for a hydrogen-oxygen fuel cell with an alkaline electrolyte.

1. 1. Oxidation (negative electrode): H₂ is oxidised to H₂O, using OH⁻ from the electrolyte:
2. $$H_2(g) + 2OH^-(aq) \rightarrow 2H_2O(l) + 2e^-$$
3. 2. Reduction (positive electrode): O₂ is reduced to OH⁻:
4. $$O_2(g) + 2H_2O(l) + 4e^- \rightarrow 4OH^-(aq)$$
5. 3. Multiply the oxidation half-equation by 2 to balance electrons, then add to the reduction half-equation:
6. $$2H_2(g) + O_2(g) \rightarrow 2H_2O(l)$$

## Common pitfalls

- **Wrong:** Swapping the sign of E° values when calculating E°cell
  - Why it fails: E° values are always reported as reduction potentials, so the formula E°cell = E°(reduction) - E°(oxidation) already accounts for the oxidation process without sign changes
  - Correct: Use E° values directly from the data booklet (all reduction) in the formula, do not flip signs for oxidation half-equations.
- **Wrong:** Assuming a positive E°cell means the reaction will definitely be observed in the lab
  - Why it fails: E° only predicts thermodynamic feasibility, not reaction rate: many feasible reactions have very high activation energy (kinetic stability) or use non-standard conditions
  - Correct: Always mention kinetic stability and non-standard conditions as limitations of E° predictions when asked in exams.
- **Wrong:** Writing cell diagrams with the stronger oxidising agent on the left side
  - Why it fails: Cell diagram convention places the oxidation half-cell (more negative E°, stronger reducing agent) on the left, reduction half-cell on the right
  - Correct: Order half-cells by E°: most negative (oxidation) left, most positive (reduction) right, separated by || for the salt bridge.
- **Wrong:** Using the wrong mole ratio for redox titration calculations (e.g. 1:1 for MnO₄⁻:Fe²⁺)
  - Why it fails: Redox reactions have electron-balanced stoichiometry: MnO₄⁻ gains 5 e⁻, each Fe²⁺ loses 1 e⁻, so 1:5 ratio is required for charge balance
  - Correct: Memorise standard ratios: MnO₄⁻:Fe²⁺ = 1:5, S₂O₃²⁻:I₂ = 2:1, or balance half-equations to find ratios for unfamiliar reactions.
- **Wrong:** Writing H⁺ ions in half-equations for alkaline fuel cells
  - Why it fails: Alkaline electrolytes have excess OH⁻, so H⁺ ions would immediately react to form H₂O and cannot exist in significant concentration
  - Correct: Use OH⁻ and H₂O to balance oxygen and hydrogen in alkaline half-equations, H⁺ and H₂O for acidic electrolytes.

## Cheatsheet

| Concept | Key Rule/Formula |
| --- | --- |
| Standard E° conditions | 298 K, 100 kPa, 1.00 mol dm⁻³ aqueous concentrations |
| E°cell calculation | $E^\circ_{\text{cell}} = E^\circ_{\text{reduction}} - E^\circ_{\text{oxidation}}$ (positive = feasible) |
| Cell diagram convention | Oxidation left \| reduction right, \| phase boundary, \|\| salt bridge |
| MnO₄⁻/Fe²⁺ titration | 1:5 mole ratio, acidic, self-indicating (purple → colourless) |
| I₂/S₂O₃²⁻ titration | 2:1 mole ratio, starch indicator (blue-black → colourless) |
| H₂-O₂ fuel cell overall | $2H_2 + O_2 \rightarrow 2H_2O$ |
| Feasibility limitations | Kinetic stability (high Eₐ), non-standard conditions |

## What's next

Now that you have mastered redox equilibria, you are ready to move on to the next topics in Edexcel IAL Chemistry Unit 5: transition metal chemistry, which relies heavily on the redox properties of d-block elements, and organic nitrogen compounds including amines, amides, and amino acids. Redox equilibria is a high-weightage topic in WCH15 exams, often combined with practical questions on electrochemical cells or redox titrations, so practice past paper questions to build speed and accuracy. Make sure you memorise the standard redox titration mole ratios and fuel cell half-equations, as these are frequently tested in 3-5 mark calculation and recall questions. You should also revise measurement uncertainty for titration practical questions, as required by the specification.

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