Cell Potential and Free Energy: Study Guide (Chemistry 9.9)

From “it works” to “how hard it pushes”

In Topic 9.8 you learned to read a cell: oxidation at the anode, reduction at the cathode, electrons through the wire. A voltmeter in a galvanic cell shows a number. That number is the cell potential, E, in volts (V). It tells you how strongly the redox reaction drives electrons through the wire.

This topic answers three questions:

1. Where does the voltage of a cell come from, and how can you calculate it from data?
2. What does its sign tell you?
3. How is it connected to the free energy change, ΔG°, that you met in Topic 9.3?

Standard reduction potentials

Every half-reaction has its own “appetite” for electrons. Chemists measure it as a standard reduction potential, E°. The half-reaction is always written as a reduction (electrons on the left).

• Each value is measured against a reference half-cell, the standard hydrogen electrode: 2H⁺(aq) + 2e⁻ → H₂(g), which is given E° = 0.00 V by definition.
• “Standard” (the ° sign) means 1 M for dissolved species, 1 atm for gases, and pure solids and liquids. Tables are usually for 25 °C (298 K).
• A more positive E° means the species on the left is more easily reduced: it is a stronger oxidising agent.
• A more negative E° means the reduction is harder, so the reverse (oxidation of the metal on the right) is easier.

Half-reaction (as a reduction)	E° (V)
Cl₂(g) + 2e⁻ → 2Cl⁻(aq)	+1.36
Br₂(l) + 2e⁻ → 2Br⁻(aq)	+1.07
Ag⁺(aq) + e⁻ → Ag(s)	+0.80
Fe³⁺(aq) + e⁻ → Fe²⁺(aq)	+0.77
I₂(s) + 2e⁻ → 2I⁻(aq)	+0.54
Cu²⁺(aq) + 2e⁻ → Cu(s)	+0.34
2H⁺(aq) + 2e⁻ → H₂(g)	0.00
Fe²⁺(aq) + 2e⁻ → Fe(s)	−0.44
Zn²⁺(aq) + 2e⁻ → Zn(s)	−0.76
Al³⁺(aq) + 3e⁻ → Al(s)	−1.66

On the exam you are given the values you need. Learn to use the table, not to memorise it.

Calculating a standard cell potential

In a cell, one half-reaction runs as a reduction (at the cathode) and the other runs backwards, as an oxidation (at the anode). The cell potential is the gap between the two reduction potentials:

E°cell = E°(cathode) − E°(anode), with both values taken straight from the table as reduction potentials.

Some teachers write the same idea as E°cell = E°(reduction) + E°(oxidation), where the oxidation potential is the reduction potential with its sign changed. Both give the same answer. Pick one method and use it every time, so you never change a sign twice.

For a zinc–copper cell, Cu²⁺ is reduced and Zn is oxidised: E°cell = (+0.34) − (−0.76) = +1.10 V.

Do not multiply E° by coefficients

To balance electrons you often multiply a half-reaction by 2 or 3. The E° value stays the same. A potential is energy per unit of charge (1 V = 1 J C⁻¹). Doubling a half-reaction doubles the energy and doubles the charge, so the ratio does not change. Think of it like the height of a waterfall: twice as much water falling does not make the waterfall taller.

The sign of E°cell tells you about favourability

• E°cell > 0: the reaction is thermodynamically favoured under standard conditions. It can run as a galvanic cell and produce a positive voltage.
• E°cell < 0: the reaction as written is thermodynamically unfavoured. It happens only if an outside power supply applies a potential at least as large as the gap: an electrolytic cell. Its reverse reaction is favoured, with the same size of E° and the opposite sign.

This matches Topic 9.8: a galvanic cell runs a favourable reaction, and an electrolytic cell uses electrical energy to force an unfavourable one.

Linking cell potential and free energy: ΔG° = −nFE°

Electrical work equals charge × potential difference. When a redox reaction pushes n mol of electrons (charge nF) through a potential E°, the most electrical work it can do is nFE°. That maximum work equals the decrease in free energy, −ΔG°:

ΔG° = −nFE°

• n = moles of electrons transferred per mole of reaction as written in the balanced equation (find it from the electrons that cancel).
• F = Faraday’s constant, 96,485 C mol⁻¹: the charge on one mole of electrons.
• E° in volts. Because 1 C × 1 V = 1 J, the answer comes out in joules (per mole of reaction). Divide by 1000 for kJ.

The minus sign does the important job: a positive E° gives a negative ΔG°, so both say “favoured”. A negative E° gives a positive ΔG°.

E°cell	ΔG°	Reaction as written	Cell type
positive	negative	thermodynamically favoured	galvanic
negative	positive	thermodynamically unfavoured	electrolytic (needs applied potential)
zero	zero	neither direction favoured at standard conditions	no net drive

ΔG° depends on how you write the equation; E° does not. If you double the equation, n doubles, so ΔG° doubles, but E° is unchanged. This is just what you would expect: ΔG° is the free energy for a stated amount of reaction, while E° is energy per unit of charge.

Worked example 1: an aluminium–iron cell

Question. A galvanic cell is made from an aluminium strip in 1 M Al(NO₃)₃ and an iron strip in 1 M Fe(NO₃)₂. Use the table to (a) identify the cathode and anode, (b) write the balanced overall equation, (c) calculate E°cell, and (d) calculate ΔG° for the reaction as written.

(a) Fe²⁺/Fe (−0.44 V) is higher on the ladder than Al³⁺/Al (−1.66 V). So Fe²⁺ is reduced: iron is the cathode. Aluminium is oxidised: aluminium is the anode.

(b)

• Cathode: Fe²⁺(aq) + 2e⁻ → Fe(s) (× 3)
• Anode: Al(s) → Al³⁺(aq) + 3e⁻ (× 2)
• Six electrons cancel: 2Al(s) + 3Fe²⁺(aq) → 2Al³⁺(aq) + 3Fe(s), so n = 6.

(c) E°cell = E°(cathode) − E°(anode) = (−0.44) − (−1.66) = +1.22 V. The multipliers 3 and 2 do not touch the E° values. (Multiplying them in would give 3(−0.44) − 2(−1.66) = +2.00 V, which is wrong.)

(d) ΔG° = −nFE° = −(6 mol e⁻)(96,485 C mol⁻¹)(1.22 V) = −706,000 J, so ΔG° = −706 kJ mol⁻¹ (per mole of reaction as written, that is, per 2 mol of Al).

Interpretation. E° is positive and ΔG° is negative, so the reaction is thermodynamically favoured: this is a working galvanic cell. If the equation were written per 1 mol of Al (n = 3), ΔG° would be −353 kJ mol⁻¹, but E° would still be 1.22 V.

Worked example 2: can bromine oxidise chloride ions?

Question. A student proposes making chlorine by adding bromine to sodium chloride solution:

Br₂(l) + 2Cl⁻(aq) → 2Br⁻(aq) + Cl₂(g)

Is this reaction thermodynamically favoured under standard conditions? Support your answer with E° and ΔG°.

1. Identify the half-reactions. Br₂ gains electrons (reduction, so the cathode would be the bromine half-cell). Cl⁻ loses electrons (oxidation, so the anode would be the chloride half-cell).
2. Cell potential. E° = E°(Br₂/Br⁻) − E°(Cl₂/Cl⁻) = (+1.07) − (+1.36) = −0.29 V.
3. Free energy. n = 2. ΔG° = −(2)(96,485 C mol⁻¹)(−0.29 V) = +56,000 J, so ΔG° = +56.0 kJ mol⁻¹.

Answer. E° is negative and ΔG° is positive, so the reaction is not thermodynamically favoured. Bromine cannot oxidise chloride ions under standard conditions. The reaction could be forced in an electrolytic cell, but only with an externally applied potential of more than 0.29 V.

Check by reasoning. Chlorine sits higher on the ladder than bromine, so chlorine is the stronger oxidising agent. The reverse reaction, Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂, has E° = +0.29 V and ΔG° = −56.0 kJ mol⁻¹: chlorine water does turn bromide solution orange-brown.

Worked example 3: working backwards to an unknown potential

Question. A fictional metal X forms X²⁺ ions. For the reaction X(s) + 2Ag⁺(aq) → X²⁺(aq) + 2Ag(s), ΔG° = −164 kJ mol⁻¹. Find E°cell and the standard reduction potential of X²⁺/X.

1. Rearrange. E° = −ΔG° ÷ (nF). Convert kJ to J first: −164 kJ = −164,000 J.
2. n = 2 (each X loses 2 electrons; 2 Ag⁺ each gain 1).
3. E°cell = −(−164,000 J) ÷ (2 × 96,485 C mol⁻¹) = +0.850 V.
4. Silver is reduced (cathode) and X is oxidised (anode): 0.850 V = (+0.80 V) − E°(X²⁺/X), so E°(X²⁺/X) = −0.05 V.

Interpretation. X²⁺/X sits just below hydrogen on the ladder, so X should be oxidised by 1 M H⁺ (E° = 0.00 − (−0.05) = +0.05 V, a small positive value) and by Cu²⁺ (E° = 0.34 − (−0.05) = +0.39 V).

Units check. J ÷ (C mol⁻¹ × mol) leaves J C⁻¹, which is volts.

Connection to K

In Topic 9.5 you saw that ΔG° = −RT ln K. Putting the two equations together shows that a positive E°cell goes with K > 1, and a negative E°cell goes with K < 1. You will use this link in the next topic, where the cell is no longer at standard conditions.

Common misconceptions

• “Multiply E° by the coefficient when you multiply a half-reaction.” No. E° is energy per unit of charge, so it never scales. Only n (and so ΔG°) changes with the coefficients.
• “Flip the sign of the anode value, then subtract it.” That changes the sign twice. Use E°(cathode) − E°(anode) with table values, or add the oxidation potential, but not both.
• “A negative standard reduction potential means that species cannot be reduced.” It can be reduced; it just needs a partner that is oxidised even more easily (lower on the ladder), or an applied potential.
• “n is the number of electrons in one half-reaction.” n is the number of electrons that cancel in the balanced overall equation (6 in Worked example 1, not 2 or 3).
• “ΔG° comes out in kJ.” With F in C mol⁻¹ and E in V, the answer is in J. Forgetting to divide by 1000 makes ΔG° a thousand times too large.
• “A larger E°cell means a faster reaction.” E° tells you about thermodynamic favourability, not rate. As in Topic 9.4, a favoured reaction can still be very slow.
• “An electrolytic cell has a positive E°cell because current flows.” The reaction it drives has a negative E°. The current flows only because the power supply pushes it.

Where this leads

Real cells rarely run at 1 M and 1 atm. Next, Topic 9.10 explains how the cell potential changes as concentrations change, why it falls to zero at equilibrium, and how a cell with the same metal on both sides can still give a voltage. Try the practice questions now, then use the revision notes and the checklist to consolidate.