OxfordAQA A-Level Chemistry: Electrode potentials and electrochemical cells (9620)

This study guide covers section 3.1.9, Electrode potentials and electrochemical cells, of the OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards. It teaches 3.1.9.1 and 3.1.9.2. The whole section is International A2 content, so it is International A-level only. The specification places 3.1.8 to 3.1.10 in Unit 3 (Inorganic 2 and Physical 2), and Unit 5 (Practical and synoptic) can draw on all content, including required practical 6.

Then use the revision notes and the practice questions. The effect of conditions on an electrode uses Le Chatelier’s principle, covered in equilibria and Le Chatelier’s principle. See also the OxfordAQA A-Level Chemistry hub, the printable checklist and the free diagnostics.

What this section covers

Every row is International A-level only.

Spec section	What you must be able to do
3.1.9.1	IUPAC half-equations; write and apply the conventional representation of a cell
3.1.9.1	Measure E against the standard hydrogen electrode; standard conditions; why conditions matter (no Nernst equation)
3.1.9.1	Use the electrochemical series to predict the direction of simple redox reactions; calculate EMF
3.1.9.1	Required practical 6: measure the EMF of an electrochemical cell
3.1.9.2	Non-rechargeable, rechargeable and fuel cells; lithium cell; alkaline hydrogen–oxygen fuel cell
3.1.9.2	Deduce reactions and EMF from given data; explain how a current is generated; benefits and risks

Inside an electrochemical cell

In a cell, electrons pass from the reducing agent to the oxidising agent through an external wire, not directly. This creates a potential difference that can drive a current. Each half-cell is either a metal in a solution of its own ions (cobalt in Co²⁺(aq)) or an inert platinum electrode in a solution of both forms (Cr³⁺ and Cr²⁺) or with a gas bubbled over it. A salt bridge lets ions move to keep each solution neutral. Electrons never travel through it.

The IUPAC convention for half-equations

Every electrode half-equation is written as a reduction: electrons on the left, a reversible arrow between the two sides.

Al³⁺(aq) + 3e⁻ ⇌ Al(s)        E⦵ = −1.66 V
Co²⁺(aq) + 2e⁻ ⇌ Co(s)        E⦵ = −0.28 V

A more positive E⦵ means the left-hand species gains electrons more readily.

The conventional representation of a cell

A cell is written on one line, with the negative electrode on the left:

negative electrode | its solution || other solution | positive electrode

• | is a phase boundary; || is the salt bridge.
• Oxidation at the left (negative) electrode; reduction at the right (positive) electrode.
• Electrodes go on the outside. Left side reads reduced form → oxidised form; right side reads oxidised form → reduced form.
• Two aqueous species: separate them with a comma and put Pt(s) on the outside. Include state symbols.

EMF = E⦵(right) − E⦵(left) = E⦵(positive electrode) − E⦵(negative electrode)

A correctly written cell has a positive EMF.

Worked example 1: an aluminium–cobalt cell

Data: Al³⁺/Al −1.66 V; Co²⁺/Co −0.28 V.

1. −1.66 V is more negative, so aluminium is the negative electrode and goes on the left.
2. Aluminium is oxidised: Al → Al³⁺ + 3e⁻. Cobalt(II) ions are reduced: Co²⁺ + 2e⁻ → Co.
3. Representation: Al(s) | Al³⁺(aq) || Co²⁺(aq) | Co(s)
4. EMF = −0.28 − (−1.66) = +1.38 V
5. Overall: multiply the aluminium equation by 2 and the cobalt equation by 3 (6e⁻ each): 2Al + 3Co²⁺ → 2Al³⁺ + 3Co. The E⦵ values are not multiplied.

Measuring electrode potentials: the standard hydrogen electrode

Only a potential difference between two half-cells can be measured, so values are measured against the standard hydrogen electrode (SHE), given 0.00 V. It is hydrogen gas at 100 kPa bubbling over a platinum electrode in a solution with [H⁺] = 1.00 mol dm⁻³, at 298 K. Platinum is inert and conducts electrons to and from the H₂/H⁺ equilibrium.

Half-equation: 2H⁺(aq) + 2e⁻ ⇌ H₂(g). In a cell diagram: Pt(s) | H₂(g) | H⁺(aq) ||

Standard electrode potential, E⦵, refers to conditions of 298 K, 100 kPa and 1.00 mol dm⁻³ solutions of ions. The E⦵ of a half-cell is the EMF when it is connected to the SHE under these conditions, with the sign of the half-cell’s electrode.

Worked example 2: reading a value against the SHE

A standard chromium half-cell, Cr in Cr³⁺(aq), is connected to the SHE. The voltmeter reads 0.74 V and the chromium electrode is negative. So E⦵(Cr³⁺/Cr) = 0.00 − 0.74 = −0.74 V. A gold half-cell reading 1.52 V with gold positive gives +1.52 V. The size comes from the voltmeter; the sign comes from the polarity.

Why conditions matter

Each electrode is an equilibrium, such as Al³⁺(aq) + 3e⁻ ⇌ Al(s). Changing conditions shifts it, so E changes (direction only; the Nernst equation is not required). Raising the concentration of the oxidised form shifts the equilibrium right and makes E more positive; lowering it makes E more negative. Temperature, and pressure for gas electrodes, also shift these equilibria, which is why E⦵ values compare fairly only at standard conditions.

Worked example 3: changing the negative electrode

In the aluminium–cobalt cell above, the Al³⁺ solution is diluted. Predict the effect on the EMF.

1. Lower [Al³⁺] shifts Al³⁺ + 3e⁻ ⇌ Al to the left, so the aluminium potential becomes more negative.
2. EMF = E(positive) − E(negative), so the EMF increases above 1.38 V.

Raising [Co²⁺] would also increase the EMF, by making the positive electrode more positive.

The electrochemical series

Arranging E⦵ values in order gives the electrochemical series. Values used on this page (in an exam, always use the values you are given):

Half-equation	E⦵ / V
Al³⁺ + 3e⁻ ⇌ Al	−1.66
Cr³⁺ + 3e⁻ ⇌ Cr	−0.74
Cr³⁺ + e⁻ ⇌ Cr²⁺	−0.41
Co²⁺ + 2e⁻ ⇌ Co	−0.28
V³⁺ + e⁻ ⇌ V²⁺	−0.26
Sn²⁺ + 2e⁻ ⇌ Sn	−0.13
2H⁺ + 2e⁻ ⇌ H₂	0.00
VO²⁺ + 2H⁺ + e⁻ ⇌ V³⁺ + H₂O	+0.34
VO₂⁺ + 2H⁺ + e⁻ ⇌ VO²⁺ + H₂O	+1.00
Au³⁺ + 3e⁻ ⇌ Au	+1.52
H₂O₂ + 2H⁺ + 2e⁻ ⇌ 2H₂O	+1.78

Strongest oxidising agent: left-hand species, most positive end (H₂O₂). Strongest reducing agent: right-hand species, most negative end (Al).

Predicting the direction of a redox reaction

A reaction is feasible under standard conditions when the oxidising agent’s half-equation is more positive than the reducing agent’s. The more positive half-equation runs forwards; the other runs backwards. Equivalently, the EMF is positive.

Worked example 4: how far will tin reduce vanadium?

Tin is added to acidic VO²⁺(aq). Does it reduce VO²⁺ to V³⁺, and then V³⁺ to V²⁺?

1. VO²⁺ → V³⁺: E⦵(VO²⁺/V³⁺) = +0.34 V is more positive than E⦵(Sn²⁺/Sn) = −0.13 V. EMF = 0.34 − (−0.13) = +0.47 V. Feasible. Equation: Sn + 2VO²⁺ + 4H⁺ → Sn²⁺ + 2V³⁺ + 2H₂O.
2. V³⁺ → V²⁺: E⦵(V³⁺/V²⁺) = −0.26 V is less positive than −0.13 V. EMF = −0.26 − (−0.13) = −0.13 V. Not feasible.
3. So under standard conditions tin stops at V³⁺.

What a prediction cannot tell you

E⦵ values say nothing about rate: a high activation energy can make a feasible reaction too slow to see. Away from standard conditions the potentials change, so a very small predicted EMF is unreliable.

Cells with platinum electrodes

Worked example 5: chromium(II) and hydrogen peroxide

Data: Cr³⁺/Cr²⁺ −0.41 V; H₂O₂/H₂O +1.78 V.

1. No metal takes part, so both electrodes are platinum.
2. Cr³⁺/Cr²⁺ is more negative: negative electrode, on the left. Cr²⁺ is oxidised.
3. H₂O₂ is reduced on the right; H⁺ takes part, so include it.
4. Representation: Pt(s) | Cr²⁺(aq), Cr³⁺(aq) || H₂O₂(aq), H⁺(aq) | Pt(s)
5. EMF = 1.78 − (−0.41) = +2.19 V
6. Overall (chromium equation × 2): H₂O₂ + 2H⁺ + 2Cr²⁺ → 2H₂O + 2Cr³⁺

Required practical 6: measuring the EMF of a cell

1. Rub each metal electrode with emery paper so no oxide coating is left on it.
2. Put each electrode in a 1.00 mol dm⁻³ solution of its ions (platinum where both species are ions).
3. Join the solutions with filter paper soaked in an unreactive salt solution such as potassium nitrate, whose ions will not precipitate ions in either half-cell.
4. Connect a high-resistance voltmeter so almost no current flows and the reading equals the EMF. Record which terminal is negative.

To evaluate results, look for oxide on electrodes, wrong concentrations, temperature away from 298 K, a dried-out salt bridge and a voltmeter that draws current.

Commercial applications of electrochemical cells

Cells are a commercial source of electrical energy, of three kinds:

Type	What it means
Non-rechargeable (irreversible)	The reactions cannot be reversed in practice; the cell is thrown away when a reactant is used up
Rechargeable	An external supply drives current the other way and reverses the electrode reactions
Fuel cell	Fuel and oxygen are fed in continuously; it generates a current and does not need to be electrically recharged

How the electrode reactions generate a current

Oxidation at the negative electrode releases electrons. They flow through the external circuit to the positive electrode, where reduction uses them, while ions move through the electrolyte to balance charge.

The lithium cell

The specification gives these simplified electrode reactions for discharge:

Positive electrode:  Li⁺ + CoO₂ + e⁻ → Li⁺[CoO₂]⁻
Negative electrode:  Li → Li⁺ + e⁻

Li⁺ ions move through the electrolyte to the positive electrode; the electrons go round the external circuit to meet them, and cobalt is reduced from +4 to +3. Recharging reverses both: Li⁺[CoO₂]⁻ → Li⁺ + CoO₂ + e⁻, and Li⁺ + e⁻ → Li.

Worked example 6: a rechargeable vanadium cell

A rechargeable cell stores energy in vanadium ions in solution. Data: VO₂⁺/VO²⁺ +1.00 V; V³⁺/V²⁺ −0.26 V.

1. Positive electrode (more positive, reduction): VO₂⁺ + 2H⁺ + e⁻ → VO²⁺ + H₂O
2. Negative electrode (less positive half-equation reversed): V²⁺ → V³⁺ + e⁻
3. Overall discharge: VO₂⁺ + 2H⁺ + V²⁺ → VO²⁺ + H₂O + V³⁺
4. EMF = 1.00 − (−0.26) = +1.26 V
5. Charging reverses both: VO²⁺ + H₂O → VO₂⁺ + 2H⁺ + e⁻, and V³⁺ + e⁻ → V²⁺.

The alkaline hydrogen–oxygen fuel cell

The electrolyte is an alkali, so the half-equations use OH⁻:

Negative electrode:  H₂ + 2OH⁻ → 2H₂O + 2e⁻
Positive electrode:  O₂ + 2H₂O + 4e⁻ → 4OH⁻
Overall:             2H₂ + O₂ → 2H₂O

Double the negative-electrode equation so 4e⁻ cancel. OH⁻ made at one electrode is used at the other, so none is used up. Hydrogen and oxygen are supplied continuously, so no electrical recharging is needed.

Benefits and risks to society

• Fuel cells: the device emits water and no carbon dioxide where it runs. But the hydrogen has to be produced first, using energy that may come from burning fossil fuels, and hydrogen mixed with air is explosive and bulky to store unless compressed.
• Rechargeable cells: one cell serves through many charge cycles, so fewer cells are made and discarded. But a lithium cell that is overcharged or crushed can overheat and ignite.
• All cells: metals such as lithium and cobalt must be mined, and dumped cells can leak harmful compounds, so collecting and recycling them matters.

Common errors

• Writing an electrode half-equation as an oxidation. IUPAC half-equations are reductions.
• Multiplying E⦵ when you multiply a half-equation.
• Taking the sign of a measured E⦵ from the reading instead of the polarity.
• Putting the more positive half-cell on the left and reporting a negative EMF.
• Leaving out Pt(s), H⁺(aq) or state symbols.
• Saying electrons move through the salt bridge, or that a fuel cell needs recharging.

Next steps

Condense the topic with the revision notes, then test yourself with the practice questions.

Official syllabus

OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards, section 3.1.9 Electrode potentials and electrochemical cells (International A2). Published by OxfordAQA.