Last updated: 14 October 2026
Redox reactions involve electrons moving from one species to another. But knowing that oxidation and reduction occur is only the beginning. In many A-Level Chemistry questions, we want to know which species is more likely to gain electrons, which species will lose them, and whether two half-reactions can combine to give a feasible overall reaction.
Electrode potentials give us a numerical way to make those predictions. Instead of simply saying that one oxidising agent is “stronger” than another, we can compare their E° values, calculate the standard cell potential and connect the result to the energetics of the reaction.
Try these five questions before reading the chapter. They test the redox ideas that electrode potentials build upon and introduce the meaning of E°.
- 5 diagnostic questions
- Redox reasoning and simple calculations
- Instant explanation for every answer option
1. What does standard electrode potential mean?
Before electrode potentials make sense, recall the basic redox language: oxidation is loss of electrons, while reduction is gain of electrons. Electrode-potential data allow us to compare the relative tendency of different species to undergo reduction.
A standard electrode potential, E°, is the potential of a half-cell measured relative to the standard hydrogen electrode under standard conditions. Because the potential of an isolated half-cell cannot be measured on its own, every E° value is ultimately a comparison with the same reference.
All standard electrode potentials are conventionally written as reduction potentials. The more positive E° is, the greater the tendency for the forward reduction half-equation to occur.
Reduction half-equations
A table of electrode potentials normally writes electrons on the left-hand side. For example:
The positive value tells us that Cu²⁺ has a greater tendency to be reduced than H⁺ under standard conditions because the Cu²⁺/Cu couple lies above the standard hydrogen electrode on the reduction-potential scale.
More positive E°: the oxidised species on the left has a greater tendency to gain electrons. It is therefore generally the stronger oxidising agent when compared with the oxidised form of a couple having a less positive E°.
More negative E°: the written reduction has a smaller tendency to occur. The reduced form on the right generally has a greater tendency to lose electrons and can therefore behave as a stronger reducing agent.
Do not interpret the sign in isolation
A negative E° does not mean that a half-reaction is impossible. Electrode potentials are relative values. What matters in a complete redox reaction is the difference between the electrode potentials of the two redox couples involved.
For Zn²⁺ + 2e⁻ ⇌ Zn, E° = −0.76 V, while for Cu²⁺ + 2e⁻ ⇌ Cu, E° = +0.34 V. Cu²⁺ therefore has the greater tendency to be reduced. When the two half-cells are connected, Cu²⁺ is reduced while Zn is oxidised.
2. The standard hydrogen electrode
If every electrode potential is a comparison, we need an agreed reference point. Electrochemistry uses the standard hydrogen electrode (SHE).
The E° value of the H⁺/H₂ couple is defined as exactly 0.00 V. Other electrode potentials can therefore be described as being more positive or more negative relative to this reference.
What is inside the standard hydrogen electrode?
The SHE contains both forms of the H⁺/H₂ redox couple. Hydrogen gas is passed over an inert platinum electrode that is in contact with a solution containing H⁺ ions.
Neither H⁺(aq) nor H₂(g) is a conducting solid that could act as the electrode itself. Platinum is therefore used to provide an electrically conducting surface on which electron transfer can occur.
| Component | Standard condition | Purpose |
|---|---|---|
| H⁺(aq) | 1.0 mol dm⁻³ | Provides the aqueous oxidised form of the H⁺/H₂ redox couple. |
| H₂(g) | 1 bar | Provides the reduced gaseous form of the redox couple. |
| Pt(s) | Inert platinum electrode | Conducts electrons and provides a surface for the electrode reaction. |
| Temperature | 298 K | Allows E° values to be compared under the same stated conditions. |
3. How different half-cells are constructed
Not every redox couple contains a solid metal. The physical states of the oxidised and reduced forms determine how the half-cell must be constructed.
Metal / metal-ion half-cells
If a redox couple contains a conducting solid metal, that metal can usually act as the electrode. For the Cu²⁺/Cu couple, a copper rod is placed in a solution containing Cu²⁺ ions.
Gas / ion half-cells
A gaseous species cannot itself provide the conducting connection to the external circuit. An inert platinum electrode is therefore required. For example, the chlorine/chloride half-cell contains Cl₂(g), Cl⁻(aq) and an inert platinum electrode.
Ion / ion half-cells
Sometimes both oxidation states are aqueous ions in the same solution. The Fe³⁺/Fe²⁺ couple is a common example. Since neither species is a conducting solid, an inert platinum electrode is again required.
If neither member of the redox couple is a suitable conducting solid, an inert electrode such as platinum is needed to provide an electrical connection to the external circuit.
4. Measuring an electrode potential
To determine the standard electrode potential of a particular half-cell, it is connected to the standard hydrogen electrode. The two half-cells are connected through an external circuit containing a high-resistance voltmeter.
Ionic movement between the half-cells is also required. Without it, charge would rapidly accumulate in the solutions and further electron flow would stop.
Why the salt bridge matters
A salt bridge completes the internal ionic circuit and helps maintain charge balance and electrical neutrality in the two half-cells. It normally contains an inert electrolyte whose ions do not react significantly with the substances in either half-cell.
Potassium nitrate, KNO₃, is commonly suitable because K⁺ and NO₃⁻ ions are often spectator ions. As oxidation occurs in the zinc half-cell, Zn²⁺ ions accumulate in solution, so anions such as NO₃⁻ move towards that half-cell. At the copper half-cell, Cu²⁺ ions are removed from solution by reduction, so cations such as K⁺ migrate towards that side.
Anode half-cell: oxidation generates additional positive ions in the Zn/Cu example, so anions from the salt bridge migrate towards the anode compartment.
Cathode half-cell: reduction removes Cu²⁺ ions from solution, so cations from the salt bridge migrate towards the cathode compartment.
5. Reading a table of standard electrode potentials
Electrode-potential tables become much easier to use once you stop viewing each E° value as an isolated number. Instead, imagine the redox couples arranged vertically according to their reduction potentials.
The oxidised forms are written on the left of the reduction half-equations, while the corresponding reduced forms appear on the right.
Oxidising agents
The species on the left-hand side of a reduction half-equation gains electrons when reduction occurs.
The more positive its E° value, the greater its tendency to accept electrons. Therefore, among the oxidised forms being compared, the species associated with the most positive E° is generally the strongest oxidising agent.
Reducing agents
The species on the right-hand side of the half-equation is the reduced form. For it to act as a reducing agent, the half-equation must operate in reverse so that the species loses electrons.
The more negative the tabulated reduction potential, the greater the relative tendency for the reduced form to undergo oxidation. Reduced forms near the bottom of the E° table therefore tend to be stronger reducing agents.
Quick visual rule: strong oxidising agents are found towards the top-left of an E° table, while strong reducing agents are found towards the bottom-right.
The diagonal rule
Consider a possible reaction between an oxidised form on the left and a reduced form on the right. If the oxidised form lies higher in the E° table, it has the more positive reduction potential and can accept electrons from a reduced form lying further down.
For example, Cl₂ has a much more positive reduction potential than the Cu²⁺/Cu, H⁺/H₂ and Zn²⁺/Zn couples. Chlorine can therefore oxidise Cu(s), H₂(g) or Zn(s) under suitable conditions.
6. Calculating standard cell potential
When two half-cells are connected, one undergoes reduction at the cathode and the other undergoes oxidation at the anode.
Since electrode-potential tables contain reduction potentials, the safest calculation is:
Both values inserted into this expression should be the tabulated reduction potentials. Do not reverse the sign of the anode value first and then subtract it again.
Worked example: the Zn/Cu cell
The Cu²⁺/Cu couple has the more positive reduction potential. Cu²⁺ therefore undergoes reduction at the cathode:
Zinc undergoes oxidation at the anode:
The standard cell potential is:
Adding the two half-equations gives:
Never multiply an E° value by a stoichiometric coefficient. Electrode potential does not change when a half-equation is multiplied.
Why E° is not multiplied
Suppose a half-equation is doubled so that its electrons cancel with those in another half-equation. Twice as much material may now appear in the equation, but the electrode potential remains unchanged.
E° is a potential difference. It is not proportional to the amount of reacting substance in the way that an extensive thermodynamic quantity such as ΔG° is.
7. Predicting whether a redox reaction is feasible
Once E°cell has been calculated for the reaction as written, its sign tells us whether that direction is thermodynamically feasible under standard conditions.
E°cell > 0: the reaction as written is thermodynamically feasible under standard conditions.
E°cell < 0: the reaction as written is not thermodynamically feasible under standard conditions. The reverse direction has a positive E°cell.
The Gibbs free energy connection
Standard cell potential is linked directly to the standard Gibbs free energy change:
Here, n is the number of moles of electrons transferred in the balanced overall reaction, while F is the Faraday constant.
Since n and F are positive, a positive E°cell gives a negative ΔG°. This is why a positive standard cell potential corresponds to a thermodynamically feasible reaction under standard conditions.
8. Why a positive E°cell does not guarantee a fast reaction
Electrode potentials provide thermodynamic information. They tell us whether a reaction is energetically favourable, but they do not tell us how quickly that reaction will occur.
A reaction may have a positive E°cell but still proceed so slowly that no obvious change is seen during an experiment. This happens when the reaction has a sufficiently large activation energy.
Thermodynamic feasibility and reaction rate are different ideas. E°cell helps predict whether a reaction is energetically favourable; kinetics determines how quickly it occurs.
Standard versus non-standard conditions
The symbol E° refers specifically to standard conditions. If the concentrations of dissolved species or the pressures of gases are changed, the actual electrode potentials can change.
Therefore, a prediction made from standard electrode potentials must be interpreted carefully when the experimental conditions differ substantially from the standard state.
9. Effect of concentration on electrode potential
At this level, concentration effects can often be understood qualitatively by considering the equilibrium represented by the reduction half-equation. Consider a simple metal-ion half-cell:
Increasing the concentration of Mⁿ⁺ favours the forward reduction direction. The tendency for Mⁿ⁺ to accept electrons therefore increases, so the electrode potential becomes more positive.
Decreasing the concentration of Mⁿ⁺ has the opposite effect. The forward reduction becomes less favourable, so the electrode potential becomes less positive, or more negative.
Higher [Mⁿ⁺]: greater tendency for Mⁿ⁺ + ne⁻ → M, so the electrode potential becomes more positive.
Lower [Mⁿ⁺]: smaller tendency for the written reduction, so the electrode potential becomes less positive.
Be careful not to turn this into the rule “increasing concentration always increases electrode potential”. For a more complicated half-equation, identify which species has changed in concentration and consider how that change affects the equilibrium represented by the reduction equation.
10. Application: the hydrogen–oxygen fuel cell
Electrochemical cells do not have to contain a fixed store of reactants. A fuel cell generates electrical energy while fuel and oxidant are continuously supplied to the electrodes.
In an alkaline hydrogen–oxygen fuel cell, hydrogen is oxidised at the anode while oxygen is reduced at the cathode. Hydroxide ions, OH⁻, move through the alkaline electrolyte from the cathode side towards the anode side.
Half-equations in alkaline conditions
At the anode, hydrogen loses electrons:
At the cathode, oxygen gains electrons:
When the two half-equations are added, the electrons and OH⁻ ions cancel:
Why a fuel cell can keep operating
A conventional battery contains a limited quantity of reactants within the cell. Once those reactants have been substantially consumed, the battery can no longer continue delivering electrical energy unless it is recharged or replaced.
A fuel cell is different. As long as hydrogen and oxygen continue to be supplied and the products are appropriately removed, the cell can continue generating electrical energy.
Common Mistakes
| Common mistake | Why it is wrong | Better approach |
|---|---|---|
| Assuming a negative E° means that the half-reaction can never occur. | E° is a relative reduction potential. Whether a redox reaction is feasible depends on the other half-cell with which it is paired. | Compare both E° values and calculate E°cell for the proposed reaction. |
| Changing the sign of the anode E° and then using E°cathode − E°anode. | This reverses the sign twice and produces the wrong cell potential. | Insert the tabulated reduction potentials directly into E°cell = E°cathode − E°anode. |
| Multiplying E° when a half-equation is multiplied. | Electrode potential does not scale with the stoichiometric amount of substance. | Multiply half-equations when balancing electrons, but leave each E° value unchanged. |
| Saying that a positive E°cell guarantees an immediate visible reaction. | E°cell describes thermodynamic feasibility rather than reaction rate. | State that the reaction is thermodynamically feasible under the specified conditions, while remembering that kinetics may make it very slow. |
| Saying that the salt bridge supplies electrons. | Electrons travel through the external metallic circuit. The salt bridge provides ionic conduction instead. | Describe the salt bridge as maintaining charge balance and electrical neutrality through ion migration. |
What you should be able to do
Interpret E° values
Explain why a more positive standard reduction potential represents a greater tendency for reduction.
Describe the standard hydrogen electrode
State its components, standard conditions and assigned potential of 0.00 V.
Construct different half-cells
Decide when a metal can act as its own electrode and when an inert platinum electrode is required.
Calculate E°cell
Use E°cell = E°cathode − E°anode correctly without multiplying E° values by stoichiometric coefficients.
Predict redox feasibility
Use E° values and the sign of E°cell to determine the thermodynamically feasible direction of a redox reaction.
Explain limitations and applications
Distinguish thermodynamics from kinetics, predict qualitative concentration effects and explain the operation of an alkaline hydrogen–oxygen fuel cell.
These ten questions test whether you can apply electrode-potential ideas to half-cells, calculations, feasibility and unfamiliar redox situations.
- 10 mastery questions
- Calculations and electrochemical reasoning
- Instant explanation for every answer option
Frequently asked questions
Questions students often ask about electrode potentials
Most difficulties with electrode potentials come from signs, conventions and deciding which half-equation should run in which direction. These questions bring those ideas together.
Does a positive E° mean that the half-reaction will definitely occur?
No. E° is the reduction potential of one redox couple relative to the standard hydrogen electrode. Whether a complete redox reaction is thermodynamically feasible depends on the second redox couple and therefore on E°cell. A positive E° for one half-cell alone is not enough to predict the direction of a complete reaction.
Why are electrode potentials written as reduction potentials?
Using one convention allows every redox couple to be compared on the same scale. Electrode-potential tables therefore write half-equations in the reduction direction. When a half-equation actually operates as an oxidation, you can still use its tabulated reduction potential directly in E°cell = E°cathode − E°anode.
Why is platinum used in some half-cells but not others?
A metal such as copper or zinc is itself a conducting solid and can provide the electrical connection for its half-cell. If both members of a redox couple are ions or gases, there is no suitable conducting solid present, so an inert electrode such as platinum is introduced. Platinum conducts electrons and provides a surface for electron transfer without being consumed in the overall reaction.
What does the salt bridge actually do?
The salt bridge allows ions to migrate between the half-cells so that excessive charge does not build up in either solution. It therefore completes the internal ionic circuit and helps maintain charge balance and electrical neutrality. A typical salt bridge may contain an inert electrolyte such as KNO₃. In a Zn/Cu cell, NO₃⁻ ions migrate towards the zinc half-cell as Zn²⁺ accumulates, while K⁺ ions migrate towards the copper half-cell as Cu²⁺ is removed from solution.
Why must I not multiply E° when I multiply a half-equation?
Multiplying a half-equation changes the amount of substance and the number of electrons involved, but it does not change the electrode potential. For example, doubling a half-equation leaves its E° exactly the same. This differs from an extensive quantity such as ΔG°, which does scale with the amount of reaction.
If E°cell is positive, why might I still see no reaction?
A positive E°cell means that the reaction is thermodynamically feasible under the specified conditions. It does not tell you how quickly the reaction occurs. If the reaction has a large activation energy, its rate may be extremely slow even though the overall reaction is energetically favourable.
Electrode potentials turn redox chemistry into a quantitative comparison. Once you remember that tabulated E° values are reduction potentials, the reasoning becomes systematic: compare the two redox couples, identify which species is reduced and which is oxidised, calculate E°cell, and then interpret what its sign tells you. The important final step is to keep thermodynamic feasibility, reaction rate and non-standard conditions as separate ideas rather than expecting one E° value to answer every electrochemical question.