1. Standard Hydrogen Electrode (SHE)
An individual half-cell cannot have its potential measured in isolation because oxidation and reduction must occur together. To establish an absolute electrochemical scale, scientists defined an arbitrary zero-point: the Standard Hydrogen Electrode (SHE), which is assigned an electrode potential of 0.00 V at all temperatures.
- Hydrogen gas pressure: 100 kPa (1 bar).
- Hydrogen ion concentration: 1.00 mol dm⁻³ H⁺(aq) (e.g. 1.0 M HCl or 0.5 M H₂SO₄).
- Temperature: 298 K (25 °C).
- Electrode: Platinum black (platinised platinum) foil.
Role of the Platinum Electrode
Platinum is chemically inert and will not take part in the redox reaction. Coated with finely divided platinum black, it provides a large catalytic surface area to facilitate the rapid attainment of dynamic equilibrium between hydrogen gas and aqueous protons:
2H⁺(aq) + 2e⁻ ⇌ H₂(g) [E° = 0.00 V]
2. Conventional IUPAC Cell Notation
By international agreement (IUPAC), electrochemical cells are depicted on paper using a standardised shorthand notation without drawing complete glassware diagrams:
- Phase boundary (|): Represented by a single vertical line between different physical states (e.g. solid metal and aqueous solution).
- Salt bridge (||): Represented by a double vertical line in the centre.
- Left half-cell: By convention, the negative electrode (oxidation / anode) is written on the left.
- Right half-cell: The positive electrode (reduction / cathode) is written on the right.
- Electrons flow: From left to right through the external circuit wire.
For example, the Daniell cell consisting of zinc and copper half-cells is written as:
Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s)
If a half-cell involves only aqueous species (such as Fe³⁺/Fe²⁺), an inert platinum electrode is included, with species in the same phase separated by a comma:
Pt(s) | H₂(g) | H⁺(aq) || Fe³⁺(aq), Fe²⁺(aq) | Pt(s)
- Use a platinum electrode whenever a half-cell has no solid metal. If the Fe²⁺/Fe³⁺ half-cell is on the left (oxidation), write it as Pt | Fe²⁺, Fe³⁺.
- In each half-cell, the oxidised forms sit next to the double line (the salt bridge).
- The EMF of the cell as written is EMF = E°(right) − E°(left). For the Daniell cell: EMF = 0.34 − (−0.76) = +1.10 V.
3. Electrochemical Cell Setup Diagram
The diagram below displays the standard experimental arrangement measuring the standard electrode potential of the copper half-cell against the SHE:
Function of the Salt Bridge
A salt bridge connects the two solutions to complete the electrical circuit while keeping the solutions physically separate. Usually prepared by soaking filter paper in saturated potassium nitrate (KNO₃), it allows the migration of ions (K⁺ towards cathode; NO₃⁻ towards anode) to prevent the build-up of net charge in either beaker, which would halt current flow.
RP6 asks you to set up two half-cells, join them with a salt bridge and a high-resistance voltmeter, and measure the EMF. See the method, results and practice questions on the RP6 page.
4. Cell EMF & Feasibility Predictions
The standard electromotive force (EMF) of a cell, E°cell, is calculated from standard electrode potentials (always tabulated as reduction potentials):
E°cell = E°right − E°left = E°reduction − E°oxidation
E°cell > 0.00 V
Beyond the OxfordAQA specification: E°cell is linked to Gibbs free energy by ΔG = −nFE°cell, where n is moles of electrons and F is the Faraday constant.
An exam question may note that a reaction with E°cell > 0 does not visibly occur in the laboratory. The reasons are:
1. Kinetic stability: The activation energy (Ea) is too high at room temperature.
2. Non-standard conditions: Concentrations deviate from 1.0 mol dm⁻³ or pressure deviates from 100 kPa, shifting equilibrium potentials.
The IUPAC Convention for Half-Equations
- Every half-equation is written as a reduction: the electrons are on the left.
- Each half-equation uses the equilibrium sign ⇌.
- The more positive the E° value, the stronger the oxidising agent on the left. It gains electrons most easily.
- The more negative the E° value, the stronger the reducing agent on the right. It loses electrons most easily.
The Electrochemical Series
An electrochemical series lists half-equations in order of E°. These are standard values at 298 K, 100 kPa and 1.00 mol dm⁻³.
| Half-equation | E° / V |
|---|---|
| Li⁺(aq) + e⁻ ⇌ Li(s) | −3.04 |
| Zn²⁺(aq) + 2e⁻ ⇌ Zn(s) | −0.76 |
| Fe²⁺(aq) + 2e⁻ ⇌ Fe(s) | −0.44 |
| 2H⁺(aq) + 2e⁻ ⇌ H₂(g) | 0.00 |
| Cu²⁺(aq) + 2e⁻ ⇌ Cu(s) | +0.34 |
| I₂(aq) + 2e⁻ ⇌ 2I⁻(aq) | +0.54 |
| Fe³⁺(aq) + e⁻ ⇌ Fe²⁺(aq) | +0.77 |
| Ag⁺(aq) + e⁻ ⇌ Ag(s) | +0.80 |
| Br₂(aq) + 2e⁻ ⇌ 2Br⁻(aq) | +1.09 |
| Cr₂O₇²⁻(aq) + 14H⁺(aq) + 6e⁻ ⇌ 2Cr³⁺(aq) + 7H₂O(l) | +1.33 |
| Cl₂(g) + 2e⁻ ⇌ 2Cl⁻(aq) | +1.36 |
| MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ ⇌ Mn²⁺(aq) + 4H₂O(l) | +1.51 |
| F₂(g) + 2e⁻ ⇌ 2F⁻(aq) | +2.87 |
- Strongest oxidising agent: F₂ (bottom left, most positive E°).
- Strongest reducing agent: Li (top right, most negative E°).
- Predicting a reaction: the species on the left of the more positive half-equation oxidises the species on the right of the less positive half-equation. The more positive half-equation goes forwards; the other goes backwards.
Why Conditions Matter
- Electrode potentials are measured against the standard hydrogen electrode (Section 1), which is 0.00 V by definition.
- The two half-cells are joined by a salt bridge, usually filter paper soaked in KNO₃(aq). It completes the circuit without letting the solutions mix.
- E depends on concentration and temperature. Each half-cell is an equilibrium. For example, increasing [Cu²⁺] moves Cu²⁺ + 2e⁻ ⇌ Cu to the right, so E becomes more positive.
- This is why standard conditions are fixed (298 K, 100 kPa, 1.00 mol dm⁻³), so that values can be compared.
- You do not need the Nernst equation.
5. Commercial Batteries & Fuel Cells
Electrochemical cells are a commercial source of electrical energy. There are three types:
- Non-rechargeable cells: used once, then thrown away.
- Rechargeable cells: the reactions are reversed by an external supply.
- Fuel cells: the reactants are supplied continuously, so they do not need recharging.
1. Primary (Non-Rechargeable) Cells
Standard zinc-carbon and alkaline cells. The chemical reactions cannot be reversed by passing an electric current because the electrode reactions are irreversible and components are consumed.
2. Secondary (Rechargeable) Lithium-Ion Cells
Used widely in mobile phones, laptops, and electric vehicles. Consists of a lithium cobalt oxide (LiCoO₂) positive electrode, a graphite negative electrode intercalated with lithium atoms, and a non-aqueous organic solvent electrolyte containing lithium salts.
The electrode reactions you need for 9620 (discharging):
- Positive electrode: Li⁺ + CoO₂ + e⁻ → Li⁺[CoO₂]⁻
- Negative electrode: Li → Li⁺ + e⁻
- Overall: Li + CoO₂ → Li⁺[CoO₂]⁻
- Recharging: An external supply reverses these reactions.
Beyond the OxfordAQA specification: the lithium sits between graphite layers (intercalation), so the reactions are often written as LixC₆ → xLi⁺ + xe⁻ + C₆ at the negative electrode and Li1−xCoO₂ + xLi⁺ + xe⁻ → LiCoO₂ at the positive electrode.
Li⁺ + CoO₂ + e⁻ ⇌ Li⁺[CoO₂]⁻ E° = +0.56 V
Li⁺ + e⁻ ⇌ Li E° = −3.04 V
Deduce the overall reaction when the cell is discharging, and the EMF.
Step 1: Find the positive electrode. The half-cell with the more positive E° is the positive electrode. Reduction happens there, so its half-equation goes forwards: Li⁺ + CoO₂ + e⁻ → Li⁺[CoO₂]⁻.
Step 2: Find the negative electrode. The half-cell with the more negative E° is the negative electrode. Oxidation happens there, so its half-equation is reversed: Li → Li⁺ + e⁻.
Step 3: Add the half-equations (balance electrons first if needed). Li⁺ cancels:
Li + CoO₂ → Li⁺[CoO₂]⁻
Step 4: EMF = E°(positive) − E°(negative):
EMF = (+0.56) − (−3.04) = 3.60 V
Electrons flow through the external circuit from the negative electrode (where Li is oxidised) to the positive electrode. This flow of electrons is the current. The same method works for any cell: use the numbers the question gives you.
3. Hydrogen-Oxygen Fuel Cells
A fuel cell operates continuously as long as fuel (hydrogen) and oxidant (oxygen) are supplied from external tanks. It does not need recharging. In an alkaline fuel cell using potassium hydroxide (KOH) electrolyte:
- Negative electrode (Anode): H₂(g) + 2OH⁻(aq) → 2H₂O(l) + 2e⁻ [E° = −0.83 V]
- Positive electrode (Cathode): O₂(g) + 2H₂O(l) + 4e⁻ → 4OH⁻(aq) [E° = +0.40 V]
- Overall reaction: 2H₂(g) + O₂(g) → 2H₂O(l) [EMF = 0.40 − (−0.83) = +1.23 V]
To get the overall equation, double the negative electrode equation so that both half-equations transfer 4 electrons, then add them. The OH⁻ ions and some H₂O cancel.
| Advantages of Hydrogen Fuel Cells | Disadvantages & Technical Challenges |
|---|---|
| Zero polluting tailpipe emissions: water is the sole product. | Hydrogen is flammable and explosive; requires high-pressure tanks. |
| Much higher thermal efficiency than internal combustion engines (60%+ vs 25%). | Most industrial hydrogen is generated by steam reforming fossil fuels (emitting CO₂). |
| Continuous generation without needing hours to recharge like batteries. | Electrodes require expensive precious metal catalysts (platinum). |
Benefits and Risks of Each Type of Cell
| Type of cell | Benefits | Risks and problems |
|---|---|---|
| Lithium (rechargeable) | Light, with a high EMF. Can be recharged many times, so less waste. | The electrolyte is flammable, so damaged or overheated cells are a fire risk. Lithium and cobalt are limited resources and are hard to recycle. |
| Non-rechargeable | Cheap and ready to use. | Used once, then thrown away. This creates waste and uses up resources. Some contain toxic metals. |
| Hydrogen fuel cell | The only product is water. No recharging: fuel is supplied continuously. | Hydrogen is usually made using fossil fuels, which releases CO₂. Hydrogen is flammable and hard to store. |
6. Worked Calculations
Zn²⁺(aq) + 2e⁻ ⇌ Zn(s) [E° = −0.76 V]
Cu²⁺(aq) + 2e⁻ ⇌ Cu(s) [E° = +0.34 V]
Calculate the standard EMF of the cell and write the conventional cell representation.
Step 1: Identify oxidation and reduction:
The more negative potential (−0.76 V) oxidises (Zn → Zn²⁺ + 2e⁻). The more positive potential (+0.34 V) reduces (Cu²⁺ + 2e⁻ → Cu).
Step 2: Calculate E°cell:
E°cell = E°reduction − E°oxidation = (+0.34) − (−0.76) = +1.10 V
Step 3: Write conventional representation:
Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s)
Cu⁺(aq) + e⁻ ⇌ Cu(s) [E° = +0.52 V]
Cu²⁺(aq) + e⁻ ⇌ Cu⁺(aq) [E° = +0.15 V]
Step 1: Write disproportionation reaction:
2Cu⁺(aq) → Cu²⁺(aq) + Cu(s)
Reduction half-cell: Cu⁺(aq) + e⁻ → Cu(s) [E° = +0.52 V]
Oxidation half-cell: Cu⁺(aq) → Cu²⁺(aq) + e⁻ [E° = +0.15 V]
Step 2: Calculate E°cell:
E°cell = E°reduction − E°oxidation = (+0.52) − (+0.15) = +0.37 V
Since E°cell > 0, the disproportionation of Cu⁺(aq) into Cu(s) and Cu²⁺(aq) is thermodynamically feasible.
Fe³⁺(aq) + e⁻ ⇌ Fe²⁺(aq) E° = +0.77 V
I₂(aq) + 2e⁻ ⇌ 2I⁻(aq) E° = +0.54 V
Step 1: Compare E° values. E°(Fe³⁺/Fe²⁺) = +0.77 V is more positive than E°(I₂/I⁻) = +0.54 V. So Fe³⁺ is the stronger oxidising agent, and it is reduced.
Step 2: Write the half-equations in the direction they go:
Fe³⁺ + e⁻ → Fe²⁺ (× 2) and 2I⁻ → I₂ + 2e⁻
Step 3: Add them:
2Fe³⁺(aq) + 2I⁻(aq) → 2Fe²⁺(aq) + I₂(aq)
Step 4: EMF = E°(more positive) − E°(less positive):
EMF = 0.77 − 0.54 = +0.23 V
Yes, Fe³⁺ oxidises I⁻. The cell is written Pt(s) | I⁻(aq), I₂(aq) || Fe³⁺(aq), Fe²⁺(aq) | Pt(s). Do not multiply E° by 2 when you double a half-equation.
Go deeper
These UK AQA pages cover the same ideas in more detail. They may include content that is not in 9620.
Exam-Style Practice Questions
Test your understanding of these core syllabus concepts with targeted questions and detailed explanations.
Question 1: What is the primary function of the platinum black coating on the hydrogen electrode in the SHE?
Show Answer & Explanation
Correct Answer: B
Finely divided platinum black provides an expansive catalytic surface that allows the dynamic equilibrium between gaseous hydrogen and aqueous protons to establish rapidly.
Question 2: Why is a high-resistance voltmeter placed between the two half-cells when measuring standard electrode potential?
Show Answer & Explanation
Correct Answer: A
If significant current flowed, concentrations of ions in the half-cells would change, disrupting standard conditions. A high-resistance voltmeter measures the maximum potential difference without drawing current.
Question 3: Which species is the strongest oxidising agent among the following standard potentials: Ag⁺ (+0.80 V), Fe³⁺ (+0.77 V), Zn²⁺ (−0.76 V), Mg²⁺ (−2.37 V)?
Show Answer & Explanation
Correct Answer: D
The more positive the standard electrode potential, the greater the tendency to gain electrons (undergo reduction). Therefore, Ag⁺ (+0.80 V) is the strongest oxidising agent.
Question 4: What is the overall reaction occurring in a hydrogen-oxygen fuel cell?
Show Answer & Explanation
Correct Answer: A
The overall cell reaction in both alkaline and acidic fuel cells is the combination of hydrogen and oxygen gas to form pure water: 2H₂(g) + O₂(g) → 2H₂O(l).
Question 5 (3 marks): Use the electrochemical series in Section 4 to decide whether bromine, Br₂, oxidises Fe²⁺ ions. Write an equation for any reaction and calculate the EMF.
Show Answer & Mark Scheme
Answer
E°(Br₂/Br⁻) = +1.09 V is more positive than E°(Fe³⁺/Fe²⁺) = +0.77 V, so Br₂ is reduced and Fe²⁺ is oxidised. Yes, it reacts.
Br₂ + 2Fe²⁺ → 2Br⁻ + 2Fe³⁺
EMF = 1.09 − 0.77 = +0.32 V
Mark points
- Br₂/Br⁻ E° more positive, so Br₂ oxidises Fe²⁺
- Balanced equation: Br₂ + 2Fe²⁺ → 2Br⁻ + 2Fe³⁺
- EMF = +0.32 V
Question 6 (3 marks): A cell is made from an Fe²⁺/Fe half-cell and an Ag⁺/Ag half-cell. Write the conventional cell representation, calculate the EMF and write the overall cell reaction.
Show Answer & Mark Scheme
Answer
Fe(s) | Fe²⁺(aq) || Ag⁺(aq) | Ag(s)
EMF = E°(right) − E°(left) = 0.80 − (−0.44) = +1.24 V
Fe + 2Ag⁺ → Fe²⁺ + 2Ag
Mark points
- Fe half-cell on the left, Ag half-cell on the right, with | and ||
- EMF = +1.24 V
- Fe + 2Ag⁺ → Fe²⁺ + 2Ag (electrons balanced)
Question 7 (4 marks): A rechargeable lithium cell has these electrode data: Li⁺ + CoO₂ + e⁻ ⇌ Li⁺[CoO₂]⁻ (E° = +0.56 V) and Li⁺ + e⁻ ⇌ Li (E° = −3.04 V). (a) Which half-cell is the positive electrode? (b) Write the overall reaction when the cell discharges. (c) Calculate the EMF. (d) What happens when the cell is recharged?
Show Answer & Mark Scheme
Answer
(a) The CoO₂ half-cell, because its E° is more positive.
(b) Li + CoO₂ → Li⁺[CoO₂]⁻
(c) EMF = 0.56 − (−3.04) = 3.60 V
(d) An external supply forces the reactions to go in reverse: Li⁺[CoO₂]⁻ → Li + CoO₂.
Mark points
- (a) CoO₂ electrode (more positive E°)
- (b) Li + CoO₂ → Li⁺[CoO₂]⁻
- (c) 3.60 V
- (d) Electrode reactions are reversed by an external supply
Question 8 (3 marks): Explain why a hydrogen-oxygen fuel cell does not need recharging. Give one benefit and one risk of using hydrogen fuel cells in cars.
Show Answer & Mark Scheme
Answer
The reactants, hydrogen and oxygen, are supplied continuously from outside the cell, so they never run out while fuel is available.
Benefit: the only product is water, so no CO₂ is released by the car.
Risk: hydrogen is flammable and hard to store (or: most hydrogen is made using fossil fuels).
Mark points
- Fuel and oxygen supplied continuously
- Benefit: only water is produced
- Risk: flammable / hard to store / made from fossil fuels