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AP Chemistry · Unit 9 of 9
Applications of Thermodynamics
7–9% of the exam6 lessons · 80 min13 terms
What this unit covers
The topics below follow the published Chemistry course framework for Unit 9. This unit is worth 7–9% of the exam, so budget your time against that rather than against how long the unit takes to teach.
ElectrochemistryGalvanic cellsElectrolysisNernst equation
Lessons in this unit
- Free Energy, Spontaneity & K13 min · 3 objectivesPredict the temperature dependence of spontaneity from the signs of ΔH and ΔS · Relate the sign of ΔG° to whether a reaction is thermodynamically favored · Connect ΔG° to the equilibrium constant K through ΔG° = −RT ln K
- Galvanic (Voltaic) Cells & Cell Potential14 min · 3 objectivesBalance redox reactions by combining oxidation and reduction half-reactions · Identify the anode, cathode, and role of the salt bridge in a galvanic cell · Compute E°cell from standard reduction potentials and link its sign to ΔG°
- Electrolysis12 min · 3 objectivesContrast electrolytic and galvanic cells in terms of ΔG and applied voltage · Apply Faraday's laws to relate charge, moles of electrons, and mass deposited · Calculate the mass or amount of product formed during an electrolysis
- The Nernst Equation & Nonstandard Cells12 min · 3 objectivesExplain qualitatively how concentration shifts a cell potential away from E° · Use the Nernst equation to calculate E under nonstandard conditions · Predict the direction a cell voltage moves as reactant and product concentrations change
- Quantitative Electrochemistry: ΔG°, E°, and K15 min · 3 objectivesLink E°, ΔG°, and K through the pair of relationships ΔG° = −nFE° and ΔG° = −RT ln K · Calculate both ΔG° and the equilibrium constant K from a single standard cell potential · Apply Faraday's laws to find the mass deposited or the time required in an electrolysis
- The Nernst Equation & Concentration Cells14 min · 3 objectivesUse the Nernst equation to calculate a cell potential under nonstandard concentrations · Predict the direction a cell potential shifts as reactant and product concentrations change · Analyze concentration cells, where E° = 0 and the voltage arises purely from a concentration gradient
Formulas in Unit 9
Gibbs free energy
ΔG = ΔH − TΔS
ΔG < 0 favored, ΔG > 0 not favored, ΔG = 0 at equilibrium. T is in kelvin, so the TΔS term always grows with temperature.
Free energy and the equilibrium constant
ΔG° = −RT ln K
R = 8.314 J·mol⁻¹·K⁻¹, T in kelvin. K > 1 ⇒ ln K > 0 ⇒ ΔG° < 0 (products favored). K < 1 ⇒ ΔG° > 0 (reactants favored).
Standard cell potential
E°cell = E°cathode − E°anode
Both values are standard *reduction* potentials read straight from the table. Do NOT multiply a potential by the number of electrons or by a balancing coefficient — potential is an intensive property.
Free energy from cell potential
ΔG° = −nFE°
n = moles of electrons transferred, F = 96 485 C·mol⁻¹ (Faraday constant). E° > 0 ⇒ ΔG° < 0 ⇒ spontaneous; E° < 0 ⇒ ΔG° > 0 ⇒ nonspontaneous.
Faraday's laws of electrolysis
moles e⁻ = Q ÷ F = (I·t) ÷ 96 485
Then scale by the half-reaction: e.g. Ag⁺ + e⁻ → Ag needs 1 mol e⁻ per mol Ag, while Al³⁺ + 3e⁻ → Al needs 3 mol e⁻ per mol Al.
The Nernst equation (at 298 K)
E = E° − (0.0592 / n) log Q
General form: E = E° − (RT/nF) ln Q. At 25 °C the constants collapse to 0.0592/n with base-10 log. Q < 1 raises E above E°; Q > 1 lowers it.
Free energy from the equilibrium constant
ΔG° = −RT ln K
R = 8.314 J·mol⁻¹·K⁻¹, T in kelvin. K > 1 ⇒ ΔG° < 0 (products favored); K < 1 ⇒ ΔG° > 0 (reactants favored).
Voltage and equilibrium (298 K shortcut)
log K = nE° / 0.0592
Derived by equating ΔG° = −nFE° with ΔG° = −RT ln K at 25 °C. A modest E° of a few tenths of a volt produces an enormous K — voltage and equilibrium are exponentially related.
The Nernst equation (298 K)
E = E° − (0.0592 / n) log Q
General form E = E° − (RT/nF) ln Q; at 25 °C the constants collapse to 0.0592/n with base-10 log. Q < 1 raises E above E°; Q > 1 lowers it; Q = 1 recovers E = E°.
Concentration cell (identical electrodes, 298 K)
E = −(0.0592 / n) log Q, with E° = 0
Q = [ion]dilute ÷ [ion]concentrated < 1, so log Q < 0 and E > 0. The concentrated half-cell is always the cathode; electrons flow to shrink the gradient.
Every term in Unit 9
All 13 terms we publish for Applications of Thermodynamics, with definitions. Reading them through is the fastest way to find the ones you cannot define — then drill those in cram mode until you can produce them without the prompt.
- Galvanic (voltaic) cell
- Spontaneous redox generating current. Oxidation at the anode, reduction at the cathode; electrons flow anode to cathode.
- Electrolytic cell
- Non-spontaneous redox driven by an external supply. E°cell is negative and ΔG is positive.
- Standard cell potential
- E°cell = E°cathode − E°anode. A positive value means the reaction is spontaneous as written.
- ΔG and cell potential
- ΔG° = −nFE°, where n is moles of electrons and F is 96,485 C/mol. Positive E° corresponds to negative ΔG.
- Nernst equation qualitatively
- Cell potential falls as reactant concentrations drop and products build; at equilibrium E = 0 and the battery is dead.
- Salt bridge
- Carries ions to keep both half-cells electrically neutral. Without it charge builds and current stops almost immediately.
- Faraday's law of electrolysis
- Moles of substance deposited = (current × time)/(n × 96,485). Charge in coulombs converts to moles of electrons.
- Standard reduction potentials
- Tabulated for reduction half-reactions. Reversing one flips the sign; scaling a half-reaction does NOT change the potential, since it is intensive.
- Corrosion as electrochemistry
- Iron oxidizes in the presence of water and oxygen. Sacrificial anodes of a more easily oxidized metal protect it by corroding first.
- Direction of ion flow in a salt bridge
- Anions migrate toward the anode and cations toward the cathode, offsetting the charge that oxidation and reduction would otherwise build up.
- Why E° is intensive
- Potential is energy per unit charge, so doubling a half-reaction doubles both and leaves the ratio unchanged. ΔG, being extensive, does double.
- Concentration cell
- Two identical electrodes in solutions of different concentration. E° is zero, but a potential exists because the system moves toward equal concentrations.
- Predicting spontaneity from a table
- The half-reaction higher in a standard reduction table proceeds as reduction; the lower one reverses and is oxidized.
What examiners penalize here
- Do not confuse ΔG° with ΔG. ΔG° = −RT ln K compares the *standard state* to equilibrium; a large positive ΔG° just means a small K. The actual ΔG (which is 0 at equilibrium) depends on the real concentrations through ΔG = ΔG° + RT ln Q.
- The AP exam frequently tests the "do not multiply potentials" trap. When you scale a half-reaction to balance electrons, you multiply the atoms and the electrons — but never the E° value. Potential is energy *per charge*, an intensive property that is independent of how much reaction you run.
- The classic Faraday problem gives current and time and asks for mass. Chain the units without skipping steps: A × s = C, then C ÷ 96 485 = mol e⁻, then ÷ electrons-per-ion = mol product, then × molar mass = grams. Carrying units through each arrow prevents the most common errors.
- A concentration cell is the purest test of the Nernst idea: identical electrodes, same E° of 0, so E depends entirely on the concentration difference through −(0.0592/n) log Q. Electrons flow to equalize the two sides, and the voltage dies the instant the concentrations match.
- Memorize the triangle: E° ↔ ΔG° ↔ K, joined by ΔG° = −nFE° and ΔG° = −RT ln K. Given any one vertex you can reach the other two. The single most common error is scaling E° by n or by a coefficient — E° is intensive and never changes; the n lives in the −nFE° expression, not inside the voltage.
- On the AP exam a concentration cell is instantly recognizable: same metal on both sides, same ion, E° = 0. Do not panic that the voltage looks like it should be zero — plug into E = −(0.0592/n) log Q and let the concentration ratio do the work. The concentrated half-cell is always the cathode.
Practice Chemistry
Our practice bank is drawn from across the whole course rather than filtered to one unit, which is closer to how the exam asks anyway — it will not tell you which unit a question is testing.
Questions about this unit
How much of the AP Chemistry exam is Unit 9?
Unit 9, Applications of Thermodynamics, is worth 7–9% of the Chemistry multiple-choice section according to the published course framework. Across all 9 units that makes it a middling share, roughly what an even split across units would give.
What topics are covered in Chemistry Unit 9?
Applications of Thermodynamics covers Electrochemistry, Galvanic cells, Electrolysis and Nernst equation. We publish 13 terms with definitions for this unit, all of them on this page.
How should I study Chemistry Unit 9?
Read the 6 lessons below first — about 80 minutes — then drill the 13 terms in cram mode until you can produce each definition from memory rather than just recognize it. Recognition is what makes a unit feel finished when it is not. Finish with practice questions and read the explanation for every one you get right by elimination as well as the ones you miss.
All 9 units of AP Chemistry
Unit names, topics and exam weights follow the published College Board course framework for AP Chemistry. AP® is a trademark registered by the College Board, which does not endorse this site.