AQA A-Level Chemistry: Electrode potentials and electrochemical cells (7405) – Revision Notes

For full explanations and worked examples, use the study guide. These notes condense section 3.1.11, Electrode potentials and electrochemical cells, of the AQA AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams from June 2016 onwards. The whole section is marked “A-level only” and is not AS content. It is listed under A-level Paper 1, and Paper 3 can assess any content. Required practical 8 belongs here.

When you have finished, try the practice questions. Half-equations are covered in the redox revision notes. See also the AQA A-Level Chemistry hub, the printable checklist and the free diagnostics.

3.1.11.1 Electrode potentials and cells (A-level only)

Key definitions

• Electrochemical cell: two half-cells joined by a wire and a salt bridge. Electrons pass from reducing agent to oxidising agent through the external circuit.
• Half-cell: an electrode in contact with a solution, such as Cu in Cu²⁺(aq), or platinum in a solution of Fe²⁺ and Fe³⁺.
• Salt bridge: usually filter paper soaked in saturated KNO₃(aq). Ions move through it to balance charge. It does not carry electrons.
• Standard electrode potential, E⦵: the EMF of a half-cell measured against the standard hydrogen electrode at 298 K, 100 kPa and 1.00 mol dm⁻³ solutions of ions.
• Electrochemical series: half-equations listed in order of E⦵.
• EMF: the potential difference of a cell when no current flows.

The standard hydrogen electrode (SHE)

Feature	Detail
Electrode	Platinum
Gas	H₂ at 100 kPa
Solution	1.00 mol dm⁻³ H⁺(aq), e.g. HCl(aq)
Temperature	298 K
Half-equation	2H⁺(aq) + 2e⁻ ⇌ H₂(g)
Potential	0.00 V by definition
Cell notation	Pt(s) | H₂(g) | H⁺(aq)

Conventions to know

Convention	Rule
IUPAC half-equation	Written as a reduction: electrons on the left, ⇌ arrow
Single line |	Phase boundary
Double line ||	Salt bridge
Left of ||	Negative electrode, oxidation
Right of ||	Positive electrode, reduction
Electrode position	On the outside of each side
Same-phase species	Separated by a comma, with Pt on the outside
EMF	E⦵(right) − E⦵(left) = E⦵(positive) − E⦵(negative)

Method: writing a cell and its EMF

1. Find the two E⦵ values.
2. The more negative half-cell is the negative electrode. Put it on the left.
3. Write the left side so it reads as an oxidation: Zn(s) | Zn²⁺(aq).
4. Write the right side so it reads as a reduction: Ag⁺(aq) | Ag(s).
5. EMF = E⦵(right) − E⦵(left). The answer must be positive.
6. Overall equation: reverse the left half-equation, scale both so electrons match, add.

Method: is a reaction feasible?

1. Find the half-equation containing the proposed oxidising agent on its left.
2. Find the one containing the proposed reducing agent on its right.
3. Feasible if the oxidising agent’s E⦵ is more positive. Check: E⦵(oxidising agent) − E⦵(reducing agent) > 0.
4. Write the equation: the more positive half-equation forwards, the other backwards.

Memory aid: the more positive half-equation goes forwards.

What E⦵ tells you, and what it does not

E⦵ tells you	E⦵ does not tell you
Whether a reaction is feasible under standard conditions	How fast it goes
Relative strength of oxidising and reducing agents	What happens far from standard conditions
The EMF of a standard cell	Whether the activation energy is too high

Effect of conditions (no Nernst equation)

Each electrode is an equilibrium, so Le Chatelier’s principle predicts the direction:

• More oxidised form (e.g. higher [Cu²⁺]) → equilibrium right → E more positive.
• Less oxidised form (lower [Cu²⁺]) → equilibrium left → E less positive.
• Temperature and gas pressure also change E. That is why E⦵ is defined at fixed conditions.

Values used on these pages

Half-equation	E⦵ / V
Mg²⁺ + 2e⁻ ⇌ Mg	−2.37
Zn²⁺ + 2e⁻ ⇌ Zn	−0.76
Fe²⁺ + 2e⁻ ⇌ Fe	−0.44
Ni²⁺ + 2e⁻ ⇌ Ni	−0.25
2H⁺ + 2e⁻ ⇌ H₂	0.00
Cu²⁺ + 2e⁻ ⇌ Cu	+0.34
I₂ + 2e⁻ ⇌ 2I⁻	+0.54
Fe³⁺ + e⁻ ⇌ Fe²⁺	+0.77
Ag⁺ + e⁻ ⇌ Ag	+0.80
Br₂ + 2e⁻ ⇌ 2Br⁻	+1.07
Cl₂ + 2e⁻ ⇌ 2Cl⁻	+1.36

Always use the data the question gives you.

• Strongest oxidising agent: left-hand species at the most positive end (Cl₂ here).
• Strongest reducing agent: right-hand species at the most negative end (Mg here).

Required practical 8: measuring EMF

• Clean metal electrodes with emery paper to remove oxide.
• Use 1.00 mol dm⁻³ solutions for standard values.
• Salt bridge of filter paper soaked in saturated KNO₃. Avoid KCl with Ag⁺ (AgCl precipitates).
• Use a high-resistance voltmeter so negligible current flows and the reading is the EMF.
• Record which electrode is negative.
• Extension suggested in the specification: change concentration or temperature in Zn | Zn²⁺ || Cu²⁺ | Cu and observe the EMF.

3.1.11.2 Commercial applications of cells (A-level only)

Types of cell

Type	Key point
Non-rechargeable (irreversible)	Reactions cannot practically be reversed; discarded when used up
Rechargeable	An external supply reverses the electrode reactions
Fuel cell	Fuel and oxygen supplied continuously; no electrical recharging needed

Deducing reactions from given data (non-rechargeable or rechargeable cell): the more negative half-equation runs backwards (oxidation) at the negative electrode; the more positive one runs forwards (reduction) at the positive electrode; EMF = E⦵(positive) − E⦵(negative). In a rechargeable cell, both reactions reverse on charging. Practice questions 4 (non-rechargeable) and 11 (rechargeable) use this method.

Lithium cell (simplified, as in the specification)

Positive electrode:  Li⁺ + CoO₂ + e⁻ → Li⁺[CoO₂]⁻
Negative electrode:  Li → Li⁺ + e⁻
Overall (discharge): Li + CoO₂ → Li⁺[CoO₂]⁻

Cobalt goes from +4 to +3 (reduced). Lithium goes from 0 to +1 (oxidised). Recharging reverses both.

Alkaline hydrogen–oxygen fuel cell

Positive electrode:  O₂ + 2H₂O + 4e⁻ → 4OH⁻        (E⦵ = +0.40 V)
Negative electrode:  H₂ + 2OH⁻ → 2H₂O + 2e⁻        (from 2H₂O + 2e⁻ ⇌ H₂ + 2OH⁻, E⦵ = −0.83 V)
Overall:             2H₂ + O₂ → 2H₂O                EMF = +1.23 V

How a cell generates a current

Oxidation at the negative electrode releases electrons. They flow through the external circuit to the positive electrode, where reduction takes them up. Ions move inside the cell to balance charge.

Benefits and risks

• Fuel cells: only water at the point of use, but hydrogen production needs energy, often from fossil fuels; hydrogen is flammable and hard to store.
• Rechargeable cells: less waste than throwaway cells; lithium is reactive and damaged cells can catch fire.
• All cells: disposal can release toxic metal compounds, so recycling matters.

Must-know distinctions

This	Is not this
Negative electrode = oxidation	Negative electrode = reduction
E⦵ stays the same when you multiply a half-equation	E⦵ doubles
Feasible	Fast
Salt bridge moves ions	Salt bridge moves electrons
Fuel cell is not recharged electrically	Fuel cell is recharged like a battery

Quick self-test

Use the values in the table above.

1. State the three standard conditions for E⦵.
2. Write the IUPAC half-equation for the silver electrode.
3. Calculate the EMF of a copper–silver cell.
4. Write the conventional representation of the copper–silver cell.
5. Calculate the EMF of an iron (Fe²⁺/Fe)–copper cell.
6. Will chlorine oxidise bromide ions? Give the EMF.
7. Will iodine oxidise Fe²⁺ ions? Explain.
8. A zinc half-cell is connected to a SHE. What does the voltmeter read, and which electrode is negative?
9. Metal X is the negative electrode when connected to a standard silver half-cell. The EMF is 1.05 V. Find E⦵ for X²⁺/X.
10. State what happens to the EMF of a magnesium–zinc cell if the Zn²⁺ concentration is raised.
11. Give the oxidation state of cobalt in Li⁺[CoO₂]⁻.
12. Why does a fuel cell not need to be electrically recharged?

Answers

1. 298 K; 100 kPa; 1.00 mol dm⁻³ solutions of ions.
2. Ag⁺(aq) + e⁻ ⇌ Ag(s)
3. 0.80 − 0.34 = +0.46 V
4. Cu(s) | Cu²⁺(aq) || Ag⁺(aq) | Ag(s)
5. 0.34 − (−0.44) = +0.78 V
6. Yes. Cl₂/Cl⁻ (+1.36) is more positive than Br₂/Br⁻ (+1.07). EMF = +0.29 V. Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂.
7. No. I₂/I⁻ (+0.54) is less positive than Fe³⁺/Fe²⁺ (+0.77), so EMF would be −0.23 V.
8. 0.76 V; zinc is the negative electrode.
9. 0.80 − 1.05 = −0.25 V (matches nickel).
10. Zn²⁺ + 2e⁻ ⇌ Zn shifts right, so the zinc (positive) electrode becomes more positive and the EMF increases above 1.61 V.
11. +3
12. Fuel and oxygen are supplied continuously, so the reactants never run out.

Where marks are usually lost

• Writing electrode half-equations as oxidations instead of IUPAC reductions.
• Multiplying an E⦵ value when the half-equation is scaled up.
• Calculating left − right and getting a negative EMF, then changing the sign without explanation.
• Missing Pt from a half-cell where both species are ions, or from the SHE.
• Leaving state symbols out of a conventional representation.
• Quoting a pressure other than 100 kPa, or leaving out temperature, for standard conditions.
• Saying a salt bridge carries electrons.
• Claiming a feasible reaction must happen quickly.
• Getting the concentration effect backwards: more oxidised form makes E more positive.
• Writing fuel cell half-equations for acid conditions when the question asks for the alkaline cell.

Official syllabus

AQA AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams June 2016 onwards, section 3.1.11 Electrode potentials and electrochemical cells (A-level only). Published by AQA.