⚡ Electrolysis – AP Chemistry Final Review

⚡ AP CHEMISTRY Electrolysis – Full Study Guide

🔄 What Happens During Electrolysis?

This section breaks down the entire process so students understand what’s really happening, step by step:

1. Breakdown of Electrolyte: The compound (electrolyte) is either molten or dissolved in water. This causes it to break into ions.
2. Ion Movement: Ions are attracted to oppositely charged electrodes when electricity is applied.
   • Cations (positive) → Cathode (–) → Gain electrons → Reduction
   • Anions (negative) → Anode (+) → Lose electrons → Oxidation
3. Redox Reaction: The ions are discharged as atoms. This may form solid metals, gases, or other elements.

🌋 Example: Molten NaCl
NaCl → Na⁺ + Cl⁻
Cathode: Na⁺ + e⁻ → Na
Anode: 2Cl⁻ → Cl₂(g) + 2e⁻

💧 Example: Aqueous CuSO₄
Cu²⁺ + SO₄²⁻ + H₂O → Cu²⁺, SO₄²⁻, H⁺, OH⁻
Cathode: Cu²⁺ + 2e⁻ → Cu
Anode: 2H₂O → O₂(g) + 4H⁺ + 4e⁻

✅ Always identify the competing ions and compare E° values in aqueous solutions.

🧪 Types of Electrolytes

• Molten Electrolytes: Simple, only compound ions present.
• Aqueous Electrolytes: Water contributes H⁺ and OH⁻ ions — multiple reactions possible.

🌊 Molten vs. Aqueous Electrolysis

Feature	Molten	Aqueous
Ions Present	Only from compound	Compound + H₂O ions
Prediction	Straightforward	Compare E° values
Complexity	Simple	Competing reactions

🧲 Ion Pathways & Product Prediction

🌋 Molten NaCl

• Na⁺ → Cathode → Gains e⁻ → Na metal
• Cl⁻ → Anode → Loses e⁻ → Cl₂ gas

💧 Aqueous NaCl

H₂O introduces H⁺ and OH⁻. Compare which ions are more easily reduced/oxidized:

• H⁺ vs. Na⁺ at cathode → H⁺ wins
• Cl⁻ vs. OH⁻ at anode → Cl⁻ wins (usually)

🧮 Faraday’s Law – How to Calculate

1. Convert time to seconds
2. Use: Q = I × t to find charge (Q)
3. Use: mol e⁻ = Q / 96485 to get moles of electrons
4. Use half-reaction to relate mol e⁻ to mol product
5. Convert mol product → grams

Example: 2.0 A for 30 min = 1800 s
Q = 2.0 × 1800 = 3600 C
mol e⁻ = 3600 / 96485 = 0.0373
Cu²⁺ + 2e⁻ → Cu → 0.0373 ÷ 2 = 0.01865 mol
Mass = 0.01865 × 63.5 = 1.19 g Cu

🚫 Common Mistakes – Topic by Topic

⚙️ Cell Setup Errors

• ❌ Assuming anode is always negative (not in electrolysis)
• ❌ Wrong electrode labeling

💧 Aqueous Ion Errors

• ❌ Forgetting H⁺ and OH⁻
• ❌ Not comparing E° values

📏 Math & Units

• ❌ Not converting minutes to seconds
• ❌ Skipping mole ratios
• ❌ Going directly from charge to grams

🧠 Flashcard Summary ℹ️ Click each card to self-test your memory. Good for spaced repetition!

• Electrolysis: Non-spontaneous redox powered by electricity
• Cathode: Reduction, gains electrons
• Anode: Oxidation, loses electrons
• Faraday’s Law: Q → mol e⁻ → mol product → grams

🔄 Electrolysis – Step-by-Step Flowchart

Step 1: Break Electrolyte Into Ions

The electrolyte (molten or aqueous) dissociates into positive and negative ions.

Example: NaCl → Na⁺ + Cl⁻

Step 2: Ions Move to Electrodes

Positive ions (cations) move to the negative cathode. Negative ions (anions) move to the positive anode.

Step 3: Redox Reactions at Electrodes

Cations gain electrons (reduction) at the cathode. Anions lose electrons (oxidation) at the anode.

Cathode: Na⁺ + e⁻ → Na (reduction)
Anode: 2Cl⁻ → Cl₂ + 2e⁻ (oxidation)

Step 4: Collect Products

Discharged atoms or gases are collected. This depends on whether the electrolyte is molten or aqueous.

Molten NaCl → Sodium metal + Chlorine gas
Aqueous CuSO₄ → Copper metal + Oxygen gas

Step 5: Optional – Use Faraday's Law

To calculate how much product forms, apply Q = It and mol e⁻ = Q / F. Use mole ratios to convert to grams.

Q = 2.0 A × 1800 s = 3600 C
mol e⁻ = 3600 / 96485 ≈ 0.0373 mol
Cu²⁺ + 2e⁻ → Cu → 0.01865 mol → 1.19 g Cu