AP Chemistry Unit 9 Study Notes
AP Chemistry 9.7: Electrolysis and Faraday’s Law
Compare cell types and convert current and time into chemical amounts using electron stoichiometry.
Aligned to Thermodynamics and Electrochemistry from the current College Board AP Chemistry course outline. Exam weighting for this unit: 7%-9% of the multiple-choice score range listed by College Board.
Study these notes
Start with each main idea, then follow the indented explanations and worked examples. Try the next calculation before reading its answer.
Organized from the provided Unit 9 study document. Further study: Khan Academy.
46. Galvanic vs. Electrolytic Cells
Galvanic Cell
-
thermodynamically favored reaction
- converts chemical energy to electrical energy
- Ecell > 0 for favored direction
- ΔG < 0
- anode negative
- cathode positive
Electrolytic Cell
-
thermodynamically unfavored reaction is forced
- converts electrical energy into stored chemical energy
- requires an external power source
- anode positive
- cathode negative
- But again:
- oxidation is always at the anode
- reduction is always at the cathode
47. Electrolysis
-
Electrolysis uses electrical energy to drive a redox reaction that would not proceed in the desired direction on its own.
- Common applications include:
- metal plating
- purification of metals
- production of substances from molten or aqueous ionic compounds
For AP Chemistry, the important part is usually the quantitative relationship between current, time, electrons, and chemical amount.
48. Electrical Current and Charge
-
Current is:
- I = q/t
- Therefore:
- q = It
- where:
- I = current in amperes
- q = charge in coulombs
- t = time in seconds
- Remember:
- 1 ampere = 1 coulomb/second
49. Faraday's Law Calculations
-
One mole of electrons carries approximately:
- 96,485 C
- So:
- mol e⁻ = charge / F
- or:
- mol e⁻ = It/F
Then use the balanced half-reaction to convert electrons into the amount of substance produced or consumed.
50. Electrolysis Workflow
-
For most quantitative electrolysis problems:
- current × time → charge → moles e⁻ → moles substance → grams
Example Structure
-
Suppose metal ion:
- M²⁺ + 2e⁻ → M
- If you determine that:
- 0.100 mol e⁻
- passed through the cell, then:
- 0.100 mol e⁻ × (1 mol M / 2 mol e⁻)
- = 0.0500 mol M
Then use molar mass to find grams.
51. Time Must Be in Seconds
-
If current is given in amperes:
- A = C/s
- so time should be converted to:
- seconds
- Example:
- 10.0 min × 60 s/min = 600 s
52. Electron Stoichiometry Matters
-
Do not automatically assume:
- 1 mol e⁻ = 1 mol metal
- Example:
- Al³⁺ + 3e⁻ → Al
- requires:
- 3 mol electrons per 1 mol Al
- while:
- Ag⁺ + e⁻ → Ag
- requires only:
- 1 mol electron per 1 mol Ag