Unit 9 combines two big concepts: thermodynamics (whether reactions are favored) and electrochemistry (reactions involving electron transfer). You'll learn about entropy and Gibbs free energy, which predict whether reactions will proceed. Then you'll apply these ideas to electrochemical cells where electrons flow and electrical energy is generated or consumed. These concepts connect everything you've learned: equilibrium, reaction rates, acid-base chemistry, and energy.
🎯 What You Need to Know for the Exam
Unit 9 makes up about 7-9% of the AP Chemistry exam. Focus your energy on these priorities:
- Entropy and energy dispersal: how entropy increases with disorder and temperature, calculating entropy change
- Gibbs free energy: predicting whether reactions are thermodynamically favored using ΔG, connecting ΔG to K and E
- Thermodynamic vs. kinetic control: why a favored reaction might not happen (activation energy), and vice versa
- Electrochemical cells: galvanic cells (spontaneous, generate electricity) vs. electrolytic cells (non-spontaneous, require electricity)
- Cell potential and electron flow: relating cell potential to spontaneity, calculating standard cell potential from half-reactions
- Faraday's law: relating charge, moles of electrons, and amounts of reactants/products in electrochemical cells
What's in this review:
- Introduction to Entropy
- Absolute Entropy and Entropy Change
- Gibbs Free Energy and Thermodynamic Favorability
- Thermodynamic and Kinetic Control
- Free Energy and Equilibrium
- Free Energy of Dissolution
- Coupled Reactions
- Galvanic and Electrolytic Cells
- Cell Potential and Free Energy
- Cell Potential Under Nonstandard Conditions
- Electrolysis and Faraday's Law
- Study Tips for Unit 9
- Summary, Review Questions & Practice
Topic 9.1: Introduction to Entropy
Entropy measures the dispersal or disorder in a system. It's a fundamental property that tells you about the number of ways particles can be arranged or the distribution of energy among them. Entropy increases when matter becomes more dispersed. For example, when a solid melts to a liquid, the particles can move more freely and occupy more space, so entropy increases. When a liquid vaporizes to a gas, the particles spread out even more and entropy increases further.
Entropy also increases with temperature. At higher temperatures, the kinetic energy of particles is more broadly distributed, increasing the number of ways energy can be distributed among the particles. Additionally, when gases are produced from liquids or solids in a reaction, or when the number of moles of gas increases, entropy typically increases.
You don't need to calculate absolute entropy values from first principles. You'll use tabulated entropy values (absolute entropies) to calculate entropy changes for reactions. The key concept is understanding what causes entropy to increase or decrease.
Key concepts to know:
- Entropy increases: When matter disperses (phase changes), when temperature increases, when gases form from liquids/solids, when the number of moles of gas increases.
- Entropy decreases: When matter becomes more ordered, when temperature decreases, when gases condense to liquids, when the number of moles of gas decreases.
- Disorder and energy dispersal: Both manifest as entropy increases.
⚠ Watch out for:
Entropy is not the same as disorder in the everyday sense. Entropy is a thermodynamic property, not just a visual measure of messiness. Focus on dispersal of matter and energy. Also, entropy change depends only on the initial and final states, not on the path taken.
Topic 9.2: Absolute Entropy and Entropy Change
Unlike enthalpy, we can define absolute entropy. Absolute entropy is the entropy relative to a perfect crystal at 0 K, where entropy = 0. Absolute entropy values are tabulated for many substances. You use these values to calculate entropy change for a reaction using:
ΔS°reaction = Σ(S° products) - Σ(S° reactants)
Just like with enthalpy, you multiply each entropy by the stoichiometric coefficient. Positive ΔS means entropy increases (disorder increases). Negative ΔS means entropy decreases (disorder decreases).
Key concepts to know:
- Absolute entropy: Measured relative to a perfect crystal at 0 K. Always positive (or zero at 0 K).
- Entropy change: ΔS°reaction = Σ(S° products) - Σ(S° reactants)
- Stoichiometry: Multiply each absolute entropy by the coefficient in the balanced equation.
- Sign interpretation: Positive ΔS = increase in disorder. Negative ΔS = decrease in disorder.
⚠ Watch out for:
Don't forget the stoichiometric coefficients. If a substance has a coefficient of 2, multiply its entropy by 2. Also, entropy is in units of J/(mol·K), not kJ/(mol·K). Watch your units.
Topic 9.3: Gibbs Free Energy and Thermodynamic Favorability
Gibbs free energy (G) combines enthalpy and entropy to predict whether a reaction is thermodynamically favored. The relationship is:
ΔG° = ΔH° - TΔS°
If ΔG° < 0, the reaction is thermodynamically favored. Products are favored at equilibrium. If ΔG° > 0, the reaction is not favored. Reactants are favored at equilibrium. If ΔG° = 0, the system is at equilibrium.
Key insight: Some reactions are favored by enthalpy (exothermic, ΔH° < 0) and entropy (ΔS° > 0). These are always favored. Some reactions are disfavored by both. These are never favored. Some reactions are favored by one but not the other. Temperature determines the outcome. For example, at low temperature, a small favorable entropy term is overcome by an unfavorable enthalpy term. At high temperature, entropy dominates.
The standard Gibbs free energy of formation (ΔGf°) can be used to calculate ΔG° for a reaction:
ΔG°reaction = Σ(ΔGf° products) - Σ(ΔGf° reactants)
Key concepts to know:
- ΔG equation: ΔG° = ΔH° - TΔS°
- Thermodynamically favored: ΔG° < 0
- Not favored: ΔG° > 0
- Equilibrium: ΔG° = 0
- Temperature dependence: At different temperatures, the same reaction may be favored or not, depending on relative magnitudes of ΔH° and TΔS°
- ΔGf° method: Calculate ΔG°reaction from formation values
⚠ Watch out for:
Temperature is in Kelvin, not Celsius. Watch units on entropy (J vs. kJ). Also, "thermodynamically favored" means ΔG° < 0, but it doesn't mean the reaction happens quickly. That's kinetics. Thermodynamics predicts where equilibrium is, not how fast you get there.
Topic 9.4: Thermodynamic and Kinetic Control
Just because ΔG° < 0 doesn't mean a reaction happens immediately. The reaction might have a very high activation energy. The reaction is thermodynamically favored (ΔG° < 0) but under kinetic control because the activation energy is too high for the reaction to proceed at a measurable rate.
For example, the conversion of diamond to graphite is thermodynamically favored at room temperature, but diamonds don't spontaneously turn into graphite because the activation energy is too high. This is kinetic control.
It's important to distinguish: if ΔG° > 0, the reaction is not thermodynamically favored and will not proceed to a significant extent regardless of the activation energy. A low activation energy just means the system reaches its (unfavorable) equilibrium faster. Kinetics determines how quickly equilibrium is reached, but thermodynamics determines where that equilibrium lies.
Key concepts to know:
- Kinetic control: High activation energy prevents a thermodynamically favored reaction from proceeding at measurable rate.
- Thermodynamic control: Low activation energy allows a thermodynamically favored reaction to proceed readily.
- Kinetics vs. thermodynamics: A negative ΔG° with high activation energy means the reaction is favored but proceeds too slowly to observe. A positive ΔG° means the reaction is not favored at equilibrium regardless of activation energy.
⚠ Watch out for:
Just because a reaction is thermodynamically favored doesn't mean it will happen. Check the activation energy. This is why catalysts are important: they lower activation energy without changing ΔG°. This allows thermodynamically favored reactions to proceed at useful rates.
Topic 9.5: Free Energy and Equilibrium
At equilibrium, ΔG = 0. Before equilibrium, ΔG < 0 if the system is shifting toward products, and ΔG > 0 if the system is shifting toward reactants.
The relationship between ΔG° and K is:
ΔG° = -RT ln K
This is powerful: it relates the standard free energy change to the equilibrium constant. If ΔG° < 0, then K > 1 (products favored). If ΔG° > 0, then K < 1 (reactants favored). If ΔG° = 0, then K = 1.
For a reaction with K >> 1, ΔG° is very negative. The equilibrium lies far toward products. For a reaction with K << 1, ΔG° is very positive. The equilibrium lies far toward reactants. For K ≈ 1, ΔG° ≈ 0. The equilibrium shows comparable amounts of reactants and products.
Key concepts to know:
- ΔG° = -RT ln K: Connects free energy to equilibrium constant
- K > 1: ΔG° < 0. Products favored.
- K < 1: ΔG° > 0. Reactants favored.
- K = 1: ΔG° = 0. Equilibrium.
- Large K: Very negative ΔG°. Reaction essentially complete.
- Small K: Very positive ΔG°. Reaction barely proceeds.
⚠ Watch out for:
This equation is different from ΔG = ΔG° + RT ln Q. The one here (ΔG° = -RT ln K) relates standard values. The other is for non-standard conditions. Know both and when to use each.
Topic 9.6: Free Energy of Dissolution
When a salt dissolves, ΔG° for dissolution reflects several competing factors: breaking intermolecular forces in the solid, reorganizing the solvent, and intermolecular interactions between dissolved ions and solvent. These can partially cancel each other out, making it difficult to predict whether a dissolution is favored without experimental data.
Some salts are very soluble because dissolution is thermodynamically favored (ΔG° < 0). Others have very low solubility because dissolution is not favored (ΔG° > 0). And some show complex behavior where solubility increases or decreases with temperature because ΔH° and TΔS° have different signs and magnitudes.
Key concepts to know:
- Dissolution factors: Breaking solid interactions (ΔH > 0), solvent reorganization (ΔH variable), hydration interactions (ΔH < 0).
- Net ΔG° determines solubility: Favorable (ΔG° < 0) means soluble; unfavorable (ΔG° > 0) means insoluble.
- Temperature dependence: Some salts are more soluble at higher T (entropy-driven); others are less soluble (heat-driven).
⚠ Watch out for:
You can estimate the sign and magnitude of ΔH° and ΔS° for dissolution, but the exam expects qualitative reasoning rather than quantitative calculations. Understand the competing factors.
Topic 9.7: Coupled Reactions
A coupled reaction pairs an unfavorable (ΔG° > 0) reaction with a favorable one (ΔG° < 0) so that the overall ΔG° < 0. External energy sources can drive unfavorable processes. In biology, ATP hydrolysis often drives unfavorable reactions. In electrochemistry, electrical energy drives unfavorable reactions in electrolytic cells.
When two reactions are coupled, their free energies add. If Reaction 1 has ΔG° = +200 kJ and Reaction 2 has ΔG° = -250 kJ, the overall process has ΔG° = -50 kJ and is favored overall.
Key concepts to know:
- Coupled reactions: Pair unfavorable with favorable reactions so overall ΔG° < 0.
- Free energies add: ΔG°overall = ΔG°1 + ΔG°2
- Common coupling agents: ATP hydrolysis, electrical energy, light energy.
⚠ Watch out for:
The reactions must be coupled through a shared intermediate. Just adding ΔG° values for independent reactions doesn't make sense. The reactions must share products and reactants or a common substrate.
Topic 9.8: Galvanic and Electrolytic Cells
An electrochemical cell is a system where redox reactions occur. In a galvanic (or voltaic) cell, the redox reaction is thermodynamically favored (ΔG° < 0), so it occurs spontaneously and produces electrical energy. In an electrolytic cell, the redox reaction is not favored (ΔG° > 0), so electrical energy must be applied to make it happen.
Both types of cells have two half-cells connected by a salt bridge. The anode is where oxidation occurs. The cathode is where reduction occurs. This is true for both galvanic and electrolytic cells. The difference is which electrode is which in terms of the overall spontaneity.
In a galvanic cell, electrons are produced at the anode and consumed at the cathode. Electrons flow through an external wire from anode to cathode. Cations flow through the salt bridge from anode to cathode. At the cathode, reduction occurs. Ions are produced and released. At the anode, oxidation occurs. Ions are consumed. The circuit is completed by ion flow through the salt bridge.
Key concepts to know:
- Galvanic cell: ΔG° < 0. Reaction is spontaneous. Produces voltage. Electrons flow anode to cathode externally.
- Electrolytic cell: ΔG° > 0. Reaction is non-spontaneous. Requires applied voltage. Electrons forced by external source.
- Anode: Oxidation occurs in both cell types.
- Cathode: Reduction occurs in both cell types.
- Salt bridge role: Allows ion flow to complete the circuit and maintain electroneutrality.
⚠ Watch out for:
The anode is where oxidation occurs, and the cathode is where reduction occurs, in both galvanic and electrolytic cells. But the polarity (positive or negative) varies between cell types. Don't confuse the definition with the polarity. The CED says not to label as positive or negative, so focus on oxidation and reduction locations.
Topic 9.9: Cell Potential and Free Energy
The cell potential (E° for standard conditions) is the voltage produced by an electrochemical cell. It's directly related to spontaneity: a positive cell potential indicates a spontaneous (thermodynamically favored) reaction. A negative cell potential indicates a non-spontaneous reaction.
The relationship is: ΔG° = -nFE°
Where n is the number of electrons transferred, F is Faraday's constant (96,485 C/mol), and E° is the standard cell potential.
You calculate the cell potential by identifying the two half-reactions and finding their standard reduction potentials (E°red). The cell potential is:
E°cell = E°cathode - E°anode
Or equivalently:
E°cell = E°reduction - E°oxidation
A positive E°cell means the reaction is spontaneous. The larger the E°cell, the more favorable the reaction.
Key concepts to know:
- Cell potential (E°): Voltage of the cell. Positive = spontaneous, Negative = non-spontaneous.
- E°cell = E°cathode - E°anode: Reduction potential of cathode minus oxidation potential of anode.
- ΔG° = -nFE°: Relates cell potential to free energy.
- Positive E°cell: ΔG° < 0. Reaction is favored.
- Negative E°cell: ΔG° > 0. Reaction is not favored.
⚠ Watch out for:
Standard reduction potentials in tables are always written as reductions. To calculate E°cell, use E°cathode - E°anode directly from the table values. You do not need to reverse the sign of a half-reaction's E° when it acts as the anode. The subtraction formula handles that automatically. Also, E° values don't change when you multiply stoichiometric coefficients, unlike ΔG°.
Topic 9.10: Cell Potential Under Nonstandard Conditions
Under nonstandard conditions (concentrations other than 1 M, for example), the cell potential differs from E°. The relationship is given by the Nernst equation:
E = E° - (RT/nF) ln Q
This shows that the cell potential depends on Q (reaction quotient). When Q = 1 (standard conditions), the log term is zero and E = E°. When Q < 1, the log is negative and E > E°, meaning the cell drives the reaction forward. When Q > 1, the log is positive and E < E°, meaning the cell is closer to equilibrium.
As the cell operates and the reaction proceeds, Q increases and E decreases. At equilibrium, Q = K and E = 0. At this point, no more electrical energy can be extracted from the cell.
The direction of spontaneous electron flow in a concentration cell can be determined by comparing concentrations. Electrons flow to dilute the more concentrated solution and concentrate the more dilute solution, seeking equilibrium.
Key concepts to know:
- Nernst equation: E = E° - (RT/nF) ln Q
- Q < 1: E > E°. Reaction driven forward.
- Q = 1: E = E°. Standard conditions.
- Q > 1: E < E°. Closer to equilibrium.
- Q = K: E = 0. At equilibrium.
- Concentration cells: Electrons flow to equalize concentration.
⚠ Watch out for:
The CED says algorithmic calculations using the Nernst equation are not assessed. Focus on qualitative understanding. Know that E depends on Q, that E decreases as Q increases, and that E = 0 at equilibrium. You don't need to plug into the equation for the exam. Also note: Le Chatelier's principle should not be used to predict changes in cell potential or voltage of electrochemical cells. Use the Nernst equation relationship (qualitatively) between Q and E instead.
Topic 9.11: Electrolysis and Faraday's Law
Faraday's law relates the amount of charge flowing through an electrochemical cell to the amount of substance reacted. It's based on the idea that each electron has a fixed charge.
The key relationships are:
Charge (C) = Current (A) × Time (s)
Where 1 Coulomb = 1 Ampere × 1 second.
Then, using Faraday's constant (96,485 C/mol), you can convert charge to moles of electrons:
Moles of electrons = Charge / Faraday's constant
Finally, using stoichiometry, you can find moles of substance reacted or produced:
Moles of substance = (Moles of electrons / n) × (molar stoichiometry)
Where n is the number of electrons involved in the half-reaction.
Common applications include calculating the mass of material deposited on an electrode, the mass of material dissolved, or the time needed for a specific amount of reaction.
Key concepts to know:
- Charge = Current × Time: Q = I × t
- Moles of electrons = Q / F: F = 96,485 C/mol
- Stoichiometry: Use the number of electrons to relate moles of electrons to moles of substance.
- Electroplating: Applies these relationships to determine mass deposited on an electrode.
⚠ Watch out for:
Make sure you use the correct number of electrons from the half-reaction. If Cu^2+ + 2e- → Cu, then 2 moles of electrons produce 1 mole of Cu. The stoichiometry is 2:1. Also, watch your units: charge in Coulombs, current in Amperes, time in seconds.
Study Tips for Unit 9
Unit 9 is conceptual and numerical. Here are strategies that work:
Understand entropy first. Entropy is the foundation of everything in Unit 9. It's not about "disorder" in everyday language. It's about matter dispersal and energy distribution. Once you understand entropy, Gibbs free energy and cell potential make sense.
Master the ΔG equation. Know what the signs mean. Know how temperature affects outcomes. Know that ΔG° < 0 means favored. This single equation explains so much in Unit 9.
Separate thermodynamic favorability from kinetics. A reaction can be thermodynamically favored (ΔG° < 0) but kinetically hindered (high activation energy). This distinction comes up often.
Relate everything to equilibrium. ΔG° = -RT ln K. This connects Unit 7 (equilibrium) to Unit 9. And cell potential at equilibrium is zero. Always link back to K and equilibrium.
Practice electrochemistry systematically. Identify the anode (oxidation) and cathode (reduction). Write half-reactions. Find E° values. Calculate E°cell. Relate to ΔG°. Predict spontaneity. Do this for 20 different examples until it's automatic.
StarSpark Practice Prompts:
- "Give me a reaction with ΔH° and ΔS°, ask me to predict favorability at different temperatures using ΔG equation"
- "Describe an electrochemical cell with two half-reactions, ask me to identify anode and cathode, calculate E°cell, and predict spontaneity"
- "Give me an electrolysis problem with current and time, ask me to calculate mass of substance deposited using Faraday's law"
Summary, Review Questions & Practice
You've covered all the topics in Unit 9. Before you move on, test yourself with these scenario-based questions. If you can answer them confidently, you're in great shape for this section of the exam.
Review Questions: Test Yourself
- For the reaction 2H2(g) + O2(g) → 2H2O(l), given ΔH° = -572 kJ and ΔS° = -327 J/K, calculate ΔG° at 25°C. Is this reaction thermodynamically favored? Explain what would happen to ΔG° if you increased the temperature.
- Identify the anode and cathode in a galvanic cell where Zn(s) + Cu^2+(aq) → Zn^2+(aq) + Cu(s). Write the half-reactions. Predict the direction of electron flow and ion flow through the salt bridge.
- Calculate the standard cell potential for the Zn/Cu galvanic cell above (E°Zn^2+/Zn = -0.76 V, E°Cu^2+/Cu = +0.34 V). Is the reaction spontaneous? Use ΔG° = -nFE° to calculate ΔG° (assume 2 electrons transferred).
- In an electrolytic cell, 5.0 A of current is passed for 2.0 hours to electroplate copper. How many grams of copper are deposited? (Cu^2+ + 2e- → Cu; 1 F = 96,485 C/mol)
- For a weak acid with Ka = 1.0 × 10^-5, calculate ΔG° for the ionization reaction at 25°C using ΔG° = -RT ln K. Is ionization favored?
Want more practice? Paste these questions into StarSpark to generate a full quiz with explanations.
Explore the Full AP Chemistry Study Guide
Unit 9 wraps up the core concepts of AP Chemistry. You've moved from atomic structure and bonding, through chemical reactions and equilibrium, to energy and electrochemistry. These three units (7, 8, and 9) test your ability to integrate ideas: equilibrium applies to acids, buffers, and solubility. Thermodynamics predicts whether reactions happen and where equilibrium lies. Electrochemistry shows reactions in action, where electrons flow and electrical energy is generated.
Check out the full AP Chemistry study plan to see how this unit connects to the rest of the course.
Other Unit Reviews:
- AP Chemistry Unit 1: Atomic Structure and Properties
- AP Chemistry Unit 2: Compound Structure and Properties
- AP Chemistry Unit 3: Properties of Substances and Mixtures
- AP Chemistry Unit 4: Chemical Reactions
- AP Chemistry Unit 5: Kinetics
- AP Chemistry Unit 6: Thermochemistry
- AP Chemistry Unit 7: Equilibrium
- AP Chemistry Unit 8: Acids and Bases
For official AP Chemistry resources, visit apcentral.collegeboard.org.
This review is aligned with the AP Chemistry Course and Exam Description. AP is a registered trademark of the College Board, which was not involved in the production of this guide.
