AP Chemistry Unit 9 — Applications of Thermodynamics.
This unit covers galvanic cells, electrolysis and free energy and equilibrium — essential concepts for AP Chemistry. Use our interactive study games to test your understanding, or review questions in traditional format below.
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This unit covers galvanic cells, electrolysis and free energy and equilibrium — essential concepts for AP Chemistry. Use our interactive study games to test your understanding, or review questions in traditional format below.
Key Concepts Breakdown
1 Galvanic Cells
Galvanic (voltaic) cells convert spontaneous chemical energy into electrical energy. Students must be able to identify the anode (oxidation) and cathode (reduction), calculate cell potential using standard reduction potentials, and predict spontaneity from the sign of E°cell. The cell potential is related to free energy by ΔG° = -nFE°.
Key Points
- Anode = oxidation (negative electrode in galvanic cell); Cathode = reduction (positive electrode)
- E°cell = E°cathode - E°anode; a positive E°cell indicates a spontaneous reaction
- Salt bridge maintains electrical neutrality by allowing ion flow between half-cells
- ΔG° = -nFE°cell; larger positive E°cell means more negative ΔG° (more spontaneous)
Given: Zn²⁺/Zn E° = -0.76 V and Cu²⁺/Cu E° = +0.34 V. Calculate E°cell for the Zn-Cu galvanic cell and determine if the reaction is spontaneous.
Zinc is the anode (lower reduction potential, gets oxidized) and copper is the cathode (higher reduction potential, gets reduced). E°cell = +0.34 V - (-0.76 V) = +1.10 V. Because E°cell is positive, the reaction is spontaneous, and ΔG° = -nFE° will be negative.
2 Electrolysis
Electrolysis uses electrical energy to drive a non-spontaneous redox reaction. Students must identify which species is oxidized and reduced at each electrode, apply Faraday's law to calculate moles of product from charge passed, and distinguish electrolytic cells from galvanic cells. In electrolytic cells, the anode is still oxidation but is now the positive electrode.
Key Points
- Electrolytic cell: non-spontaneous (E°cell < 0), requires external power source
- Anode = oxidation (positive); Cathode = reduction (negative) — opposite sign convention from galvanic
- Faraday's law: moles of electrons = charge (C) ÷ 96,485 C/mol; then use stoichiometry to find moles of product
- In aqueous electrolysis, water can compete: O₂ produced at anode, H₂ at cathode if ion reduction/oxidation is unfavorable
A current of 2.00 A is passed through a solution of CuSO₄ for 965 seconds. How many grams of Cu are deposited at the cathode? (Cu²⁺ + 2e⁻ → Cu, M = 63.55 g/mol)
First calculate total charge: q = It = 2.00 A × 965 s = 1930 C. Convert to moles of electrons: 1930 ÷ 96485 = 0.02000 mol e⁻. Since 2 mol e⁻ deposits 1 mol Cu, moles of Cu = 0.01000 mol, and mass = 0.01000 × 63.55 = 0.636 g.
3 Free Energy And Equilibrium
The standard free energy change ΔG° is directly related to the equilibrium constant K by the equation ΔG° = -RT ln K. Students must be able to predict the direction of spontaneity, relate ΔG° to K (sign and magnitude), and use ΔG = ΔG° + RT ln Q to determine spontaneity at non-standard conditions. These relationships connect thermodynamics to equilibrium and electrochemistry.
Key Points
- ΔG° = -RT ln K: if K > 1, ΔG° < 0 (products favored); if K < 1, ΔG° > 0 (reactants favored)
- ΔG = ΔG° + RT ln Q: when Q < K, ΔG < 0 (forward spontaneous); when Q > K, ΔG > 0 (reverse spontaneous)
- At equilibrium: ΔG = 0 and Q = K
- All three equations link: ΔG° = -nFE° = -RT ln K; know how to convert between E°, K, and ΔG°
For a reaction at 298 K, E°cell = +0.592 V and n = 2. Calculate K for this reaction.
Use ΔG° = -nFE° = -(2)(96485)(0.592) = -114,271 J/mol. Then apply ΔG° = -RT ln K: ln K = -ΔG°/RT = 114271 ÷ (8.314 × 298) = 46.1, so K = e^46.1 ≈ 10^20. Alternatively, use the shortcut log K = nE° / 0.0592 = (2)(0.592)/0.0592 = 20, giving K = 10^20.
Questions, answered.
What is Applications of Thermodynamics?
Applications of Thermodynamics is Unit 9 of AP Chemistry, covering galvanic cells, electrolysis and free energy and equilibrium.
How to study for AP Chemistry Unit 9?
Start with the Quick Summary above, review the Key Concepts, then test yourself with our interactive study games. Aim for 80%+ accuracy before moving on.
How many questions are in this unit?
This unit has 25+ review questions across 5 different game modes.