Electrochemical Cells: Principles and Applications Revision Notes | A Level AQA Chemistry

1. From Redox to Electricity: The Core Idea

Every battery you have ever used — from the AA cell in a remote control to the lithium-ion pack in a phone — is an electrochemical cell. The chemistry inside is a redox reaction, but instead of letting electrons transfer directly between reactants, the cell forces them to travel through an external wire. That flow of electrons is the electric current.

Think of it this way: in a beaker, zinc metal dropped into copper sulphate solution transfers electrons directly and you get heat. In an electrochemical cell, you physically separate the two half-reactions so electrons must travel through a circuit to get from one to the other. That detour through the wire is what powers a device.

The two half-cells are connected by two pathways: an external wire (for electron flow) and a salt bridge (for ion flow). The salt bridge — typically a gel or solution of an inert electrolyte such as $\text{KNO}_3\text{(aq)}$ — completes the circuit by allowing ions to migrate and maintain electrical neutrality in each half-cell without mixing the solutions.

2. Standard Electrode Potentials and Deducing Cell EMF

Every half-reaction has a tendency to proceed as a reduction — gaining electrons. Standard electrode potential ($E^\circ$) quantifies that tendency on a scale anchored to the standard hydrogen electrode (SHE), which is assigned $E^\circ = 0.00$ V by convention. The more positive the $E^\circ$ value, the greater the tendency to be reduced, and therefore the stronger the oxidising agent.

When two half-cells are connected, the half-cell with the more positive $E^\circ$ acts as the cathode (reduction occurs there) and the half-cell with the less positive (or more negative) $E^\circ$ acts as the anode (oxidation occurs there). The cell EMF is calculated as:

Worked example: Deducing Cell EMF: Zinc-Copper Cell

A cell is constructed using a zinc half-cell ($\text{Zn}^{2+}\text{(aq)}/\text{Zn(s)}$, $E^\circ = -0.76$ V) and a copper half-cell ($\text{Cu}^{2+}\text{(aq)}/\text{Cu(s)}$, $E^\circ = +0.34$ V). Deduce the overall cell reaction and calculate the standard EMF.

1. Copper: $E^\circ = +0.34$ V (more positive) → cathode (reduction) Zinc: $E^\circ = -0.76$ V (less positive) → anode (oxidation) — The half-cell with the higher $E^\circ$ is always the cathode. This is where the stronger oxidising agent ($\text{Cu}^{2+}$) gains electrons.
2. Cathode (reduction): $\text{Cu}^{2+}\text{(aq)} + 2e^- \rightarrow \text{Cu(s)}$ Anode (oxidation): $\text{Zn(s)} \rightarrow \text{Zn}^{2+}\text{(aq)} + 2e^-$ — The anode half-equation is the reverse of the standard reduction half-equation for zinc. Electrons are released at the anode and consumed at the cathode — this is the electron flow through the external circuit. [M1]
3. $\text{Zn(s)} + \text{Cu}^{2+}\text{(aq)} \rightarrow \text{Zn}^{2+}\text{(aq)} + \text{Cu(s)}$ — The electrons cancel (2 released, 2 consumed). Check: atoms and charges are balanced on both sides. [M1]
4. $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$ $E^\circ_{\text{cell}} = (+0.34) - (-0.76)$ $E^\circ_{\text{cell}} = +0.34 + 0.76 = +1.10$ V — Subtracting a negative number adds it. The positive result confirms this cell reaction is thermodynamically feasible under standard conditions. [A1]
5. Electrons flow from Zn (anode) → through external circuit → to Cu (cathode) — Conventional current flows in the opposite direction to electron flow, but AQA questions typically ask about electron flow. Always state: from anode to cathode through the external circuit.

Answer: Overall reaction: $\text{Zn(s)} + \text{Cu}^{2+}\text{(aq)} \rightarrow \text{Zn}^{2+}\text{(aq)} + \text{Cu(s)}$; $E^\circ_{\text{cell}} = +1.10$ V

3. How Electrode Reactions Generate an Electric Current

Now that you can calculate the EMF, it is worth being precise about the mechanism by which the cell actually drives a current. At the anode, oxidation releases electrons into the electrode. Those electrons experience a potential difference — the EMF — that drives them through the external circuit towards the cathode, where they are consumed by the reduction reaction.

Simultaneously, ions migrate through the salt bridge: cations move towards the cathode compartment and anions move towards the anode compartment. This ion movement maintains electrical neutrality in both solutions. Without this ionic current through the salt bridge, charge would build up, the potential difference would collapse, and current would cease.

The rate at which electrons are produced at the anode must equal the rate at which they are consumed at the cathode — the two half-reactions are coupled. This is why the stoichiometry of the half-equations must be balanced before combining them: if one half-reaction produces 2 electrons and the other consumes only 1, you must double the second half-equation before adding them together.

When interpreting cell diagrams or emf data, keep the physical picture clear: electrons flow through the external circuit because oxidation and reduction are separated into different half-cells, while ions in the salt bridge move only enough to prevent charge build-up. That distinction is what allows the redox reaction to generate a measurable voltage instead of simply reacting in one container.

4. Non-Rechargeable and Rechargeable Cells

Electrochemical cells are commercially important because they store chemical energy and release it on demand as electrical energy. The key distinction between cell types lies in whether the electrode reactions can be reversed by applying an external electrical current.

In a non-rechargeable cell, the spontaneous redox reaction proceeds in one direction only. The reactants are consumed and the products accumulate. Because the reaction is not practically reversible, attempting to force current back through the cell does not regenerate the original reactants in a useful way — it may cause dangerous decomposition instead.

During discharge of a rechargeable cell, the spontaneous redox reaction generates current in the normal way. During charging, an external power supply drives the reverse overall reaction, so the species formed during discharge are converted back into the original reactants. In practice the same electrode may act as the anode during discharge and as the cathode during charging because the reaction direction has been reversed.

When using electrode data to deduce the reactions in a specific cell, the approach is identical to the general method: identify the half-cell with the more positive $E^\circ$ as the cathode (reduction) and the other as the anode (oxidation). For a rechargeable cell during charging, these roles are reversed by the external supply — the cathode during discharge becomes the anode during charging.

5. Fuel Cells: Continuous Electrochemical Energy

Both non-rechargeable and rechargeable cells store a finite quantity of reactants. A fuel cell operates differently: fuel and oxidant are supplied continuously from external reservoirs, so the cell can generate electricity for as long as fuel is provided. The hydrogen-oxygen fuel cell is the most commonly examined example at A-level.

In the hydrogen-oxygen fuel cell, hydrogen gas is fed to the anode and oxygen gas is fed to the cathode. Both electrodes are typically made of porous platinum, which acts as a catalyst and allows the gases to contact the electrolyte. The electrolyte can be acidic or alkaline, and the half-equations differ accordingly — AQA requires you to know both versions.

Notice that the overall reaction is the same regardless of electrolyte: hydrogen is oxidised and oxygen is reduced to form water. The electrolyte type only changes the ionic species involved in the half-equations. The fuel cell does not store energy — it converts chemical energy from the fuel directly into electrical energy at high efficiency, with water as the only product when pure hydrogen is used.

6. Electrochemical Cells as Commercial Energy Sources

The commercial appeal of electrochemical cells rests on their ability to convert chemical energy directly into electrical energy without combustion. This gives them several practical advantages over thermal power generation, but also some significant limitations that AQA expects you to evaluate.

• Non-rechargeable cells: portable, reliable, no maintenance — but single-use and generate chemical waste
• Rechargeable cells: reusable over hundreds of cycles, reducing waste per unit of energy — but require a charging infrastructure and contain metals (e.g. lithium, cobalt) that are resource-intensive to mine
• Hydrogen fuel cells: high efficiency, zero harmful emissions at point of use, continuous operation — but require a hydrogen supply infrastructure, and hydrogen production currently relies largely on fossil fuels (steam methane reforming), which offsets the environmental benefit

A key point for evaluation questions: the environmental credentials of a fuel cell depend entirely on how the hydrogen is produced. If hydrogen comes from electrolysis powered by renewable electricity ('green hydrogen'), the overall process is genuinely low-carbon. If it comes from steam methane reforming without carbon capture, the net carbon footprint is comparable to burning the fossil fuel directly.

7. Exam Technique and Common Mistakes

The most reliable way to pick up marks on electrochemical cell questions is to follow a consistent method: identify cathode and anode from $E^\circ$ values, write balanced half-equations with state symbols, combine them correctly, and calculate EMF using $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$. Examiners award method marks at each stage, so showing clear working protects you even if you make an arithmetic slip.

For 'explain' questions about how a cell generates current, the mark scheme typically requires three linked ideas: oxidation at the anode releases electrons, electrons flow through the external circuit to the cathode, and ions migrate through the salt bridge to maintain electrical neutrality. Missing the salt bridge explanation is one of the most common reasons students lose marks on this type of question.

Bringing the whole lesson together: electrochemical cells harness spontaneous redox reactions by separating the two half-reactions so that electrons must travel through an external circuit, generating a measurable current. The EMF depends on the difference in standard electrode potentials between the two half-cells. Whether the cell is non-rechargeable, rechargeable, or a fuel cell determines how long it can operate and whether the electrode reactions can be reversed — all of which are deducible from electrode data and the principles covered here.