Lesson 3 of 7

Cell direction and real conditions

Use the potential difference and recognise its limits.

A-Level 9476 (2026-2027)

I can define standard electrode and standard cell potentials.Syllabus
Syllabus 9476, 12(b). Define standard electrode and standard cell potentials.
I can calculate standard cell potentials.Syllabus
Syllabus 9476, 12(e). Calculate standard cell potentials.
I can use cell potentials to predict electron flow and spontaneity.Syllabus
Syllabus 9476, 12(f). Use cell potentials to predict electron flow and spontaneity.
I can recognise limits of standard-potential predictions.Syllabus
Syllabus 9476, 12(g). Recognise limits of standard-potential predictions.
I can predict concentration effects on electrode potentials.Syllabus
Syllabus 9476, 12(j). Predict concentration effects on electrode potentials.

The spontaneous cell sends electrons from anode to cathode

Use reduction potentials consistently, then balance the chemical equation.

The standard cell potential E°_cell is the cell electromotive force, or maximum potential difference measured at negligible current, when all components are in standard states at the stated temperature. For a chosen reaction, E°_cell = E°_cathode - E°_anode, with both tabulated values kept as reduction potentials.

At the cathode, reduction consumes electrons. At the anode, oxidation releases them. In a spontaneous galvanic cell, electrons flow through the wire from the negative anode to the positive cathode. Ions carry charge through the electrolyte and salt bridge; electrons do not flow through the bridge.

A zinc-copper galvanic cell

For standard concentrations at 298 K, the Zn/Cu cell has E°_cell = 1.10 V. The sketch shows discharge; an ideal potential measurement draws negligible current.

Worked examplePredict the direction and voltage

Use E°(Zn²⁺/Zn) = -0.76 V and E°(Cu²⁺/Cu) = +0.34 V.

Try it first. Then show the worked stepsHide the worked steps

The copper reduction is more favourable, so Cu²⁺ is reduced at the cathode.
Reverse the zinc half-equation: Zn → Zn²⁺ + 2e^-. Zinc is the anode and reducing agent.
E°_cell = 0.34 - (-0.76) = +1.10 V.
The half-equations already transfer two electrons each: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s).

Answer

Electrons flow from zinc to copper. The written reaction is thermodynamically spontaneous under standard conditions. Zinc mass falls and copper mass rises.

Try it step by step

Work it out in steps: The cathode, then E cell = E(cathode) - E(anode).

Problem 1. A cell joins Mg²⁺/Mg, E = -2.38 V, with Ag⁺/Ag, E = +0.80 V. Find the standard cell potential in V to 2 decimal places.
Problem 2. A cell joins Fe²⁺/Fe, E = -0.44 V, with Cu²⁺/Cu, E = +0.34 V. Find the standard cell potential in V to 2 decimal places.
Problem 3. A cell joins Zn²⁺/Zn, E = -0.76 V, with Cl₂/Cl^-, E = +1.36 V. Choose the cathode, then find the standard cell potential in V to 2 decimal places.

Show the answer

Problem 1.

The cathode: Ag⁺/Ag has the more positive E, so silver ions are reduced: that electrode is the cathode. Magnesium is oxidised at the anode.

E cell = E(cathode) - E(anode): E cell = +0.80 - (-2.38) = +3.18 V.

Problem 2.

The cathode: Cu²⁺/Cu is more positive, so it is the cathode; iron is oxidised.

E cell = E(cathode) - E(anode): E cell = +0.34 - (-0.44) = +0.78 V.

Problem 3.

The cathode: Cl₂/Cl^- is more positive, so chlorine is reduced at the cathode and zinc is oxidised at the anode.

E cell = E(cathode) - E(anode): E cell = +1.36 - (-0.76) = +2.12 V.

Interpret the sign for the reaction as written
E°_cell	Standard thermodynamic conclusion
Positive	Forward reaction is favourable: ΔG° < 0.
Negative	Forward reaction is unfavourable; the reverse is favoured under standard conditions.
Zero	ΔG° = 0; there is no standard thermodynamic driving force in either direction.

Watch out for this

A student says: "Ag⁺ + e^- -> Ag is doubled to balance electrons, so its E value of +0.80 V must be doubled too." What is wrong with this?

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Potential is energy per unit charge. Doubling the reaction doubles both the energy and charge transferred, leaving the potential unchanged. Balance electrons using equation coefficients, then subtract the two unscaled reduction potentials.

A standard prediction is not a guarantee of a rapid real reaction

Concentration affects equilibrium; activation barriers affect the time taken.

Make a guess. You are not marked.

In the half-cell Cu²⁺(aq) + 2e^- <=> Cu(s), the Cu²⁺ concentration is raised above 1 mol dm^-3. What happens to the electrode potential?

It stays at +0.34 V, as the metal is the same
It becomes more negative, as more ions crowd the electrode
It becomes more positive, as reduction is favoured

Show the answer

It becomes more positive, as reduction is favoured

More Cu²⁺ shifts the equilibrium to the right, so the half-cell gains electrons more readily and its potential rises.

Electrode potentials depend on the actual composition. Think of each half-equation as a reduction equilibrium: a change favouring reduction generally makes that electrode potential more positive. A change favouring the reverse makes it less positive. These are qualitative predictions at fixed temperature, with other conditions held constant.

Concentration changes with other conditions fixed
Half-equation	Change	Effect on reduction potential
Cu²⁺ + 2e^- ⇌ Cu	Increase [Cu²⁺].	More positive: reduction of Cu²⁺ is favoured.
Fe³⁺ + e^- ⇌ Fe²⁺	Increase [Fe³⁺], or decrease [Fe²⁺].	More positive; the oxidised-to-reduced concentration ratio rises.
2H⁺ + 2e^- ⇌ H₂	Decrease [H⁺] at fixed hydrogen pressure.	Less positive: hydrogen-ion reduction is less favoured.

In the Zn/Cu cell, diluting Cu²⁺ lowers the copper reduction potential. Increasing Zn²⁺ raises the zinc reduction potential. Either change reduces E_cell = E_Cu - E_Zn. Far enough from standard conditions, an E° prediction can cease to describe the actual direction. Complex formation and precipitation matter because they change the free ion concentrations.

Even a positive actual cell potential says nothing about how quickly a reaction proceeds. Large activation barriers, slow electrode reactions or a protective surface film can make a thermodynamically favoured change difficult to observe. Different temperature, gas pressure, pH or chemical species can also invalidate a direct application of tabulated E° values. Use the exact relevant half-equation and the conditions, not only the names of the elements.

Check your understanding

Excess ammonia complexes most of the free Cu²⁺ ions. What happens to the electrode potential of the Cu²⁺/Cu couple?

It becomes less positive, because the concentration of free Cu²⁺ has fallen.
It stays the same, because the total amount of dissolved copper has not changed.
It becomes more positive, because the complex holds more copper in the solution.

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It becomes less positive, because the concentration of free Cu²⁺ has fallen.

Lowering [Cu²⁺] shifts Cu²⁺ + 2e^- <=> Cu to the left, so reduction is less favoured. The free-ion concentration matters, not the total copper dissolved.