AP Chemistry · Handsworth Secondary 2026–27

Unit 9 · Study Guide

Dr. Ras Mulinta
Thermodynamics & Electrochemistry
Exam-focused review

A one-page map of what the AP exam expects from Unit 9. We give this unit no separate test (only 5 blocks), so this guide is your safety net, it is fully fair game on the AP exam. This is a checklist, not a re-teach: if a line doesn't click, return to that section of the notes package. A periodic table, equation sheet (R, F, ΔG = −nFE, Nernst), and a table of standard reduction potentials are provided on the exam.

Must be able to do

The big idea that ties it together

One question, three lenses. "Which way does a change go, and how far?" is answered by ΔG° = ΔH° − TΔS°, which links to equilibrium through ΔG° = −RT ln K and to voltage through ΔG° = −nFE°. Favored ⇔ ΔG° < 0 ⇔ K > 1 ⇔ E°cell > 0, three faces of the same coin. Kinetics (Eₐ) is a separate axis: it sets the speed, never the destination.

Don't waste time on (excluded by the CED)

Labeling electrodes "+/−" (not assessed, identify by oxidation/reduction) · plug-and-chug algorithmic Nernst calculations (qualitative reasoning only) · applying Le Châtelier's principle to a running electrochemical cell (it isn't at equilibrium). Know the directions and the relationships, skip the rote arithmetic on Nernst.

Quick self-check (answer in your head, then verify)

  1. Sign of ΔS for CaCO₃(s) → CaO(s) + CO₂(g)?
  2. ΔG° at 298 K if ΔH° = −571.6 kJ and ΔS° = −326.4 J·K⁻¹? Favored?
  3. ΔG° at 298 K if K = 1.0 × 10⁴? (R = 8.314 J·mol⁻¹·K⁻¹.)
  4. E°cell for Zn | Zn²⁺ ‖ Ag⁺ | Ag, given E°(Ag⁺/Ag) = +0.80 V, E°(Zn²⁺/Zn) = −0.76 V?
  5. Using that cell (n = 2), what is ΔG°? (F = 96485 C·mol⁻¹.)
  6. Mass of Cu deposited by 2.00 A for 965 s? (Cu²⁺ + 2e⁻ → Cu, M = 63.55 g/mol.)
  7. As a galvanic cell discharges, which way does its voltage drift, and what is E at equilibrium?

Check yourself: 1) ΔS > 0 (a gas is produced from a solid)   2) ΔG° = −571.6 − (298)(−0.3264) = −474.3 kJ → favored   3) ΔG° = −(8.314)(298)(ln 10⁴) = −(2477.6)(9.210) = −22.8 kJ   4) E°cell = 0.80 − (−0.76) = +1.56 V   5) ΔG° = −(2)(96485)(1.56) = −3.01 × 10⁵ J = −301 kJ   6) q = 2.00 × 965 = 1930 C → 1930/96485 = 0.0200 mol e⁻ → ÷2 = 0.0100 mol Cu → ×63.55 = 0.636 g   7) E falls toward zero; at equilibrium (Q = K) E = 0 (a dead battery).

AP Chemistry · Unit 9 Study Guide · 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. Constants: R = 8.314 J·mol⁻¹·K⁻¹, F = 96485 C·mol⁻¹.