ΔG° = -nFE° is the equation connecting standard Gibbs free energy change to standard cell potential, where n is moles of electrons transferred and F is Faraday's constant (96,485 C/mol e⁻). The negative sign means a positive E° gives a negative ΔG°, so the redox reaction is thermodynamically favored.
ΔG° = -nFE° is the bridge between thermodynamics and electrochemistry. It says that the standard Gibbs free energy change of a redox reaction is directly proportional to the standard cell potential. Here n is the number of moles of electrons transferred in the balanced redox equation, F is Faraday's constant (96,485 coulombs per mole of electrons), and E° is the standard cell potential you get from adding the oxidation and reduction half-reaction potentials.
The most important part of the equation is the negative sign. It flips the relationship so that the two favorability tests agree with each other. A positive E° produces a negative ΔG°, and both signal a thermodynamically favored reaction (that's your galvanic cell). A negative E° produces a positive ΔG°, meaning the reaction is unfavored and needs an outside push, like an applied voltage in electrolysis. In other words, voltage is just free energy per electron transferred, repackaged. One equation, two languages for the same idea.
This equation lives in Topic 9.9 (Cell Potential Under Nonstandard Conditions) in Unit 9: Thermodynamics and Electrochemistry, and it directly supports learning objective AP Chem 9.9.A, which asks you to explain whether an electrochemical cell is thermodynamically favored based on its standard cell potential. The essential knowledge (9.9.A.3) states that ΔG° is proportional to E°, and this equation is that proportionality written out. It's also the payoff of all of Unit 9. You spent the first half of the unit learning that negative ΔG° means favored, and now electrochemistry hands you a voltmeter that measures favorability directly. The equation is on the AP Chem reference sheet, so you don't memorize it, but you absolutely have to know what each variable means and what the signs tell you.
Keep studying AP® Chemistry Unit 9
E°red and Standard Cell Potential (Unit 9)
You can't use ΔG° = -nFE° without first finding E°. That means identifying the oxidation and reduction half-reactions, pulling their standard reduction potentials from a table, and combining them. E° feeds directly into this equation.
Gibbs Free Energy and Thermodynamic Favorability (Unit 9)
Earlier in Unit 9 you learned that ΔG° < 0 means a reaction is thermodynamically favored. This equation imports that entire framework into electrochemistry, so a battery's voltage is really a free-energy measurement in disguise.
Stoichiometric Coefficients (Unit 9)
The n in the equation comes from balancing the redox reaction. Doubling the coefficients doubles n and doubles ΔG° (free energy is extensive), but E° stays the same (potential is intensive). This distinction is a classic AP trap.
Nonstandard Conditions and the Nernst Equation (Unit 9)
ΔG° = -nFE° only applies at standard conditions (1 M, 1 atm, 25°C). When concentrations drift away from standard, Topic 9.9 extends the same logic to E and Q, explaining qualitatively why a battery's voltage drops as it runs down.
Multiple-choice questions usually test the sign logic rather than the arithmetic. A typical stem gives you a cell potential like +0.34 V and asks what that implies about ΔG° or about thermodynamic favorability. The correct chain is positive E° → negative ΔG° → favored (galvanic). You'll also see the reverse, where a ΔG° value like +85 kJ/mol tells you the reaction is unfavored and would have a negative cell potential. On FRQs, expect to calculate ΔG° from E° (watch your units, since F gives joules, not kilojoules) or to justify whether a cell is galvanic or electrolytic using the relationship. Always state the sign connection explicitly in your reasoning; "E° is positive, so ΔG° is negative, so the reaction is thermodynamically favored" is the full-credit sentence.
ΔG° depends on how much reaction happens, so if you double the balanced equation, n doubles and ΔG° doubles. E° is a per-electron quantity, like a price per item, so it does NOT change when you scale the equation. Multiplying a half-reaction's coefficients never multiplies its standard potential. The equation handles the scaling for you through n.
ΔG° = -nFE° connects standard Gibbs free energy to standard cell potential, where n is moles of electrons transferred and F is Faraday's constant, 96,485 C/mol e⁻.
A positive E° means a negative ΔG°, so the reaction is thermodynamically favored and the cell is galvanic.
A negative E° means a positive ΔG°, so the reaction is unfavored and needs an external applied potential, which is electrolysis.
Scaling the balanced equation changes n and ΔG° but never changes E°, because potential is an intensive property.
The equation gives ΔG° in joules, so divide by 1,000 if the answer choices or your comparison values are in kJ.
This relationship only holds at standard conditions; under nonstandard conditions, concentration changes shift the actual cell potential E away from E°.
It's the equation relating standard Gibbs free energy change to standard cell potential. n is the moles of electrons transferred in the balanced redox reaction, F is Faraday's constant (96,485 C/mol e⁻), and E° is the standard cell potential. It tells you whether a redox reaction is thermodynamically favored.
No, and this is the most common mistake. The negative sign in ΔG° = -nFE° means a positive E° gives a negative ΔG°. Both of those indicate a thermodynamically favored reaction, like the +0.34 V galvanic cell that shows up in practice questions.
They both indicate favorability, but ΔG° is extensive (it scales with the amount of reaction) while E° is intensive (it doesn't). Double the balanced equation and ΔG° doubles, but E° stays exactly the same.
No, it's printed on the AP Chemistry equations and constants sheet along with Faraday's constant. What you do need is to know what n, F, and E° mean, find n from the balanced redox equation, and interpret the signs correctly.
Balance the two half-reactions so the electrons cancel, then n is the number of moles of electrons that were transferred. For example, if Zn loses 2 electrons and Cu²⁺ gains 2, then n = 2. Getting n wrong is the most common calculation error with this equation.
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