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ΔG = −nFE

Learn and apply ΔG = −nFE in the published Chemistry course sequence.

  • GCE A-Level H2 Chemistry 9476-2027
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ΔG° = −nFE°: Orientation

This lesson connects electrochemistry to energetics: once you know E⦵_cell and the electron count n, you can use Δ G⦵ = -nF E⦵_cell to decide feasibility and calculate energy changes.

Before the harder applications, review Writing Redox Equations from Half-Equations and keep the Electrochemistry hub as your route map.

This page is really testing:

  • Can you extract the correct n from a balanced redox equation?
  • Can you do a clean units chain (J mol⁻¹ vs kJ mol⁻¹) without sign errors?
  • Can you interpret signs correctly: positive E⦵_cell and negative Δ G⦵?

Definitions (Must Know)

A. Relationship between free energy and cell potential

Under standard conditions: Δ G⦵ = -nF E⦵_cell

Where:

  • Δ G⦵ is the standard Gibbs free energy change (J mol⁻¹)
  • n is the number of moles of electrons transferred per mole of reaction
  • F is the Faraday constant (≈ 9.65 × 10⁴ C mol⁻¹)
  • E⦵_cell is the standard cell potential (V)

Detailed Explanations

A. How to find n reliably

  1. Write the two half-equations.
  2. Balance electrons by multiplying.
  3. The number of electrons that cancel is n.

Mini example: For Zn + Cu²⁺ → Zn²⁺ + Cu, n = 2 (2 electrons transferred).

B. What the equation is telling you (in words)

Cell potential measures the “push” for electron flow. Δ G⦵ measures the maximum useful electrical work available per mole of reaction. The minus sign links a positive cell potential to a negative free energy change.

C. Worked calculation method (exam-safe order)

  1. Find or calculate E⦵_cell first.
  2. Determine n from the balanced electron transfer.
  3. Substitute into Δ G⦵ = -nFE⦵_cell with F = 9.65 × 10⁴ C mol⁻¹.
  4. Keep units in J mol⁻¹ during substitution.
  5. Convert to kJ mol⁻¹ only at the end if required.
  6. Add a sign interpretation sentence about feasibility under standard conditions.

Worked Examples

Modelled example 1

Calculate Standard Gibbs Energy

Core

Problem

A cell has E⦵_cell = +1.10 V and transfers n = 2 electrons. Calculate Δ G⦵.
Study the worked solution

  1. Apply the sign convention

    Method

    Use Δ G⦵ = -nFE⦵_cell.

    Reason

    A positive cell potential corresponds to a negative standard Gibbs-energy change.

    Working

    -(2)(9.65 × 10⁴)(1.10)

  2. Evaluate

    Method

    Calculate in joules, then convert if desired.

    Reason

    Faraday’s constant is in C mol⁻¹ and V·C = J.

    Working

    -2.12 × 10⁵ J mol⁻¹ = -212 kJ mol⁻¹

Guided practice 2

Recover Cell Potential

About 6 min

Problem

For a reaction, Δ G⦵ = -145 kJ mol⁻¹ and n = 3. Calculate E⦵_cell.

Try this before viewing the solution

Hints

Hint 1: joules
Convert −145 kJ mol⁻¹ to −145000 J mol⁻¹.
Hint 2: subject
Use E⦵ = -Δ G⦵/(nF).
View solution step by step

  1. Align units

    Method

    Convert Gibbs energy to joules per mole.

    Reason

    F is expressed using coulombs, so the energy result pairs with joules.

    Working

    Δ G⦵ = -145000 J mol⁻¹

  2. Rearrange

    Method

    Divide -Δ G⦵ by nF.

    Reason

    The two negative signs give a positive potential.

    Working

    E⦵ = 145000/(3)(9.65 × 10⁴) = 0.501 V

Common misconception 3

Correct the Sign Inference

Find and correct the mistake

Learner claim

A learner says positive Δ G⦵ requires positive E⦵_cell. Correct the sign reasoning.

Try this before viewing the solution

If ΔG° is positive, E°cell is

View solution step by step

  1. Fix signs of constants

    Method

    Recognise that n > 0 and F > 0.

    Reason

    Only the leading minus sign reverses the sign of E⦵.

    Working

    Δ G⦵ = -(positive)E⦵

  2. Infer direction

    Method

    Give E⦵_cell < 0 when Δ G⦵ > 0.

    Reason

    The two quantities have opposite signs.

    Working

    Δ G⦵ > 0 ⇒ E⦵_cell < 0

Examiner practice 4

Handle a Scaled Cell Reaction

3 marks

Problem

A cell reaction as written transfers 4 mol of electrons and has E⦵_cell = 0.80 V. Calculate Δ G⦵. [3 marks]

Try this before viewing the solution

View solution step by step

  1. Use reaction electron count

    1 mark

    Method

    Set n = 4 for the equation as written.

    Reason

    Gibbs energy is extensive and scales with stoichiometric electron transfer.

    Working

    Δ G⦵ = -nFE⦵

  2. Substitute

    1 mark

    Method

    Use the unchanged intensive cell potential.

    Reason

    Potential is not multiplied when the equation is scaled.

    Working

    -(4)(9.65 × 10⁴)(0.80)

  3. State result

    1 mark

    Method

    Report energy per mole of reaction as written.

    Reason

    The calculation gives the extensive energy change for that stoichiometry.

    Working

    -3.09 × 10⁵ J mol⁻¹

Challenge 5

Infer the Electron Count

Minimal support

Problem

A reaction has Δ G⦵ = -193 kJ mol⁻¹ and E⦵_cell = +1.00 V. Determine n.

Try this before viewing the solution

Hints

Hint 1: convert
Use −193000 J mol⁻¹.
Hint 2: rearrange
n = -Δ G⦵/(FE⦵).
View solution step by step

  1. Solve numerically

    Method

    Divide the positive magnitude by FE⦵.

    Reason

    The signs cancel for a spontaneous forward reaction.

    Working

    n = 193000/(9.65 × 10⁴)(1.00) = 2.00

  2. Interpret

    Method

    State two electrons per reaction as written.

    Reason

    n is a stoichiometric electron count and should agree with the balanced redox equation.

    Working

    n = 2

Mind Stretchers

Mind stretcher 1Extension

Why is E⦵_cell independent of how you balance the overall equation, but n changes when you scale the equation?

Show Hint

Classify each quantity as intensive or extensive before thinking about equation scaling.

Show Answer

Mark scheme:

  • E⦵_cell depends on the difference between two intensive half-cell potentials, so it does not scale with amount of substance.
  • n is tied to the stoichiometric amount of electrons transferred per mole of reaction as written, so it scales when you multiply the equation.

Mind stretcher 2: Scaling a cell reaction without scaling its voltageExtension

Question. A balanced cell reaction has E⦵_cell = +0.62 V, n = 2 and Δ G⦵ = -120 kJ mol⁻¹ to the stated precision. The entire equation is doubled. State the new values of E⦵_cell, n and Δ G⦵, and explain the pattern.

Show Hint

Voltage is intensive; electron amount and free-energy change follow the stoichiometric scale of the reaction as written.

Show Answer

E⦵_cell remains + 0.62 V because electrode potential is intensive. The electron count becomes n = 4, and Δ G⦵ becomes -240 kJ mol⁻¹ because the amount of reaction and maximum electrical work have doubled. This keeps Δ G⦵ = -nFE⦵_cell consistent.