Electrochemical Cells: EMF and Representation Revision Notes | A Level AQA Chemistry

1. Why Electrochemical Cells Matter

Every battery you have ever used is an electrochemical cell. The voltage printed on its side is the EMF — and that number comes directly from the electrode potentials of the two half-cells inside it.

At A-level, you need to do three things with electrochemical cells: calculate their EMF, write them in a standardised shorthand notation, and explain how electrode potentials are measured in the first place. These three skills connect tightly — master the notation and the calculation becomes straightforward.

2. Half-Cells, Half-Equations, and the IUPAC Convention

An electrochemical cell is made of two half-cells joined by a salt bridge. Each half-cell hosts one redox couple — a species and its oxidised or reduced form. The IUPAC convention is to write every half-equation as a reduction, with electrons on the left-hand side.

Here are two examples written correctly in IUPAC reduction form, with state symbols included as required:

• $\text{Cu}^{2+}(\text{aq}) + 2\text{e}^- \rightarrow \text{Cu}(\text{s})$ $\quad E^\circ = +0.34$ V
• $\text{Zn}^{2+}(\text{aq}) + 2\text{e}^- \rightarrow \text{Zn}(\text{s})$ $\quad E^\circ = -0.76$ V

Notice that a more positive $E^\circ$ means the species is more readily reduced — it is a stronger oxidising agent. Copper(II) ions are far more easily reduced than zinc(II) ions, which is why copper sits below zinc in the electrochemical series (where more negative $E^\circ$ values appear at the top).

3. The Standard Hydrogen Electrode — The Universal Reference

You cannot measure the potential of a single half-cell in isolation — you always need a second half-cell to complete the circuit. Chemists solved this by agreeing on one universal reference: the standard hydrogen electrode (SHE), assigned $E^\circ = 0.00$ V by definition.

To measure the standard electrode potential of any half-cell, you connect it to the SHE under standard conditions and record the cell voltage. The sign of $E^\circ$ tells you which direction the half-cell drives the reaction relative to hydrogen.

For example, connecting a Cu$^{2+}$/Cu half-cell to the SHE gives a reading of +0.34 V. This means copper(II) ions are reduced more readily than H$^+$ ions — the copper half-cell acts as the cathode (site of reduction) and the SHE acts as the anode (site of oxidation).

Half-cells with negative $E^\circ$ values are oxidised more readily than hydrogen — they act as anodes when paired with the SHE. Half-cells with positive $E^\circ$ values are reduced more readily than hydrogen — they act as cathodes.

4. Conventional Cell Representation

Writing out a full cell diagram every time would be impractical. The conventional cell representation is a compact, internationally agreed shorthand that encodes the same information in a single line.

For the zinc–copper cell, the representation is written as follows, with the more negative electrode (zinc, the anode) on the left:

Reading left to right: zinc metal is oxidised at the anode ($\text{Zn}(\text{s}) \rightarrow \text{Zn}^{2+}(\text{aq}) + 2\text{e}^-$), electrons flow through the external circuit, and copper(II) ions are reduced at the cathode ($\text{Cu}^{2+}(\text{aq}) + 2\text{e}^- \rightarrow \text{Cu}(\text{s})$).

When a gas electrode is involved — such as the SHE — platinum is used as the inert electrode and appears in the representation. For example, the SHE half-cell is written as: $\text{Pt}(\text{s}) \mid \text{H}_2(\text{g}) \mid \text{H}^+(\text{aq})$, with the platinum on the outside and the gas phase shown between the two boundaries.

5. Calculating the EMF of a Cell

Now that you can identify which half-cell is the anode and which is the cathode, calculating the EMF is a single subtraction. The formula uses the standard electrode potentials directly from the data table.

The cathode always has the more positive (or less negative) $E^\circ$ value. If the calculated $E^\circ_{\text{cell}}$ is positive, the reaction is thermodynamically feasible under standard conditions. A negative value means the reaction is not feasible in that direction — it would proceed spontaneously in reverse.

Worked example: Calculating the EMF of a Zinc–Copper Cell

Calculate the standard EMF of the cell formed between a Zn$^{2+}$/Zn half-cell ($E^\circ = -0.76$ V) and a Cu$^{2+}$/Cu half-cell ($E^\circ = +0.34$ V). State which electrode is the anode and which is the cathode.

1. Cu$^{2+}$/Cu: $E^\circ = +0.34$ V (more positive) → cathode (reduction) Zn$^{2+}$/Zn: $E^\circ = -0.76$ V (more negative) → anode (oxidation) — The half-cell with the more positive $E^\circ$ is always reduced — it is the cathode. The half-cell with the more negative $E^\circ$ is oxidised — it is the anode. [B1]
2. $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$ — Write the formula before substituting — this earns the method mark. [M1]
3. $E^\circ_{\text{cell}} = (+0.34) - (-0.76)$ — Take care with the double negative here: subtracting a negative number adds it. This is the most common arithmetic slip in this topic. [M1]
4. $E^\circ_{\text{cell}} = +0.34 + 0.76 = +1.10$ V — Verify: $0.34 + 0.76 = 1.10$. The positive value confirms the reaction is thermodynamically feasible. [A1]
5. $\text{Zn}(\text{s}) \mid \text{Zn}^{2+}(\text{aq}) \| \text{Cu}^{2+}(\text{aq}) \mid \text{Cu}(\text{s})$ — Anode (Zn) on the left, cathode (Cu) on the right, salt bridge as double vertical line. [B1]

Answer: $E^\circ_{\text{cell}} = +1.10$ V. Zinc is the anode; copper is the cathode. The cell representation is: $\text{Zn}(\text{s}) \mid \text{Zn}^{2+}(\text{aq}) \| \text{Cu}^{2+}(\text{aq}) \mid \text{Cu}(\text{s})$

Worked example: Calculating the EMF of a Cell Involving the SHE

A silver half-cell ($\text{Ag}^+(\text{aq}) + \text{e}^- \rightarrow \text{Ag}(\text{s})$, $E^\circ = +0.80$ V) is connected to the standard hydrogen electrode ($E^\circ = 0.00$ V). Calculate the standard EMF and write the conventional cell representation.

1. Ag$^+$/Ag: $E^\circ = +0.80$ V → cathode SHE: $E^\circ = 0.00$ V → anode — Silver ions are more readily reduced than H$^+$ ions, so the silver half-cell is the cathode. [B1]
2. $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$ — State the formula explicitly. [M1]
3. $E^\circ_{\text{cell}} = (+0.80) - (0.00) = +0.80$ V — Subtracting zero leaves the value unchanged. The positive result confirms feasibility. [A1]
4. $\text{Pt}(\text{s}) \mid \text{H}_2(\text{g}) \mid \text{H}^+(\text{aq}) \| \text{Ag}^+(\text{aq}) \mid \text{Ag}(\text{s})$ — The SHE (anode) is on the left. Platinum is the inert electrode; the gas phase appears between the two phase boundaries on the left half-cell. [B1]

Answer: $E^\circ_{\text{cell}} = +0.80$ V. Conventional representation: $\text{Pt}(\text{s}) \mid \text{H}_2(\text{g}) \mid \text{H}^+(\text{aq}) \| \text{Ag}^+(\text{aq}) \mid \text{Ag}(\text{s})$

6. Feasibility and Its Limitations

A positive $E^\circ_{\text{cell}}$ tells you the reaction is thermodynamically feasible — meaning it is energetically downhill. This is an important distinction from saying the reaction will actually occur at a measurable rate.

Kinetic barriers can prevent a thermodynamically feasible reaction from proceeding at an observable rate. High activation energy or the absence of a suitable reaction pathway can slow a reaction to the point where it appears not to occur, even when $E^\circ_{\text{cell}}$ is positive. Note that low concentrations are a separate, non-standard-conditions issue: they alter the actual cell EMF away from the calculated $E^\circ_{\text{cell}}$ value rather than acting as a kinetic barrier.

Equally, $E^\circ$ values are measured under standard conditions (298 K, 1 mol dm$^{-3}$, 1 atm). Real cells often operate under non-standard conditions, so the actual cell voltage will differ from the calculated $E^\circ_{\text{cell}}$. AQA examiners regularly test whether you can articulate this caveat.

7. Exam Technique and Common Mistakes

The calculation and notation questions in this topic are highly predictable — examiners use the same structures repeatedly. The marks are there to be taken, but a small number of recurring errors cost students dearly.

Before the next quiz, connect the notation to the chemistry: the cell representation shows which species belong to each half-cell and helps you identify the electron-flow direction and the overall cell reaction.