AP Chemistry · Handsworth Secondary 2026–27

Unit 9 · Thermodynamics & Electrochemistry

Dr. Ras Mulinta
Handsworth Secondary
Notes Package

CED Unit 9, 7–9% of the AP exam, but we have only 5 blocks (the College Board budgets 10–13), so this unit gets no separate unit test. It is still firmly on the AP exam, and last year it was under-covered, so we front-load it well. This unit ties up the redox you met in Unit 4 and connects entropy, free energy, equilibrium (Unit 7), and electrochemistry into one story: which way does a change go, and how far? BC link: extends the redox & cells of BC Chemistry 12 and the energy-change ideas of BC Chemistry 11.

Blocks:  1 → A–B (entropy)  ·  2 → C–D (Gibbs & control)  ·  3 → E–F (free energy ↔ K, coupling)  ·  4 → G–H (cells, E°cell, ΔG=−nFE)  ·  5 → I–J (Nernst qualitatively, electrolysis)

Name:Block:Date:

A · Introduction to Entropy CED 9.1 · LO 9.1.A

Entropy (S) measures how dispersed matter and energy are. Nature drifts toward more spread-out arrangements, more ways to be. Your first job is just to read the sign of ΔS by inspection.

Sign rules: ΔS > 0 when matter disperses, solid→liquid→gas, dissolving a solid, or more moles of gas on the product side. ΔS also rises with temperature (KMT: kinetic energy spreads over a broader distribution). Fewer gas moles or a gas condensing → ΔS < 0.
Worked example
Predict the sign of ΔS for:  2 SO₂(g) + O₂(g) → 2 SO₃(g).
Count gas moles: reactants 2 + 1 = 3 mol gas → products 2 mol gas. Gas moles drop (3 → 2).
ΔS < 0 (matter becomes less dispersed)

Practice A

  1. Predict the sign of ΔS for H₂O(l) → H₂O(g).
  2. Predict the sign of ΔS for N₂(g) + 3 H₂(g) → 2 NH₃(g).
  3. A gas is warmed from 25 °C to 200 °C at constant volume. Does its entropy rise or fall? Why (use KMT)?

B · Absolute Entropy & Entropy Change CED 9.2 · LO 9.2.A

Unlike enthalpy, every substance has an absolute molar entropy S° (in J·mol⁻¹·K⁻¹), even elements, whose S° is not zero. You combine these the same "products minus reactants" way.

Core: ΔS°rxn = ΣS°(products) − ΣS°(reactants), each term weighted by its coefficient. Watch the units, entropy is in J/K, while ΔH is usually in kJ.
Worked example
For N₂(g) + 3 H₂(g) → 2 NH₃(g), find ΔS°.  S°: N₂ = 191.6, H₂ = 130.7, NH₃ = 192.8 J·mol⁻¹·K⁻¹.
Products: 2 × 192.8 = 385.6  ·  Reactants: 191.6 + 3 × 130.7 = 191.6 + 392.1 = 583.7
ΔS° = 385.6 − 583.7 = −198.1 J·K⁻¹
ΔS° = −198.1 J·K⁻¹ (negative, as the gas-mole count predicted in A)

Practice B

  1. For 2 H₂(g) + O₂(g) → 2 H₂O(g), find ΔS°. S°: H₂ = 130.7, O₂ = 205.2, H₂O(g) = 188.8 J·mol⁻¹·K⁻¹.
  2. Your calculated ΔS° in Q1 is negative. Does that match the gas-mole sign rule from section A? Explain.
  3. Why is the absolute entropy of a perfect crystal at 0 K equal to zero, while at 298 K it is positive?

C · Gibbs Free Energy & Thermodynamic Favorability CED 9.3 · LO 9.3.A

Enthalpy and entropy compete. Gibbs free energy combines them into one number that tells you whether a process is thermodynamically favored. (We say "favored," not "spontaneous", spontaneous wrongly suggests "instant.")

Core: ΔG° = ΔH° − TΔS°. Favored when ΔG° < 0. Also ΔG°rxn = ΣΔG°f(products) − ΣΔG°f(reactants).
Sign table: ΔH° < 0, ΔS° > 0 → favored at all T. ΔH° > 0, ΔS° < 0 → favored at no T. ΔH° > 0, ΔS° > 0 → favored at high T. ΔH° < 0, ΔS° < 0 → favored at low T. (Same / opposite signs are read by inspection, no calculation needed.)
Worked example
A reaction has ΔH° = +178 kJ and ΔS° = +161 J·K⁻¹. Is it favored at 298 K? At what temperature does it become favored?
Convert ΔS° to kJ: 161 J·K⁻¹ = 0.161 kJ·K⁻¹. At 298 K: ΔG° = 178 − (298)(0.161) = 178 − 47.98 = +130 kJ → not favored.
Crossover where ΔG° = 0:  T = ΔH°/ΔS° = 178 / 0.161 = 1106 K.
Not favored at 298 K (ΔG° ≈ +130 kJ); becomes favored above ≈ 1.11 × 10³ K (high-T case).

Practice C

  1. ΔH° = −92.2 kJ, ΔS° = −198.1 J·K⁻¹. Calculate ΔG° at 298 K. Is it favored?
  2. Using only signs: ΔH° < 0 and ΔS° > 0. Favored at which temperatures? Why no calculation needed?
  3. Find the temperature at which ΔG° = 0 for ΔH° = +44.0 kJ, ΔS° = +118.8 J·K⁻¹.

D · Thermodynamic vs. Kinetic Control CED 9.4 · LO 9.4.A

"Favored" (ΔG° < 0) tells you a reaction can go, not that it will go fast. A reaction can be thermodynamically favored yet effectively frozen by a high activation energy.

Key distinction: thermodynamics (ΔG°) sets the destination; kinetics (Eₐ) sets the speed. A favored reaction stuck at an unmeasurable rate is under kinetic control: and a system that isn't reacting is not necessarily at equilibrium.
Worked example
Diamond → graphite has ΔG° < 0 at room conditions, yet diamonds last forever. Explain.
ΔG° < 0 means the conversion is thermodynamically favored, graphite is the lower-free-energy form.
But breaking the rigid carbon network has a huge activation energy, so the rate is essentially zero at 25 °C.
The reaction is favored but under kinetic control (high Eₐ); the unreacted diamond is not at equilibrium.

Practice D

  1. A gasoline–air mixture has ΔG° ≪ 0 for combustion but sits unburned in a tank. What controls it, and what removes the block?
  2. True or false: if a favored reaction shows no measurable change, the system must be at equilibrium. Explain.
  3. Does a catalyst change ΔG° for a reaction? Does it change the rate? Explain the difference.

E · Free Energy & Equilibrium CED 9.5 · LO 9.5.A

ΔG° is the bridge to the equilibrium constant K you learned in Unit 7. The sign of ΔG° tells you which side equilibrium favors; its size tells you how lopsided.

Standard state: ΔG° = −RT ln K, so K = e^(−ΔG°/RT). ΔG° < 0 → K > 1 (products favored); ΔG° > 0 → K < 1; ΔG° ≈ 0 → K ≈ 1.
Any state: ΔG = ΔG° + RT ln Q. Compare Q to K: the reaction runs toward equilibrium and stops where ΔG = 0 (Q = K). Use R = 8.314 J·mol⁻¹·K⁻¹.
Worked example
At 298 K, a reaction has ΔG° = −16.0 kJ·mol⁻¹. Find K.
ln K = −ΔG°/(RT) = −(−16000 J·mol⁻¹) / [(8.314)(298)] = 16000 / 2477.6 = 6.458
K = e^6.458 = 6.4 × 10²
K ≈ 6.4 × 10² (> 1, so products are favored, consistent with ΔG° < 0)

Not tested: equilibrium arguments like Le Châtelier's principle are not applied to operating electrochemical cells (section I), those systems are not at equilibrium.

Practice E

  1. At 298 K, K = 1.0 × 10⁻⁵. Find ΔG°. Is it positive or negative? (R = 8.314 J·mol⁻¹·K⁻¹.)
  2. Without calculating, what is K (roughly) if ΔG° = 0? Justify with the equation.
  3. A reaction has ΔG° = +25 kJ·mol⁻¹. Are reactants or products favored at equilibrium? Explain.

F · Free Energy of Dissolution & Coupled Reactions CED 9.6–9.7 · LO 9.6.A, 9.7.A

Two qualitative ideas. First, why some salts dissolve and others don't is a tug-of-war of enthalpy and entropy terms. Second, an unfavorable reaction can be driven by an outside energy source or by coupling to a favorable one.

Dissolution: ΔG° for dissolving reflects breaking lattice forces, reorganizing solvent, and solvating ions. The competing enthalpy/entropy pieces nearly cancel, so the overall sign is hard to predict, reason about it qualitatively, don't expect a clean number.
Coupling: an unfavorable step (ΔG° > 0) can be pushed by an external source (electricity in electrolysis; light in photosynthesis) or by adding a favorable step that shares an intermediate so the summed ΔG° < 0 (e.g. ATP → ADP driving biosynthesis).
Worked example
Step 1: A → B, ΔG° = +20 kJ (unfavorable). Step 2: B → C, ΔG° = −35 kJ. Can the overall A → C proceed?
B is the shared intermediate; add the steps: ΔG°total = (+20) + (−35) = −15 kJ.
Yes, coupled, A → C has ΔG° = −15 kJ < 0, so the favorable step drags the unfavorable one along.

Practice F

  1. Step 1 has ΔG° = +30 kJ; step 2 has ΔG° = −18 kJ. Is the coupled overall reaction favored? Show the sum.
  2. Name the external energy source that drives (a) an electrolytic cell and (b) photosynthesis.
  3. Why is predicting the overall sign of ΔG° for dissolving a salt harder than for a gas-phase reaction?

G · Galvanic (Voltaic) & Electrolytic Cells CED 9.8 · LO 9.8.A

An electrochemical cell separates a redox reaction into two half-cells so the electrons must travel through a wire, that flow is electricity. This is the Unit 4 redox you already balanced, now put to work.

Anatomy: oxidation at the anode, reduction at the cathode (mnemonic: An Ox, Red Cat). Electrons flow anode → cathode through the wire; the salt bridge carries ions to keep each half-cell neutral. Galvanic = favored reaction makes current (E°cell > 0); electrolytic = an external supply forces an unfavored reaction (E°cell < 0).
Cell notation: anode | anode soln ‖ cathode soln | cathodeoxidation on the left, reduction on the right, ‖ is the salt bridge. Example: Zn | Zn²⁺ ‖ Cu²⁺ | Cu.
Worked example
In a Zn/Cu galvanic cell, Zn is oxidized and Cu²⁺ is reduced. Write the half-reactions, identify anode/cathode, and give the cell notation.
Oxidation (anode): Zn → Zn²⁺ + 2e⁻  ·  Reduction (cathode): Cu²⁺ + 2e⁻ → Cu
Electrons leave the Zn anode, flow through the wire to the Cu cathode.
Cell notation: Zn | Zn²⁺ ‖ Cu²⁺ | Cu  (anode left, cathode right)

Not tested: labeling an electrode as "positive" or "negative" is not assessed (it flips between galvanic and electrolytic cells). Identify electrodes by oxidation/reduction instead.

Practice G

  1. In a cell, Fe → Fe²⁺ + 2e⁻ occurs in one half-cell and Ag⁺ + e⁻ → Ag in the other. Which is the anode? Write the cell notation.
  2. State the direction of electron flow (through the wire) in any galvanic cell, in terms of anode and cathode.
  3. How does an electrolytic cell differ from a galvanic cell in (a) sign of E°cell and (b) energy source?

H · Cell Potential & Free Energy CED 9.9 · LO 9.9.A

Each half-reaction has a standard reduction potential E°. Combine the two to get the cell's voltage E°cell, which is directly linked to ΔG°.

Core: E°cell = E°cathode − E°anode (both taken as reduction potentials; do NOT multiply E° by coefficients). A positive E°cell means a favored (galvanic) reaction.
Link to free energy: ΔG° = −nFE°, where n = moles of electrons transferred and F = 96485 C·mol⁻¹. E°cell > 0 ⇔ ΔG° < 0 ⇔ favored.
Worked example
For Zn | Zn²⁺ ‖ Cu²⁺ | Cu: E°(Cu²⁺/Cu) = +0.34 V, E°(Zn²⁺/Zn) = −0.76 V. Find E°cell, then ΔG°.
Cathode = Cu (reduction), anode = Zn (oxidation): E°cell = E°cathode − E°anode = 0.34 − (−0.76) = +1.10 V.
n = 2 electrons. ΔG° = −nFE° = −(2)(96485 C·mol⁻¹)(1.10 V) = −212267 J·mol⁻¹.
E°cell = +1.10 V (favored); ΔG° = −2.12 × 10⁵ J·mol⁻¹ = −212 kJ·mol⁻¹.

Practice H

  1. Given E°(Ag⁺/Ag) = +0.80 V and E°(Cu²⁺/Cu) = +0.34 V, find E°cell for a Cu/Ag galvanic cell. Which metal is the anode?
  2. A cell has E°cell = +0.46 V with n = 2. Calculate ΔG° (F = 96485 C·mol⁻¹).
  3. A proposed cell gives E°cell = −0.25 V. Is the reaction galvanic or electrolytic? What is the sign of ΔG°?

I · Cell Potential Under Nonstandard Conditions (Nernst, Qualitative) CED 9.10 · LO 9.10.A

Real cells rarely sit at 1 M / 1 atm. As a cell runs, concentrations change and the voltage drifts, toward zero, reaching exactly zero at equilibrium (a dead battery). Reason about this with the Nernst equation qualitatively.

Nernst (qualitative): E = E° − (RT/nF) ln Q. At standard conditions Q = 1 and E = E°. As the reaction proceeds, Q rises toward K and E falls; at equilibrium Q = K and E = 0. Pushing the cell farther from equilibrium than Q = 1 raises |E|; moving it closer lowers |E|.
Worked example
A Zn/Cu cell (E° = +1.10 V) runs until [Cu²⁺] has dropped well below [Zn²⁺]. Does E rise or fall from +1.10 V? Reason with Nernst, no number needed.
Q = [Zn²⁺]/[Cu²⁺]. Lower [Cu²⁺] and higher [Zn²⁺] make Q > 1, so ln Q > 0.
In E = E° − (RT/nF) ln Q, subtracting a positive term lowers E.
E falls below +1.10 V, the cell drifts toward equilibrium (E → 0), consistent with a battery running down.

Not tested: plug-and-chug algorithmic Nernst calculations are insufficientand Le Châtelier reasoning does not apply to a running cell (it is not at equilibrium). You need the qualitative direction of the voltage change.

Practice I

  1. For a cell at Q = 1, what is E in terms of E°? Justify from the Nernst equation.
  2. As any galvanic cell discharges, Q moves toward K. What happens to E, and what is E at equilibrium?
  3. A concentration cell has the same metal/ion in both half-cells but different concentrations. Is E° zero? Why can E still be nonzero?

J · Electrolysis & Faraday's Law CED 9.11 · LO 9.11.A

In electrolysis we push current through a cell to force a reaction (e.g. electroplating). Faraday's law is a stoichiometry chain: charge → moles of electrons → moles of substance → mass.

The chain: q = I·t (charge in C) → mol e⁻ = q / F (F = 96485 C·mol⁻¹) → mol substance via the half-reaction's electron ratio → mass via molar mass. Also I = q/t.
Worked example
A current of 2.00 A flows for 30.0 min through molten/aqueous Cu²⁺. What mass of copper plates out?  Cu²⁺ + 2e⁻ → Cu, M(Cu) = 63.55 g/mol.
t = 30.0 min × 60 = 1800 s.  q = I·t = 2.00 × 1800 = 3600 C.
mol e⁻ = 3600 / 96485 = 0.03731 mol.  mol Cu = 0.03731 / 2 = 0.018655 mol (2 e⁻ per Cu).
mass Cu = 0.018655 × 63.55 = 1.19 g

Practice J

  1. How long (s) must 1.50 A flow to deposit 0.500 g of silver? Ag⁺ + e⁻ → Ag, M = 107.87 g/mol.
  2. A current of 3.00 A runs for 1.00 h. How many moles of electrons pass? (F = 96485 C·mol⁻¹.)
  3. Why does plating one mole of Al from Al³⁺ require more charge than plating one mole of Ag from Ag⁺?
AP Chemistry · Unit 9, Thermodynamics & Electrochemistry · Dr. Ras Mulinta · Handsworth Secondary 2026–27. Pegged to the College Board AP Chemistry CED (topics 9.1–9.11) and BC Chemistry 11/12. This unit is not given a separate unit test but is on the AP exam, front-loaded here. Constants: R = 8.314 J·mol⁻¹·K⁻¹, F = 96485 C·mol⁻¹. Atomic masses from the IUPAC periodic table.