What is AP Chem unit 9?
Unit 9 asks you to predict and explain energy changes in chemical systems using two connected frameworks: thermodynamics and electrochemistry. You will calculate entropy changes, evaluate thermodynamic favorability using Gibbs free energy, connect ΔG° to equilibrium constants, and then apply all of that reasoning to galvanic and electrolytic cells.
Unit 9 is about predicting whether chemical processes are thermodynamically favored (ΔG° < 0), how that connects to equilibrium constants and cell potentials, and how to use Faraday's law to relate charge, current, and mass in electrochemical cells.
Entropy and Gibbs Free Energy
Entropy (S) measures dispersal of matter and energy. The sign of ΔS° tells you whether disorder increases or decreases. Gibbs free energy combines enthalpy and entropy: ΔG° = ΔH° - TΔS°. When ΔG° < 0, the process is thermodynamically favored. Temperature determines which sign of ΔH° and ΔS° combinations are favored.
Free Energy, Equilibrium, and Coupled Reactions
ΔG° and K are linked by ΔG° = -RT ln K. A negative ΔG° means K > 1 and products are favored at equilibrium. Thermodynamically unfavorable processes can be driven by coupling them to favorable reactions (shared intermediates) or by supplying external energy such as electricity or light.
Electrochemical Cells and Calculations
Galvanic cells convert thermodynamically favored redox reactions into electrical energy; electrolytic cells use external power to drive unfavorable reactions. Standard cell potential E°cell = E°cathode - E°anode, and ΔG° = -nFE°. The Nernst equation describes how cell potential changes under nonstandard conditions, and Faraday's law connects charge, current, time, and mass deposited.
Energy drives the direction of changeEvery topic in Unit 9 returns to one question: will this process occur, and how much energy is involved? Entropy measures dispersal, Gibbs free energy combines enthalpy and entropy into a single favorability criterion, and electrochemistry translates that criterion into measurable voltage and calculable mass. The equation ΔG° = -nFE° = -RT ln K is the backbone of the entire unit, linking thermodynamics, equilibrium, and electrochemistry into one coherent framework.
Unit 9 review notes
9.1
Introduction to Entropy
Entropy (S) measures how dispersed matter and energy are in a system. You need to predict the sign and relative magnitude of ΔS for physical and chemical processes without calculating exact values. Focus on the direction of change: does the process increase or decrease disorder?
- Matter dispersal: Entropy increases when particles gain freedom of movement: solid to liquid to gas, or a gas expanding into a larger volume.
- Moles of gas rule: For reactions involving gases, ΔS° is positive when the total moles of gaseous products exceed the total moles of gaseous reactants.
- Energy dispersal: Entropy increases with temperature because the distribution of kinetic energy among particles broadens, creating more microstates.
- Microstates: Entropy is related to the number of distinct molecular arrangements available to a system; more microstates means higher entropy.
- Phase change direction: Melting and vaporization increase entropy; freezing and condensation decrease entropy.
For the reaction 2H2(g) + O2(g) → 2H2O(g), predict the sign of ΔS° and explain your reasoning using moles of gas.
| Process | ΔS sign | Reason |
|---|
| Solid melting to liquid | Positive | Particles gain translational freedom |
| Liquid vaporizing to gas | Positive (large) | Large increase in volume and freedom |
| Gas compressed to smaller volume | Negative | Fewer accessible positions |
| 3 mol gas → 1 mol gas (reaction) | Negative | Fewer moles of gas means fewer microstates |
| Increasing temperature of a gas | Positive | Broader kinetic energy distribution |
9.2
Absolute Entropy and Entropy Change
Unlike enthalpy, entropy has a true absolute value anchored at zero for a perfect crystal at 0 K (Third Law). Standard molar entropy values S° are tabulated and used to calculate ΔS° for reactions using a products-minus-reactants summation, just like Hess's law for enthalpy.
- ΔS°reaction = ΣS°products - ΣS°reactants: Multiply each S° by its stoichiometric coefficient, sum the products, then subtract the sum of the reactants.
- Units: Standard molar entropy S° has units of J/(mol·K), not kJ. Watch for unit mismatches when combining with ΔH° in kJ.
- State matters: Always include the physical state when using tabulated S° values; S°(H2O, g) is much larger than S°(H2O, l).
- Absolute reference: Because entropy has an absolute zero, S° values are always positive for real substances above 0 K.
Using tabulated S° values, calculate ΔS° for N2(g) + 3H2(g) → 2NH3(g) and explain whether the sign matches your prediction from moles of gas.
9.3
Gibbs Free Energy and Kinetic vs. Thermodynamic Control
Gibbs free energy (ΔG°) combines enthalpy and entropy into a single criterion for thermodynamic favorability. A negative ΔG° means the process is thermodynamically favored under standard conditions. However, a favorable ΔG° does not guarantee the reaction occurs at a measurable rate. When a high activation energy prevents a favorable reaction from proceeding, the system is under kinetic control.
- ΔG° = ΔH° - TΔS°: Temperature determines which sign combinations of ΔH° and ΔS° produce a negative ΔG°. When ΔH° and ΔS° have opposite signs, temperature controls favorability.
- Thermodynamically favored: ΔG° < 0 means products are favored; ΔG° > 0 means reactants are favored. Avoid the word spontaneous on the AP exam.
- ΔG°reaction = ΣΔG°f products - ΣΔG°f reactants: An alternative calculation route using standard free energies of formation from a data table.
- Kinetic control: A thermodynamically favored reaction with a high activation energy will not proceed at a measurable rate. The system is not at equilibrium; it is kinetically trapped.
- Activation energy: The energy barrier that must be overcome for a reaction to proceed. A catalyst lowers activation energy without changing ΔG°.
A reaction has ΔH° = +50 kJ/mol and ΔS° = +200 J/(mol·K). At what temperature does the reaction become thermodynamically favored? If the reaction still does not occur at that temperature, what explains this?
| ΔH° | ΔS° | Favored at... |
|---|
| Negative | Positive | All temperatures |
| Positive | Negative | No temperature (never favored) |
| Negative | Negative | Low temperatures only |
| Positive | Positive | High temperatures only |
9.5
Free Energy, Equilibrium, and Dissolution
ΔG° and K describe the same equilibrium position from two different angles. A large negative ΔG° corresponds to a large K (products strongly favored); a large positive ΔG° corresponds to a small K (reactants strongly favored). Dissolution applies this framework to salts, where ΔG° of dissolution reflects competing enthalpy and entropy contributions.
- ΔG° = -RT ln K: Negative ΔG° gives K > 1; positive ΔG° gives K < 1; ΔG° near zero gives K near 1.
- K = e^(-ΔG°/RT): Equivalent form; useful for estimating K magnitude when ΔG° is much larger or smaller than RT (~2.5 kJ/mol at 298 K).
- Qualitative estimation: When |ΔG°| >> RT, K deviates strongly from 1. When ΔG° is near zero, K is close to 1 and neither products nor reactants are strongly favored.
- Free energy of dissolution: ΔG° for dissolving a salt reflects three factors: breaking the lattice (endothermic), reorganizing the solvent (entropy cost), and forming ion-dipole interactions (exothermic). These often partially cancel, making predictions difficult.
- Solubility and temperature: If dissolution is endothermic overall, solubility increases with temperature. If exothermic, solubility decreases with temperature.
A reaction has ΔG° = -17.1 kJ/mol at 298 K. Estimate K qualitatively and then calculate it using ΔG° = -RT ln K.
9.7
Coupled Reactions
A thermodynamically unfavorable reaction (ΔG° > 0) can be driven by coupling it to a favorable reaction (ΔG° < 0) that shares a common intermediate. The total ΔG° is the sum of the individual ΔG° values. If the sum is negative, the overall coupled process is thermodynamically favored.
- Common intermediate: Coupled reactions must share at least one intermediate so that the individual reactions add to give the desired overall reaction.
- ATP coupling: In biological systems, ATP hydrolysis (ΔG° < 0) is coupled to unfavorable biosynthetic reactions. The shared phosphate intermediate links the two processes.
- External energy sources: Electrical energy drives electrolytic cells; light drives photosynthesis (CO2 to glucose). Both are examples of external energy making unfavorable processes occur.
- Adding ΔG° values: Write out both reactions, confirm they share an intermediate, add them to get the net reaction, and add their ΔG° values to determine overall favorability.
Reaction A has ΔG° = +25 kJ/mol and reaction B has ΔG° = -40 kJ/mol. If they share a common intermediate, is the coupled process thermodynamically favored? What is the overall ΔG°?
9.8
Electrochemical Cells and Cell Potential
Galvanic cells use thermodynamically favored redox reactions to generate voltage. Electrolytic cells use an external power source to drive unfavorable reactions. In both cell types, oxidation occurs at the anode and reduction occurs at the cathode. Standard cell potential E°cell connects directly to ΔG° through ΔG° = -nFE°.
- Galvanic vs. electrolytic: Galvanic cells: ΔG° < 0, E°cell > 0, no external power needed. Electrolytic cells: ΔG° > 0, E°cell < 0, external power required.
- Anode and cathode: Oxidation always occurs at the anode; reduction always occurs at the cathode. Electrons flow through the external wire from anode to cathode.
- Salt bridge: Maintains electrical neutrality by allowing ion flow between half-cells without mixing the solutions. Without it, the cell stops working.
- E°cell = E°cathode - E°anode: Use tabulated standard reduction potentials. The half-reaction with the higher reduction potential is the cathode.
- ΔG° = -nFE°: n is the moles of electrons transferred in the balanced redox equation; F is 96,485 C/mol. A positive E°cell gives a negative ΔG°, confirming thermodynamic favorability.
For a Zn-Cu galvanic cell (E°(Zn2+/Zn) = -0.76 V, E°(Cu2+/Cu) = +0.34 V), identify the anode and cathode, calculate E°cell, and determine ΔG° for the transfer of 2 moles of electrons.
| Feature | Galvanic Cell | Electrolytic Cell |
|---|
| ΔG° | Negative | Positive |
| E°cell | Positive | Negative |
| Energy flow | Produces electrical energy | Consumes electrical energy |
| Anode charge | Negative | Positive |
| Example | Zn-Cu Daniell cell | Electroplating, charging a battery |
9.10
Nonstandard Conditions and Faraday's Law
Under nonstandard conditions, cell potential shifts from E°cell depending on the reaction quotient Q. The cell is always moving toward equilibrium, so E approaches zero as Q approaches K. Faraday's law connects the charge passed through an electrolytic cell to the mass of material deposited or removed at an electrode.
- Nernst equation (qualitative): When Q < 1 (reactants favored), the cell is farther from equilibrium than standard conditions and |E| > |E°|. When Q > 1, |E| < |E°|. At Q = K, E = 0.
- Le Chatelier does not apply: Electrochemical systems are not at equilibrium, so Le Chatelier's principle cannot be used to predict shifts. Use Q vs. K reasoning instead.
- Concentration cell: A cell where both electrodes are the same material but differ in electrolyte concentration. Electrons flow from the lower-concentration side (anode) to the higher-concentration side (cathode) to reach equilibrium.
- I = q/t: Current (amperes) equals charge (coulombs) divided by time (seconds). Rearrange to find charge from current and time.
- Faraday's law calculation chain: Convert current and time to charge (q = I x t), then to moles of electrons (divide by 96,485 C/mol), then use stoichiometry of the half-reaction to find moles of substance, then convert to mass using molar mass.
A current of 2.00 A is passed through a solution of CuSO4 for 965 seconds. How many grams of copper are deposited at the cathode? (Cu2+ + 2e- → Cu, M = 63.55 g/mol)
Practice AP Chem unit 9 questions
Try stimulus-based AP practice questions and written prompts after you review the notes.
Standard cell potentials are measured for the reaction M2+(aq)+H2(g)→M(s)+2H+(aq) with several unknown metals paired with a standard hydrogen electrode. The bar chart shows Ecell∘ values for M1, M2, M3, and M4.
QuestionWhich claim is best supported by the experimental data?
Reaction with M2 is favored because its Ecell∘ is +0.80 V, so ΔG∘ is negative.
Reaction with M4 is favored because its Ecell∘ is -0.76 V, so ΔG∘ is negative.
Reaction with M1 is unfavored because its Ecell∘ is +0.45 V, so ΔG∘ is positive.
Reaction with M3 is unfavored because its Ecell∘ is -0.25 V, so ΔG∘ is negative.
A galvanic cell is constructed with Cu(s) and Ag(s) electrodes in 1.0 M solutions of their nitrate salts. The mass of each electrode is measured after the cell operates for 30 minutes. A claim is made that the copper electrode is the anode.
QuestionWhich reasoning best justifies the claim?
Solid Cu atoms lose electrons to form aqueous ions, causing the mass to decrease.
Aqueous Cu ions gain electrons to form solid atoms, causing the mass to decrease.
Solid Cu atoms gain electrons to form aqueous ions, causing the mass to decrease.
Aqueous Cu ions lose electrons to form solid atoms, causing the mass to decrease.
4. A student constructs a galvanic cell to investigate the thermodynamic properties of the reaction between aluminum and nickel. The experimental setup is shown in Figure 1, and the relevant standard reduction potentials are provided in Table 1.
Table 1. Standard Reduction Potentials at 25°C
Half-Reaction | Standard Reduction Potential E° (V) |
|---|
Al³⁺(aq) + 3e⁻ → Al(s) | −1.66 |
Ni²⁺(aq) + 2e⁻ → Ni(s) | −0.25 |
Figure 1. Galvanic cell setup (Al/Al³⁺ || Ni²⁺/Ni) with 1.0 M solutions and KNO₃ salt bridge; voltmeter shows a positive reading.
i. Propose a change to the concentration of Ni²⁺(aq) (increase or decrease relative to 1.0 M) that would result in a cell potential greater than the standard cell potential.
ii. Calculate the standard Gibbs free energy change, ΔG∘, in kJ/mol, for the thermodynamically favored reaction that occurs in the cell. (Assume Faraday's constant F=96,485 C/mol e−)
2. Answer the following questions about an electrochemical cell involving zinc and nickel.
A student constructs a galvanic cell at 298 K using a zinc electrode immersed in a 1.0 M Zn(NO₃)₂ solution and a nickel electrode immersed in a 1.0 M Ni(NO₃)₂ solution, as shown in Figure 1. A salt bridge containing KNO₃(aq) connects the two half-cells.
Table 1. Standard Reduction Potentials at 25°C
Half-Reaction | Standard Reduction Potential, E° (V) |
|---|
Zn²⁺(aq) + 2e⁻ → Zn(s) | −0.76 |
Ni²⁺(aq) + 2e⁻ → Ni(s) | −0.25 |
Figure 1. Galvanic cell at 298 K: Zn(s)|Zn2+(1.0 M) || Ni2+(1.0 M)|Ni(s) with a KNO3(aq) salt bridge and an external voltmeter
The student prepares a second galvanic cell at 298 K with the same electrodes but different ion concentrations. In this new cell, [Zn²⁺] = 1.50 M and [Ni²⁺] = 0.010 M.
The original galvanic cell (with 1.0 M concentrations) is allowed to operate for 45.0 minutes. During this time, the cell produces a constant average current of 0.750 A.
i. Calculate the mass, in grams, of solid nickel deposited on the cathode during this 45.0-minute period.
ii. Does the mass of the zinc electrode increase, decrease, or remain the same during this operation? Justify your answer.
1. Answer the following questions about an electrochemical cell constructed using silver and chromium electrodes.
The standard reduction potentials for the half-reactions related to the cell are given in the table in Figure 1.
Figure 1. Standard Reduction Potentials at 25°C
i. Write the balanced net ionic equation for the thermodynamically favored reaction that occurs in the cell.
ii. Calculate the standard cell potential, E°cell, for the reaction.
The galvanic cell is constructed as shown in Figure 2.
Figure 2. Galvanic cell setup
The cell operates for a period of time, and the concentrations of the ions change. The new concentration of Ag⁺(aq) is 0.010 M, and the concentration of Cr³⁺(aq) is 1.0 M.
i. Calculate the value of the reaction quotient, Q, under these new conditions.
ii. Is the cell potential, Ecell, under these conditions greater than, less than, or equal to the standard cell potential, E°cell, calculated in part A? Justify your answer.
In a separate experiment, a student uses an external power source to plate silver metal onto an object from a solution of AgNO₃(aq). The student applies a constant current of 2.00 amperes for 30.0 minutes.
i. Calculate the mass, in grams, of silver metal deposited on the object.
ii. If the student used a solution of Cu(NO₃)₂ instead of AgNO₃ and applied the same current for the same amount of time to plate copper metal, would the mass of copper produced be greater than, less than, or equal to the mass of silver calculated in part F(i)? Justify your answer.