Electrochemical Cells: EMF and Representation Revision Notes | A Level AQA Chemistry
In cell notation, the anode half-cell is written on the left and the cathode half-cell on the right. Free A Level AQA Chemistry notes and 13 questions answered.
Updated 5 Sept 2026 · Revise Right Team
Electrochemical Cells: EMF and Representation is part of Electrode potentials and electrochemical cells (A-level only) in A Level AQA Chemistry.
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What this lesson covers
In cell notation, the anode half-cell is written on the left and the cathode half-cell on the right. Free A Level AQA Chemistry notes and 13 questions answered.
Calculate the EMF of an electrochemical cell using standard electrode potentials
Write and apply the conventional cell representation using IUPAC notation
Write half-equations for electrode reactions following IUPAC convention
Explain how cells are used to measure electrode potentials relative to the standard hydrogen electrode
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Use this lesson to stay inside Electrode potentials and electrochemical cells (A-level only) while connecting the detail back to the wider Physical chemistry topic.
Standard electrode potential ($E^\circ$): The EMF of a half-cell measured under standard conditions (298 K, 1 mol dm$^{-3}$ solution, 1 atm gas pressure) relative to the standard hydrogen electrode, which is assigned $E^\circ = 0.00$ V.
EMF of a cell ($E^\circ_{\text{cell}}$): The potential difference between two half-cells under standard conditions, calculated as $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$. A positive value indicates the reaction is thermodynamically feasible.
Standard hydrogen electrode (SHE): The universal reference half-cell consisting of hydrogen gas at 1 atm bubbled over a platinum electrode in 1 mol dm$^{-3}$ H$^+$(aq) at 298 K, assigned $E^\circ = 0.00$ V by convention.
Conventional cell representation: A standardised shorthand for an electrochemical cell written as: anode half-cell | salt bridge | cathode half-cell, with phase boundaries shown as single vertical lines and the salt bridge as a double vertical line.
Revision notes
Read the core explanations from Electrochemical Cells: EMF and Representation before testing yourself from memory.
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.
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]
$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$ — Write the formula before substituting — this earns the method mark. [M1]
$E^\circ_{\text{cell}} = (+0.34) - (-0.76)$ — Take care with the double negative here: subtracting a negative number adds it. This is the most common arithmetic slip in this topic. [M1]
$E^\circ_{\text{cell}} = +0.34 + 0.76 = +1.10$ V — Verify: $0.34 + 0.76 = 1.10$. The positive value confirms the reaction is thermodynamically feasible. [A1]
$\text{Zn}(\text{s}) \mid \text{Zn}^{2+}(\text{aq}) \| \text{Cu}^{2+}(\text{aq}) \mid \text{Cu}(\text{s})$ — 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.
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]
$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$ — State the formula explicitly. [M1]
$E^\circ_{\text{cell}} = (+0.80) - (0.00) = +0.80$ V — Subtracting zero leaves the value unchanged. The positive result confirms feasibility. [A1]
$\text{Pt}(\text{s}) \mid \text{H}_2(\text{g}) \mid \text{H}^+(\text{aq}) \| \text{Ag}^+(\text{aq}) \mid \text{Ag}(\text{s})$ — 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]
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.
Sources and content standards
Published by Revise Right. Use the official specification reference to check the requirements for your course. Content updated: 2026-09-05.
How do you write a conventional cell representation?
In a conventional cell representation, the anode half-cell is written on the left and the cathode on the right, a single vertical line shows a phase boundary and a double vertical line shows the salt bridge. For the zinc–copper cell this is Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s).
How do you write the standard hydrogen electrode in cell notation?
When a gas electrode such as the standard hydrogen electrode is used, platinum is the inert electrode and is written on the outside: Pt(s) | H₂(g) | H⁺(aq). Connected to a silver half-cell, the cell is Pt(s) | H₂(g) | H⁺(aq) || Ag⁺(aq) | Ag(s), with an EMF of +0.80 V.
What is a redox couple?
A redox couple is a species and its oxidised or reduced form hosted in one half-cell, such as Cu²⁺/Cu or Zn²⁺/Zn. Each half-cell of an electrochemical cell hosts one redox couple, and the two half-cells are joined by a salt bridge.
What is the IUPAC convention for writing half-equations?
The IUPAC convention writes every half-equation as a reduction, with electrons on the left-hand side: oxidised form + electrons → reduced form, for example Cu²⁺(aq) + 2 e⁻ → Cu(s), E° = +0.34 V. The electrode potential quoted in data tables always corresponds to this reduction direction.
Does a positive cell EMF mean the reaction will happen?
A positive E°cell means a reaction is thermodynamically feasible, but it may not occur at an observable rate: a high activation energy or no suitable reaction pathway can make it appear not to happen. Real cells also often run under non-standard conditions, so the actual voltage differs from the calculated E°cell.
How is a half-cell's electrode potential measured against the hydrogen electrode?
A half-cell's standard electrode potential is measured by connecting it to the standard hydrogen electrode under standard conditions and recording the voltage. A reading of +0.34 V for Cu²⁺/Cu means copper(II) ions are reduced more readily than H⁺, so copper is the cathode; half-cells with negative values, such as zinc at −0.76 V, act as anodes.
What does a single vertical line mean in cell notation?
A single vertical line in conventional cell notation represents a phase boundary between different states, such as between Zn(s) and Zn²⁺(aq). A double vertical line represents the salt bridge.
Which side is the anode written on in a cell diagram?
In conventional cell representation, the anode, where oxidation occurs, is always written on the left and the cathode, where reduction occurs, on the right, with the salt bridge shown as a double vertical line between them.
What is the cell representation for a zinc-copper cell?
The conventional cell representation for a zinc-copper cell is Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s). Zinc, the anode with the more negative E° of −0.76 V, is on the left, and copper, the cathode at +0.34 V, is on the right; E°cell = +1.10 V.
What is the EMF of a silver half-cell connected to the standard hydrogen electrode?
The EMF of a silver half-cell connected to the standard hydrogen electrode is +0.80 V, since E°cell = (+0.80) − (0.00). Silver ions are more readily reduced than H⁺, so silver is the cathode; the cell is written Pt(s) | H₂(g) | H⁺(aq) || Ag⁺(aq) | Ag(s).
Why is the actual voltage of a cell different from its calculated EMF?
The actual voltage of a real cell differs from the calculated E°cell because E° values are measured under standard conditions of 298 K, 1 mol dm⁻³ and 1 atm, while real cells often operate under non-standard conditions, such as low concentrations, which alter the EMF.
How do you read a conventional cell representation?
A conventional cell representation is read by taking the left-hand half-cell as the oxidation half-equation and the right-hand half-cell as the reduction half-equation. Single lines locate phase boundaries and the double line is the salt bridge; combining the two half-equations gives the overall cell reaction.
Why is a copper half-cell positive against the hydrogen electrode?
A copper half-cell gives +0.34 V against the standard hydrogen electrode because copper(II) ions are reduced more readily than H⁺ ions. The copper half-cell acts as the cathode and the hydrogen electrode as the anode; half-cells with negative E° values act as anodes instead.