IB Chemistry HLTopic 6 — Electron TransferPaper 1 & 2Core skill~10 min read
Gibbs Energy & Standard Cell Potential
You have met two different tests for “will this reaction go?” — a negative ΔGθ from energetics, and a positive Eθcell from electrochemistry. They are not two tests. They are the same test in different clothes, and one short equation joins them up.
📚 What you need to know
The link is ΔGθ = −nFEθ (given in Section 1 of the data booklet).
n = moles of electrons transferred in the balanced overall equation.
F = the Faraday constant, 9.65 × 104 C mol−1 (Section 2).
Because of the minus sign, the two signs are always opposite: positive Eθ means negative ΔGθ.
Volts × coulombs = joules, so the answer comes out in J mol−1. Divide by 1000 for kJ mol−1.
At equilibrium both are zero — that is a flat battery.
Changing n changes ΔGθ but never changes Eθ.
Where the equation comes from
ΔGθ is the maximum useful work a reaction can do. In a voltaic cell that work is electrical: you are pushing charge through a circuit.
Electrical work is charge multiplied by voltage. The charge carried by one mole of electrons is F, so n moles of electrons carry nF coulombs. Multiply by the voltage and you have the energy:
charge moved = nF
energy released = nFEθ
energy released means ΔG is negative, hence the minus sign
The bridge equation
ΔGθ = −nFEθ
The minus sign is not a trick to catch you out. It is just bookkeeping: chemists count energy leaving the system as negative, but a cell that releases energy has a positive voltage. Something has to flip, and it is the sign.
Reading the signs
If a reaction is non-spontaneous as written, the reverse reaction is spontaneous. That is exactly what electrolysis forces you to pay for.
Getting n right
This is where most marks are lost. n is not the number of electrons in one half-equation — it is the number that actually cross over in the balanced overall equation.
🧩 Finding n every time
Write both half-equations the way they run: one oxidation, one reduction.
Scale them so the electrons match.
n is that matched number. For 2Al + 3Cu2+ you needed 6 electrons, so n = 6.
Do not touch Eθ. Scaling changes n and therefore ΔGθ, but the voltage is fixed.
Why voltage does not scale: volts are joules per coulomb. Double the reaction and you double both the joules and the coulombs, so the ratio — the voltage — is unchanged. ΔGθ is a total, so it does double.
Worked examples
WORKED EXAMPLE
Magnesium (−2.37 V) is used with a copper half-cell (+0.34 V). Calculate ΔGθ for the cell reaction.
Step 1: Find EθcellEθcell = (+0.34) − (−2.37) = +2.71 VStep 2: Write the overall equation and count electronsMg(s) + Cu2+(aq) → Mg2+(aq) + Cu(s), so n = 2Step 3: SubstituteΔGθ = −2 × 96500 × 2.71= −523030 J mol−1ΔGθ = −523 kJ mol−1big negative number, big positive voltage — exactly what you expect
WORKED EXAMPLE
Calculate ΔGθ for 2Al(s) + 3Cu2+(aq) → 2Al3+(aq) + 3Cu(s), given Al3+/Al = −1.66 V and Cu2+/Cu = +0.34 V.
Step 1: Cell potential firstEθcell = (+0.34) − (−1.66) = +2.00 VStep 2: Count the electrons in the equation you were given2Al gives away 6 electrons; 3Cu2+ takes 6. So n = 6, not 2 or 3.Step 3: SubstituteΔGθ = −6 × 96500 × 2.00 = −1158000 J mol−1ΔGθ = −1.16 × 103 kJ mol−1the voltage stayed at 2.00 V — only n did the scaling
WORKED EXAMPLE
A cell has ΔGθ = −213 kJ mol−1 and transfers 2 moles of electrons. Find Eθcell.
Step 1: Convert to joules before anything elseΔGθ = −213 × 1000 = −213000 J mol−1Step 2: Rearrange the equationEθ = −ΔGθ ÷ nFStep 3: SubstituteEθ = 213000 ÷ (2 × 96500) = 1.1036…Eθcell = +1.10 Vforget the kJ to J step and you get 1.10 millivolts — a classic lost mark
WORKED EXAMPLE
For Ag+(aq) + Fe2+(aq) → Ag(s) + Fe3+(aq), Eθcell = +0.03 V. Calculate ΔGθ and comment.
Step 1: Count electronsAg+ gains one; Fe2+ loses one. n = 1.Step 2: SubstituteΔGθ = −1 × 96500 × 0.03 = −2895 J mol−1ΔGθ = −2.90 kJ mol−1negative, so it goes — but barely. The position of equilibrium will not be far to the right.
The flat battery
As a cell runs, the reactants get used up and the concentrations drift away from 1.00 mol dm−3. The voltage falls. When the system reaches equilibrium, there is no longer any tendency to push electrons in either direction:
Eθcell = 0
ΔGθ = 0
no net electron flow — the battery is dead
This is also why cell potentials connect to equilibrium constants. A large positive Eθcell means a very negative ΔGθ, which means a very large K — the reaction goes essentially to completion.
💡 Exam tip
Units are the biggest trap. F is in C mol−1, so ΔGθ comes out in J mol−1. Convert at the end, not the middle.
Use 96500 (or 9.65 × 104) — both are the booklet value and both are accepted.
Write the overall balanced equation before you decide n. Guessing n from one half-equation is the most common error in this topic.
State the sign in your final answer and add the units. “−523 kJ mol−1” scores; “523” does not.
If a question gives ΔGθ and asks for Eθ, rearrange first, substitute second. Doing it the other way round invites sign errors.
Both the equation and F are in the data booklet, so do not waste revision time memorising them — practise using them instead.
⚠ Common mix-up
Using n from a single half-equation. For 2Al + 3Cu2+, n is 6, not 3.
Multiplying Eθ when scaling. ΔGθ scales, Eθ does not. Ever.
Losing the minus sign. A positive voltage must give a negative ΔGθ. If your signs match, you have dropped one.
Answering in J when the question asked for kJ (or the reverse). Read the units in the question stem.
Thinking ΔGθ = 0 means nothing is happening. At equilibrium both reactions are still going, just at equal rates.
Assuming a big ΔGθ means a fast reaction. It does not. Thermodynamics predicts; kinetics controls.
Up next: Electrolysis of Aqueous Solutions — if a reaction has a negative Eθcell, you can still force it to happen by paying for it with electricity. That is what electrolysis is.
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