AQA A-Level Chemistry (7405)

## Learning Outcomes

- Define electrode potential and standard electrode potential precisely
- Describe how electrochemical cells produce electricity from redox reactions
- Explain the role of the standard hydrogen electrode and standard conditions
- Calculate standard cell e.m.f. using reduction potentials
- Predict which species are oxidised and reduced in an electrochemical cell
- Write correct conventional cell diagrams, including inert electrodes where needed
- Evaluate what a positive E°cell does and does not tell you about a reaction

## AQA A-Level Chemistry (7405) Syllabus

For the AQA A-Level Chemistry (7405), you are expected to understand electrode potentials and cells, with a focus on the following syllabus points:

- Know that a half-cell contains a species in two oxidation states that can establish a redox equilibrium
- Define a **standard electrode potential, E°**, as the e.m.f. of a half-cell measured against the standard hydrogen electrode under standard conditions
- Recall the standard conditions for electrode potentials: **298 K**, solutions at **1.00 mol dm⁻³**, and gases at **100 kPa**
- Describe the standard hydrogen electrode, including **H₂(g)**, **H⁺(aq)**, and an inert **Pt** electrode with **E° = 0.00 V**
- Use tabulated standard electrode potentials, written as reduction half-equations, to predict the direction of redox reactions
- Calculate **E°cell = E°(right-hand electrode) − E°(left-hand electrode)**
- Write conventional cell notation using **|** for phase boundaries and **||** for the salt bridge
- Explain the function of the salt bridge and the direction of electron flow in an electrochemical cell

## Test Your Knowledge

_Attempt these questions before reading this article. If you find some difficult or cannot remember the answers, look more closely at that area during your revision._

1. What is meant by a standard electrode potential?
2. Why can a single half-cell potential not be measured on its own?
3. In a Zn/Cu cell, which electrode is oxidised and which is reduced?
4. How do you calculate E°cell from two tabulated reduction potentials?
5. Why is a platinum electrode needed in some half-cells?

## Introduction

Electrode potentials link redox chemistry to electricity. In this topic, you are using redox ideas you already know, but now in a more precise way: which species has the greater tendency to gain electrons, which will lose electrons, and how that difference can be measured as a voltage.

This matters in Paper 3 because the questions are often synoptic. You may need to combine redox reasoning, equilibrium ideas, practical set-up, and careful use of data-book values. The marks are usually won or lost on conventions: knowing that tabulated values are **reduction** potentials, using **right minus left**, and writing the cell diagram correctly.

> **Key Term: electrode potential**
> A measure of the tendency of a species to gain electrons and be reduced.

### How half-cells and electrochemical cells work

A **half-cell** is a system where a redox equilibrium can be established between two forms of a species. The simplest example is a metal in a solution containing its own ions:

Cu²⁺(aq) + 2e⁻ ⇌ Cu(s)

At the metal surface, some copper atoms can form Cu²⁺ ions, releasing electrons, while some Cu²⁺ ions can gain electrons and deposit as copper metal. Both processes are possible, so an equilibrium exists.

> **Key Term: half-cell**
> A system containing a species in two forms that can interconvert by gain or loss of electrons.

A single half-cell does not have an absolute measurable voltage by itself. You can only measure a **potential difference** between two half-cells. When two half-cells are connected by a wire and a salt bridge, they form an **electrochemical cell**.

> **Key Term: electrochemical cell**
> A cell that produces electrical energy from a redox reaction occurring in separate half-cells.

In a working cell, one half-cell undergoes oxidation and the other reduction.

- **Oxidation** happens at the **anode**
- **Reduction** happens at the **cathode**

For a chemical cell operating spontaneously, electrons flow through the external wire from the more negative half-cell to the more positive half-cell.

Take the classic zinc-copper cell:

- Zn²⁺(aq) + 2e⁻ ⇌ Zn(s) E° = **−0.76 V**
- Cu²⁺(aq) + 2e⁻ ⇌ Cu(s) E° = **+0.34 V**

Copper has the more positive reduction potential, so copper ions are more likely to be reduced. Zinc therefore acts in the opposite direction: zinc metal is oxidised.

Half-equations in the operating cell:

- Anode: Zn(s) → Zn²⁺(aq) + 2e⁻
- Cathode: Cu²⁺(aq) + 2e⁻ → Cu(s)

Overall equation:

Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

Why do electrons flow from zinc to copper? Because zinc has a greater tendency to lose electrons than copper does. That is what its more negative E° value is telling you.

What happens to the solutions?

- In the zinc half-cell, Zn²⁺ concentration increases because zinc atoms enter the solution
- In the copper half-cell, Cu²⁺ concentration decreases because ions are removed from solution

The salt bridge is essential. Without it, charge would quickly build up and electron flow would stop.

> **Key Term: salt bridge**
> A connection containing an inert electrolyte that allows ions to move between half-cells, maintaining electrical neutrality.

The ion movement is often tested. In a Zn/Cu cell:

- The anode solution becomes more positive as Zn²⁺ forms, so **anions** from the salt bridge move towards the anode
- The cathode solution becomes less positive as Cu²⁺ is removed, so **cations** from the salt bridge move towards the cathode

A very common error is saying electrons move through the salt bridge. They do not. Electrons move through the wire; ions move through the salt bridge.

> **Test Tip:** If a question asks for the role of the salt bridge, give both ideas: it allows ion movement and maintains electrical neutrality. Do not say it carries electrons.

### Standard electrode potentials and the standard hydrogen electrode

Because you cannot measure a single half-cell on its own, chemists need a reference half-cell. The reference used is the **standard hydrogen electrode**, often abbreviated to **SHE**.

> **Key Term: standard electrode potential (E°)**
> The e.m.f. of a half-cell measured against the standard hydrogen electrode under standard conditions.

The standard hydrogen electrode consists of:

- hydrogen gas, H₂(g)
- hydrogen ions, H⁺(aq)
- a platinum electrode

The half-equation is:

2H⁺(aq) + 2e⁻ ⇌ H₂(g)

Its potential is defined as **0.00 V**.

> **Key Term: standard hydrogen electrode (SHE)**
> The reference half-cell assigned a potential of 0.00 V, using H₂(g), H⁺(aq), and a platinum electrode under standard conditions.

The platinum electrode is needed because there is no solid conducting substance in the H⁺/H₂ system. Platinum is used because it is inert and conducts electricity.

Standard conditions for electrode potentials are:

- temperature = **298 K**
- solution concentration = **1.00 mol dm⁻³**
- gas pressure = **100 kPa**

These conditions matter because electrode potentials change with concentration, pressure, and temperature. A standard electrode potential is only valid under those standard conditions.

Now the most important interpretation rule:

All tabulated E° values are written as **reductions**.

This is why a more positive E° means a greater tendency for the left-hand species in the half-equation to gain electrons. So:

- more positive E° = reduction more favourable = stronger oxidising agent on the left
- more negative E° = reduction less favourable, so the reverse oxidation is easier = stronger reducing agent on the right

For example:

Ag⁺(aq) + e⁻ ⇌ Ag(s) E° = +0.80 V

Ag⁺ is a reasonably strong oxidising agent because it is quite readily reduced to Ag.

Compare that with:

Ca²⁺(aq) + 2e⁻ ⇌ Ca(s) E° = −2.87 V

This very negative value means reduction of Ca²⁺ is not favourable compared with hydrogen. The reverse process, oxidation of Ca(s), is much easier. So calcium metal is a very strong reducing agent.

This distinction is where many students slip. The table does not tell you what actually happens in the cell until you compare two half-cells together.

> **Test Tip:** Never rewrite the data table as oxidation potentials in your head. Keep all values as reduction potentials, decide which one stays as reduction, reverse the other only when writing the reaction.

### Calculating E°cell and predicting feasibility

For a conventional cell diagram, the oxidation half-cell is written on the left and the reduction half-cell on the right. The standard cell e.m.f. is:

E°cell = E°(right-hand electrode) − E°(left-hand electrode)

This is the same as:

E°cell = E°(cathode) − E°(anode)

using reduction potentials.

> **Key Term: electromotive force (e.m.f.)**
> The potential difference between two half-cells in an electrochemical cell.

#### Worked example 1: Zn/Cu cell

Half-equations from the data table:

Zn²⁺(aq) + 2e⁻ ⇌ Zn(s) E° = −0.76 V  
Cu²⁺(aq) + 2e⁻ ⇌ Cu(s) E° = +0.34 V

Step 1: Identify the more positive E°

- Cu²⁺/Cu is more positive, so it is reduced at the cathode

Step 2: Identify the more negative E°

- Zn²⁺/Zn is more negative, so zinc is oxidised at the anode

Step 3: Write cell diagram
Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s)

Step 4: Calculate E°cell
E°cell = +0.34 − (−0.76)  
E°cell = +1.10 V

Step 5: Write overall equation
Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

A positive E°cell means the reaction is feasible under standard conditions.

#### Worked example 2: Fe³⁺/Fe²⁺ with Mg²⁺/Mg

Data:

Fe³⁺(aq) + e⁻ ⇌ Fe²⁺(aq) E° = +0.77 V  
Mg²⁺(aq) + 2e⁻ ⇌ Mg(s) E° = −2.37 V

Step 1: More positive half-equation is Fe³⁺/Fe²⁺, so reduction occurs there.
Step 2: Magnesium is oxidised.

Oxidation half-equation:
Mg(s) → Mg²⁺(aq) + 2e⁻

Reduction half-equation must be multiplied by 2 for electron balance:
2Fe³⁺(aq) + 2e⁻ → 2Fe²⁺(aq)

Overall equation:
Mg(s) + 2Fe³⁺(aq) → Mg²⁺(aq) + 2Fe²⁺(aq)

Now calculate E°cell:

E°cell = +0.77 − (−2.37) = +3.14 V

Notice something crucial: although the Fe³⁺/Fe²⁺ half-equation was multiplied by 2, the E° value stayed at **+0.77 V**.

This is because electrode potential is not multiplied when you multiply a half-equation.

Common student errors here:

- adding the E° values instead of subtracting
- doing left minus right instead of right minus left
- multiplying E° values by stoichiometric coefficients
- changing the sign in the table before the calculation
  > **Test Tip:** A safe exam routine is: write both reduction potentials exactly as given, place the more negative half-cell on the left, then use right minus left. This avoids most sign errors.

A positive E°cell suggests thermodynamic feasibility under standard conditions. But you must be careful with the wording.

It does **not** tell you:

- how fast the reaction is
- whether the conditions in the question are standard
- whether side reactions or kinetic barriers matter

For example, a reaction may have E°cell > 0 but still appear not to happen quickly because the activation energy is high.

### Cell diagrams, inert electrodes, and exam technique

Cell diagrams are heavily marked for conventions. A correct cell diagram shows phases, species order, and the salt bridge.

Rules:

- **|** separates different phases
- **||** shows the salt bridge
- species in the same phase are separated by commas
- oxidised form goes next to the salt bridge in each half-cell
- left-hand side is the oxidation half-cell
- right-hand side is the reduction half-cell

Example with metal/metal ion half-cells:

Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s)

Example with ions only, needing an inert electrode:

Pt(s) | Fe²⁺(aq), Fe³⁺(aq) || Ag⁺(aq) | Ag(s)

Why is Pt needed on the left? Because Fe²⁺ and Fe³⁺ are both ions in solution, so there is no solid conductor already present. The platinum provides a surface for electron transfer.

Another common type uses gases:

Pt(s) | H₂(g) | H⁺(aq) || Pb²⁺(aq) | Pb(s)

In questions involving the SHE, students often forget platinum. If there is no conducting solid in the half-cell, include an inert electrode.

You should also be able to identify agents from the half-equations.

For a reduction half-equation:
Ox + e⁻ ⇌ Red

- the left-hand species is the **oxidising agent** because it gains electrons
- the right-hand species is the **reducing agent** because it can lose electrons in the reverse direction

Example:

Cl₂(g) + 2e⁻ ⇌ 2Cl⁻(aq)

- Cl₂ is the oxidising agent
- Cl⁻ is the reducing agent

This helps in prediction questions. If iron filings are added to a solution, you compare Fe²⁺/Fe with the other half-cell values. If the other reduction potential is more positive than Fe²⁺/Fe, then iron can be oxidised and the other species reduced.

Finally, remember the wording on feasibility. AQA often expects: **feasible under standard conditions**. That phrase matters because changing concentration, pressure, or temperature changes the actual cell potential.

For example:

- increasing ion concentration can shift the electrode equilibrium
- changing gas pressure affects gas electrodes
- changing temperature alters the potential because equilibrium position changes with temperature

So the measured e.m.f. in the lab may differ from the tabulated E°cell.

## Key Point Checklist

_This article has covered the following key knowledge points:_

- Electrode potential measures the tendency of a species to gain electrons and be reduced
- A half-cell contains a redox pair that can establish an equilibrium
- A standard electrode potential is measured relative to the standard hydrogen electrode
- The standard hydrogen electrode uses H₂(g), H⁺(aq), and a platinum electrode and has E° = 0.00 V
- Standard conditions are 298 K, 1.00 mol dm⁻³ solutions, and gases at 100 kPa
- All tabulated E° values are written as reduction half-equations
- More positive E° values show a greater tendency for reduction and a stronger oxidising agent
- More negative E° values mean the reduced form is a stronger reducing agent
- In a spontaneous electrochemical cell, oxidation occurs at the anode and reduction occurs at the cathode
- Electrons flow through the wire from the more negative half-cell to the more positive half-cell
- The salt bridge allows ion movement and maintains electrical neutrality
- Standard cell e.m.f. is calculated using E°cell = E°(right) − E°(left)
- A positive E°cell indicates a reaction is feasible under standard conditions, not that it is fast
- Electrode potentials are not multiplied when half-equations are scaled to balance electrons
- Inert electrodes such as platinum are needed when no conducting solid is present

## Key Terms and Concepts

- electrode potential
- half-cell
- electrochemical cell
- salt bridge
- standard electrode potential (E°)
- standard hydrogen electrode (SHE)
- electromotive force (e.m.f.)


Electrode potentials and electrochemical cells - Electrode potentials and cells