Spontaneity from E°_cell and ΔG = −nFE

Core Question

How do we predict whether a redox reaction will occur spontaneously in an electrochemical cell, and what is the bridge between electrochemistry and thermodynamics?

[!intuition] The Big Picture

Imagine pushing a boulder: if it rolls downhill by itself, that's spontaneous (releases energy). If you must push it uphill, that's non-spontaneous (requires energy input).

In redox reactions, electrons flow from reducing agent to oxidizing agent. If this flow releases energy (like the boulder rolling down), the reaction is spontaneous. The cell potential E° measures the "voltage hill" — positive E° means downhill (spontaneous), negative means uphill (forced). The Gibbs free energy ΔG is thermodynamics' measure of spontaneity: negative ΔG = spontaneous, positive = non-spontaneous.

The equation ΔG = −nFE links these two worlds: electrochemistry's voltage and thermodynamics' free energy.

[!definition] Key Terms

• E°_cell (Standard Cell Potential): The voltage difference between cathode and anode under standard conditions (1 M, 1 atm, 25°C), measured in volts (V).
• ΔG (Gibbs Free Energy Change): Energy available to do work; tells us spontaneity. Units: joules (J) per mole of reaction as written (or kJ divided by 1000). Because n is "moles of electrons per mole of reaction," the result is energy per mole of reaction.
• n: Number of moles of electrons transferred in the balanced redox equation.
• F (Faraday's Constant): Charge of 1 mole of electrons 96,485 C/mol ≈ 96,500 C/mol.
• Spontaneous Reaction: Proceeds on its own without external energy (ΔG < 0, E° > 0).

[!formula] The Bridge Equation — Derived from First Principles

WHY does this equation exist?

Because electrical work and chemical free energy are two views of the same energy change.

DERIVATION (with consistent sign conventions):

Step 1: Electrical work done BY the cell When the cell operates spontaneously, it pushes charge through the external circuit, doing useful electrical work on the surroundings.

• Charge transported = q = nF (n moles of e⁻, each mole carries charge F coulombs)
• Work done by the cell on the surroundings through potential E: $w_{by\;cell} = qE = nFE$

Why positive? For a spontaneous cell, E > 0 and the cell delivers energy to the surroundings, so the work output is positive.

Step 2: Connect to thermodynamics At constant temperature and pressure, the maximum non-PV (useful) work a system can do on the surroundings equals the decrease in Gibbs free energy: $w_{by\;system,\;max} = −ΔG$

Why? ΔG represents the "useful" energy budget. When the reaction proceeds spontaneously, ΔG decreases (ΔG < 0), and that lost free energy becomes usable work. So the work the system delivers is +(−ΔG) = −ΔG.

Step 3: Equate the two (both are "work done BY the system") The electrical work the cell delivers = maximum useful work thermodynamics allows: $nFE = −ΔG$

Multiply both sides by −1: $\boxed{ΔG = −nFE}$

The key was defining both quantities as work done by the system on the surroundings, so the algebra is consistent.

Under Standard Conditions:

$\boxed{ΔG° = −nFE°_{cell}}$

Where:

• ΔG° = standard free energy change (per mole of reaction as written)
• E°_cell = standard cell potential

[!example] Example 1: Predicting Spontaneity from E°

Given: $Zn(s) + Cu^{2+}(aq) → Zn^{2+}(aq) + Cu(s)$ Standard reduction potentials:

• E°(Cu²⁺/Cu) = +0.34 V
• E°(Zn²⁺/Zn) = −0.76 V

Find: Is this reaction spontaneous? Calculate ΔG°.

Solution:

Step 1: Calculate E°_cell $E°_{cell} = E°_{cathode} − E°_{anode}$ Cu²⁺ is reduced (cathode), Zn is oxidized (anode): $E°_{cell} = (+0.34) − (−0.76) = +1.10 \text{ V}$

Why this step? E°_cell > 0 means the reaction is spontaneous.

Step 2: Find n Balanced equation shows Zn loses 2e⁻, Cu²⁺ gains 2e⁻: $n = 2$

Step 3: Calculate ΔG° $ΔG° = −nFE°_{cell}$ $ΔG° = −(2)(96,485)(1.10)$ $ΔG° = −212,267 \text{ J} = −212.3 \text{ kJ (per mole of reaction)}$

Why this step? ΔG° < 0 confirms spontaneity thermodynamically.

Answer: Reaction is spontaneous; ΔG° = −212.3 kJ per mole of reaction.

[!example] Example 2: Non-Spontaneous Reaction

Given: $\text{Cu(s) + Zn}^{2+}\text{(aq) → Cu}^{2+}\text{(aq) + Zn(s)}$

Find: Is this spontaneous? Calculate ΔG°.

Solution:

Step 1: Calculate E°_cell Now Zn²⁺ is reduced (cathode), Cu is oxidized (anode): $E°_{cell} = (−0.76) − (+0.34) = −1.10 \text{ V}$

Why this step? E°_cell < 0 immediately tells us: non-spontaneous.

Step 2: n = 2 (same electron transfer)

Step 3: ΔG° $ΔG° = −(2)(96,485)(−1.10) = +212.3 \text{ kJ (per mole of reaction)}$

Answer: Non-spontaneous; requires external energy (like a battery) to proceed.

KEY INSIGHT: Reversing a reaction flips the sign of both E° and ΔG°.

[!example] Example 3: Multi-Electron Transfer

Given: $2\text{Al}(s) + 3\text{Cu}^{2+}(aq) \rightarrow 2\text{Al}^{3+}(aq) + 3\text{Cu}(s)$

• E°(Al³⁺/Al) = −1.66 V
• E°(Cu²⁺/Cu) = +0.34 V

Find: ΔG°.

Solution:

Step 1: E°_cell $E°_{cell} = (+0.34) − (−1.66) = +2.00 \text{ V}$

Step 2: Find n carefully

• Each Al loses 3e⁻ → 2 Al atoms lose 6e⁻
• Each Cu²⁺ gains 2e⁻ → 3 Cu²⁺ gain 6e⁻ $n = 6$

Why this step? Must count total electrons transferred in the balanced equation.

Step 3: ΔG° $ΔG° = −(6)(96,485)(2.00) = −1,157,820 \text{ J} ≈ −1158 \text{ kJ (per mole of reaction)}$

Answer: Highly spontaneous (large negative ΔG°).

[!mistake] Common Mistakes & How to Steel-Man Them

Mistake 1: "I used n = 1 when the equation had 2e⁻ transferred"

Why it feels right: You saw "2 electrons" and thought "n = 2" without checking the balanced equation's stoichiometry.

Steel-man: You're trying to identify the electron flow, which is correct. The error is not tracking the total electrons across all stoichiometric coefficients.

Fix: After balancing the half-reactions, count electrons in one half-reaction (they're equal). If the equation is multiplied to balance atoms, electrons multiply too. In 2Al → 2Al³⁺ + 6e⁻, n = 6, not 3.

Mistake 2: "E° = +0.5 V but I got ΔG° positive"

Why it feels right: You forgot the negative sign in ΔG = −nFE, so calculated ΔG = +nFE.

Steel-man: You correctly used the formula structure but missed that spontaneous E° must give negative ΔG, which requires the minus sign.

Fix: Memorize the sign: ==ΔG = **−**nFE==. Positive E° → negative ΔG → spontaneous.

Mistake 3: "I used F = 96.5 instead of 96,485"

Why it feels right: Saw "96,500" rounded and dropped units, thinking F ≈ 96.5 kJ.

Steel-man: You tried to simplify, which is good instinct, but confused the magnitude.

Fix: F = 96,485 C/mol, always in coulombs. If you want kJ, you must convert: 1 V·C = 1 J, so divide the final answer by 1000 to get kJ.

Mistake 4: "Used E°_anode − E°_cathode"

Why it feels right: Confused the formula for E°_cell.

Fix: ==E°_cell = E°_cathode − E°_anode==. Cathode is reduction (electron gain), always subtract the anode (oxidation) potential. Mnemonic: Cathode Comes first, Reduction is Right.

Mistake 5: "Sign confusion in the derivation"

Why it feels right: People sometimes write electrical work as −nFE (work done on the system) and then equate it to w_max = −ΔG (work done by the system), mixing two different sign conventions.

Fix: Keep both quantities on the same side of the ledger. Define both as work done BY the system: electrical work delivered = nFE, and max useful work = −ΔG. Then nFE = −ΔG gives ΔG = −nFE cleanly.

[!recall]- Feynman Explanation (Explain to a 12-year-old)

Okay, imagine you have two metals, like zinc and copper, sitting in water with their ions. Zinc really wants to give away electrons, and copper ions really want to grab electrons.

When you connect them with a wire, electrons flow from zinc to copper—that's electricity! The voltage (E°) is like how steep the "electron slide" is. If the slide goes downhill (positive E°), the electrons slide on their own—that's spontaneous. If it's uphill (negative E°), you'd need to push them (use a battery).

Now, scientists also measure energy using "Gibbs free energy" (ΔG)—think of it as "how much useful energy we can get." The magic formula ΔG = −nFE connects the two: the voltage (E) tells us exactly how much energy (ΔG) we get per electron flowing. The "n" is how many electrons move, and "F" is just a number that converts everything to the right units.

Bottom line: Positive voltage = electrons flow freely = reaction happens = negative ΔG. They're the same thing, just measured differently!

[!mnemonic] Memory Aid

"Negative G, Positive E, Reaction Runs Free"

• ΔG < 0 → spontaneous
• E° > 0 → spontaneous
• Formula: ΔG = −nFE → negative sign flips the relationship

For n: Count electrons in the balanced half-reaction.

F = 96,485: Think "F for Faraday, 96 thousand coulombs for a mole of charge."

Connections

• Standard Electrode Potentials — where E° values come from
• Nernst Equation — how E and ΔG change with concentration (non-standard)
• Gibbs Free Energy Fundamentals — thermodynamic basis of spontaneity
• Galvanic vs Electrolytic Cells — spontaneous (ΔG < 0) vs driven (ΔG > 0)
• Relationship between K_eq and ΔG° — ΔG° = −RT ln K links to equilibrium
• Faraday's Laws of Electrolysis — quantitative electrochemistry using F

Flashcards

Concept Map

Hinglish (regional understanding)