🎯⭐ INTERACTIVE LESSON

Electrolytic Cells and Quantitative Electrolysis

Learn step-by-step with interactive practice!

Electrolytic Cells and Quantitative Electrolysis - Complete Interactive Lesson

Part 1: Electrolysis Basics

⚡ Electrolysis — Driving Non-Spontaneous Reactions

Part 1 of 7 — Electrolytic Cells and External Voltage

Topics in This Part

Section
🔧 How Electrolysis Works
The Key Idea
Requirements
Electrode Conventions in Electrolytic Cells
⚡ Energy Considerations

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 1

Understanding the core concepts covered in Part 1
Applying these ideas to solve practice problems
Building toward AP exam readiness for this topic

🔧 How Electrolysis Works

The Key Idea

An external voltage source (battery or power supply) pushes electrons in the opposite direction from what they would naturally go, driving a non-spontaneous reaction forward.

Requirements

An external power source providing voltage > $|E°_{\text{cell}}|$
An electrolyte (molten salt or aqueous solution) to carry current via ions
Two electrodes (often inert — Pt or graphite)

Electrode Conventions in Electrolytic Cells

Property	Galvanic Cell	Electrolytic Cell
Anode	Oxidation ✓	Oxidation ✓
Cathode	Reduction ✓	Reduction ✓
Anode sign	− (negative)	+ (positive)
Cathode sign	+ (positive)	− (negative)
Spontaneous?	Yes	No

AN OX and RED CAT still apply! Oxidation is always at the anode, reduction at the cathode — regardless of cell type.

⚡ Energy Considerations

The Thermodynamic Reality

For any electrolysis reaction, the numbers tell the story:

Quantity	Value	Meaning
$\Delta G$	$> 0$	Non-spontaneous — needs energy input
$E_{\text{cell}}$	$< 0$	Negative cell potential
External voltage required	$\geq \lvert E_{\text{cell}} \rvert$	Must overcome the thermodynamic barrier

🔑 Bottom line: You have to pay with electrical energy to make an electrolysis reaction go.

⚠️ Overpotential — The Hidden Cost

In practice, the actual voltage needed is higher than the thermodynamic minimum. This extra voltage is called overpotential — it overcomes kinetic barriers at the electrode surfaces.

$\boxed{V_{\text{applied}} = |E_{\text{cell}}| + \eta_{\text{overpotential}}}$

💡 Overpotential depends on the electrode material, current density, and which gases are being produced. It's why real electrolysis always costs more energy than theory predicts.

🧪 Example: Electrolysis of Water

$2\text{H}_2\text{O}(l) \rightarrow 2\text{H}_2(g) + \text{O}_2(g)$

Parameter	Value
$E°$	$-1.23$ V (non-spontaneous)
Minimum applied voltage	$1.23$ V
Typical actual voltage	$\sim 1.8 - 2.0$ V
Overpotential	$\sim 0.6 - 0.8$ V

📏 This reaction is how we produce hydrogen gas for fuel cells — electrolysis and fuel cells are reverse processes of each other!

Electrolysis Concept Quiz 🎯

Electrolytic Cell Basics 🔽

Electrolysis Energy 🧮

1) The electrolysis of water has $E° = -1.23$ V. What minimum voltage must be applied? (in V, positive value)

2) If the overpotential is 0.5 V, what is the actual applied voltage needed? (in V)

3) Is the ΔG for electrolysis positive or negative? (type "positive" or "negative")

Round all answers to 3 significant figures.

Exit Quiz — Electrolysis Basics ✅

Part 2: Electrolytic vs Galvanic Cells

🔄 Galvanic vs. Electrolytic Cells

Part 2 of 7 — A Detailed Comparison

Topics in This Part

Section
⚖️ Complete Comparison
What STAYS THE SAME
What CHANGES
🔋 Recharging: Galvanic → Electrolytic
⬇️ Discharging (Galvanic Mode)

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 2

Understanding the core concepts covered in Part 2
Applying these ideas to solve practice problems
Building toward AP exam readiness for this topic

⚖️ Complete Comparison

Feature	Galvanic Cell	Electrolytic Cell
Spontaneous?	Yes ( $\Delta G < 0$ )	No ( $\Delta G > 0$ )
$E_{\text{cell}}$	Positive	Negative
Energy conversion	Chemical → Electrical	Electrical → Chemical
External power?	No (produces power)	Yes (requires power)
Anode	Oxidation (−)	Oxidation (+)
Cathode	Reduction (+)	Reduction (−)
Electron flow	Anode → Cathode	Anode → Cathode
Salt bridge	Usually present	Often not needed
Example	Battery, fuel cell	Electroplating, electrolysis

What STAYS THE SAME

Oxidation at the anode (AN OX)
Reduction at the cathode (RED CAT)
Electrons flow from anode to cathode
Cations migrate toward cathode, anions toward anode

🔑 AP Must-Know: AN OX / RED CAT applies to ALL electrochemical cells. This never changes.

What CHANGES

Sign of anode/cathode (reversed!)
Direction of energy flow (chemical ↔ electrical)
Spontaneity (spontaneous vs. forced)

💡 Memory Aid: In galvanic cells the anode is (−) and cathode is (+). In electrolytic cells, it flips: anode is (+) and cathode is (−).

🔋 Recharging: Galvanic → Electrolytic

Every rechargeable battery lives a double life — it's a galvanic cell when discharging and an electrolytic cell when charging. The chemistry literally runs in reverse!

⬇️ Discharging (Galvanic Mode)

$\text{Pb}(s) + \text{PbO}_2(s) + 2\text{H}_2\text{SO}_4(aq) \rightarrow 2\text{PbSO}_4(s) + 2\text{H}_2\text{O}(l)$

Property	Value
Spontaneous?	✅ Yes
$E$	Positive ( $> 0$ )
Energy	Chemical → Electrical (powers your car)

⬆️ Charging (Electrolytic Mode)

$2\text{PbSO}_4(s) + 2\text{H}_2\text{O}(l) \rightarrow \text{Pb}(s) + \text{PbO}_2(s) + 2\text{H}_2\text{SO}_4(aq)$

Property	Value
Spontaneous?	❌ No
$E$	Negative ( $< 0$ )
Energy	Electrical → Chemical (from the charger)

🔀 What Swaps During Charging?

	Discharging	Charging
Anode	Electrode A	Electrode B
Cathode	Electrode B	Electrode A
Electron flow	A → B	B → A
Reaction direction	Forward	Reverse

⚠️ AP Trap: The electrodes that were anode/cathode during discharge swap roles during charging. The chemistry reverses, and so do the labels!

Galvanic vs. Electrolytic Quiz 🎯

Cell Comparison 🔽

Quick Comparison 🧮

Answer with "galvanic" or "electrolytic":

1) ΔG < 0 and E > 0 describes a _____ cell.

2) Requires an external power source: _____ cell.

3) The anode is positive in a _____ cell.

Exit Quiz — Galvanic vs. Electrolytic ✅

Part 3: Electrolysis of Molten Salts

🧪 Electrolysis of Molten Salts and Aqueous Solutions

Part 3 of 7 — Predicting Products

Topics in This Part

Section
🔋 Electrolysis of Molten Salts
Why Molten?
Simple Case: Molten NaCl
Molten Salt Rule
Examples

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 3

Understanding the core concepts covered in Part 3
Applying these ideas to solve practice problems
Building toward AP exam readiness for this topic

🔋 Electrolysis of Molten Salts

Why Molten?

Ionic compounds must be in a molten (liquid) state or dissolved in water to conduct electricity. In the solid state, ions are locked in place and cannot migrate.

Simple Case: Molten NaCl

At the cathode (reduction): $\text{Na}^+(l) + e^- \rightarrow \text{Na}(l)$

At the anode (oxidation): $2\text{Cl}^-(l) \rightarrow \text{Cl}_2(g) + 2e^-$

Molten Salt Rule

In a molten salt, there are only two ions present. The prediction is straightforward:

Cation is reduced at the cathode → metal forms
Anion is oxidized at the anode → nonmetal forms

Examples

Salt	Cathode Product	Anode Product
NaCl	Na(l)	$Cl_{2}(g)$
$MgCl_{2}$	Mg(l)	$Cl_{2}(g)$
$Al_{2}O_{3}$	Al(l)	$O_{2}(g)$
$CaBr_{2}$	Ca(l)	$Br_{2}(g)$

🧪 Electrolysis of Aqueous Solutions

The Complication: Water Competes!

In aqueous solutions, water can be oxidized or reduced instead of the dissolved ions. You must compare the reduction potentials to predict which reaction occurs.

📋 Key Reduction Potentials to Know

Half-Reaction	$E°$ (V)
$\text{Au}^{3+} + 3e^- \rightarrow \text{Au}$	$+1.50$
$\text{Ag}^+ + e^- \rightarrow \text{Ag}$	$+0.80$
$\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$	$+0.34$
$2\text{H}_2\text{O} + 2e^- \rightarrow \text{H}_2 + 2\text{OH}^-$	$-0.83$
$\text{Ni}^{2+} + 2e^- \rightarrow \text{Ni}$	$-0.26$
$\text{Zn}^{2+} + 2e^- \rightarrow \text{Zn}$	$-0.76$
$\text{Al}^{3+} + 3e^- \rightarrow \text{Al}$	$-1.66$
$\text{Na}^+ + e^- \rightarrow \text{Na}$	$-2.71$
$\text{K}^+ + e^- \rightarrow \text{K}$	$-2.93$

🔑 The decision rule: Whichever half-reaction has the more positive (less negative) $E°$ is easier to reduce and wins the competition at the cathode.

⬇️ At the Cathode — Which Gets Reduced?

Water's reduction potential is $E° = -0.83$ V. Compare the metal ion to this benchmark:

Metal Ion $E°$	What Happens	Examples
$> -0.83$ V	Metal deposits	$Cu^{2+}$ , $Ag^{+}$ , $Au^{3+}$ , $Ni^{2+}$
$< -0.83$ V	$H_{2}$ gas forms	$Na^{+}$ , $K^{+}$ , $Al^{3+}$

⬆️ At the Anode — Which Gets Oxidized?

Water's oxidation potential is $E° = +1.23$ V.

Anion Type	What Happens	Examples
Simple halides	Anion is oxidized	$Cl^{-}$ → $Cl_{2}$ , $Br^{-}$ → $Br_{2}$ , $I^{-}$ → $I_{2}$
Oxyanions or $F^{-}$	Water is oxidized → $O_{2}$	$SO_{4}^{2-}$ , $NO_{3}^{-}$ , $F^{-}$

⚠️ Why do halides win even though their $E°$ is less favorable? Overpotential! The kinetic barrier for $O_{2}$ production is high, so in practice, halides get oxidized first.

🗺️ Quick Decision Flowchart

Step	Question	If YES	If NO
1	Is it a molten salt?	Cation → metal, Anion → nonmetal	Go to step 2
2	Cathode: Is metal $E°$ above $-0.83$ V?	Metal deposits	$H_{2}$ forms
3	Anode: Is anion a simple halide?	Halide is oxidized	$O_{2}$ forms

Electrolysis Product Quiz 🎯

📋 Reference: Water cathode $E° = -0.83$ V | Water anode $E° = +1.23$ V

Reduction potentials:

Ion $E°$ (V) Ion $E°$ (V)
$Au^{3+}$ +1.50 $Zn^{2+}$ −0.76
$Ag^{+}$ +0.80 $Al^{3+}$ −1.66
$Cu^{2+}$ +0.34 $Na^{+}$ −2.71
$Ni^{2+}$ −0.26 $K^{+}$ −2.93

Oxidation potentials (anode):

Half-reaction $E°$ (V)
$2Cl^{-}$ → $Cl_{2}$ + $2e^{-}$ +1.36
$2Br^{-}$ → $Br_{2}$ + $2e^{-}$ +1.07
$2I^{-}$ → $I_{2}$ + $2e^{-}$ +0.54
$2H_{2}O$ → $O_{2}$ + $4H^{+}$ + $4e^{-}$ +1.23

Predicting Electrolysis Products 🔽

📋 Reduction potentials:

Ion $E°$ (V) Ion $E°$ (V)
$Au^{3+}$ +1.50 $Zn^{2+}$ −0.76
$Ag^{+}$ +0.80 $Al^{3+}$ −1.66
$Cu^{2+}$ +0.34 $Na^{+}$ −2.71
$Ni^{2+}$ −0.26 $K^{+}$ −2.93

Anode: $2Cl^{-}$ → $Cl_{2}$ (+1.36 V) | $2Br^{-}$ → $Br_{2}$ (+1.07 V) | $2I^{-}$ → $I_{2}$ (+0.54 V) | $H_{2}O$ → $O_{2}$ (+1.23 V)

Rules: Metal $E°$ above −0.83 V → metal deposits. Below → $H_{2}$ . Simple halides → oxidized. Oxyanions → $O_{2}$ .

Product Identification 🧮

📋 Reference: $Ag^{+}$ $E°$ = +0.80 V | $Cu^{2+}$ = +0.34 V | $Ni^{2+}$ = −0.26 V | $Zn^{2+}$ = −0.76 V | Water cathode = −0.83 V | $Na^{+}$ = −2.71 V | $K^{+}$ = −2.93 V | $Mg^{2+}$ = −2.37 V

Anode: Halides ( $Cl^{-}$ , $Br^{-}$ , $I^{-}$ ) → oxidized. Oxyanions ( $NO_{3}^{-}$ , $SO_{4}^{2-}$ ) → $O_{2}$ forms.

What gas or metal is produced at the cathode during electrolysis of:

1) Molten $MgCl_{2}$ (cathode product)?

2) Aqueous $AgNO_{3}$ (cathode product — is $Ag^{+}$ or $H_{2}O$ reduced)?

3) Aqueous KI (anode product — is $I^{-}$ or $H_{2}O$ oxidized)?

Exit Quiz — Electrolysis Products ✅

Part 4: Electrolysis of Aqueous Solutions

⚖️ Faraday's Laws of Electrolysis

Part 4 of 7 — Quantitative Electrolysis: mol = It/(nF)

Topics in This Part

Section
📏 Faraday's Laws
The Key Equation
Step-by-Step Problem Solving
Important: What Is n?
🧪 Worked Example

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 4

Understanding the core concepts covered in Part 4
Applying these ideas to solve practice problems
Building toward AP exam readiness for this topic

📏 Faraday's Laws

The Key Equation

$\boxed{\text{mol of substance} = \frac{It}{nF}}$

Symbol	Meaning	Units
$I$	Current	Amperes (A) = C/s
$t$	Time	Seconds (s)
$n$	Electrons per ion in the half-reaction	—
$F$	Faraday's constant	$96{,}485$ C/mol $e^-$
$It$	Total charge	Coulombs (C)

Step-by-Step Problem Solving

Calculate total charge: $q = It$ (coulombs)
Find moles of electrons: $\text{mol } e^- = q/F = It/F$
Use stoichiometry: relate moles of electrons to moles of substance using $n$
Convert to mass if needed: $m = \text{mol} \times M$

Important: What Is n?

$n$ = number of electrons in the balanced half-reaction

🔑 Key Point: Always write the half-reaction first to determine $n$ . Getting $n$ wrong is the most common Faraday’s law mistake.

Half-Reaction	$n$
$\text{Ag}^+ + e^- \rightarrow \text{Ag}$	1
$\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$	2
$\text{Al}^{3+} + 3e^- \rightarrow \text{Al}$	3
$2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^-$	2

🧪 Worked Example — Faraday's Law

Problem: How many grams of Cu are deposited by passing a current of $2.00$ A through $\text{CuSO}_4$ solution for $1.00$ hour?

Given

Quantity	Value
Current ( $I$ )	2.00 A
Time ( $t$ )	1.00 h = 3600 s
Half-reaction	$\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$
$n$ (electrons per ion)	2
$M_{\text{Cu}}$	63.55 g/mol
$F$	96,485 C/mol $e^-$

Step-by-Step Solution

Step	Action	Calculation	Result
1	Total charge	$q = It = (2.00)(3600)$	7200 C
2	Moles of electrons	$q / F = 7200 / 96{,}485$	0.07462 mol $e^-$
3	Moles of Cu	$\text{mol } e^- / n = 0.07462 / 2$	0.03731 mol Cu
4	Mass of Cu	$(0.03731)(63.55)$	2.37 g

Alternative: One-Step Formula

$\boxed{m = \frac{It \cdot M}{nF}} = \frac{(2.00)(3600)(63.55)}{(2)(96{,}485)} = 2.37 \text{ g}$

🔑 Tip: The one-step formula combines all four steps. Use it for speed on the AP exam, but understand each step for conceptual questions.

Faraday's Law Quiz 🎯

Faraday's Law Calculations 🧮

Use $F = 96{,}485$ C/mol, $M_{\text{Ag}} = 107.87$ g/mol

1) A current of $5.00$ A flows for $1000$ s. Total charge = ? (in C)

2) Using the charge from (1), how many moles of electrons? (to 3 significant figures)

3) How many grams of Ag are deposited? ( $\text{Ag}^+ + e^- \rightarrow \text{Ag}$ , $n = 1$ ) (to 3 significant figures)

Faraday's Law Concepts 🔽

Exit Quiz — Faraday's Laws ✅

Part 5: Faraday's Laws of Electrolysis

🏭 Electroplating and Industrial Applications

Part 5 of 7 — Real-World Electrolysis

Topics in This Part

Section
🔋 Electroplating
Setup
How It Works
Controlling Thickness
Common Plating Metals

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 5

Understanding the core concepts covered in Part 5
Applying these ideas to solve practice problems
Building toward AP exam readiness for this topic

🔋 Electroplating

Electroplating is the process of coating an object with a thin layer of metal using electrolysis.

Setup

Cathode: the object to be plated (e.g., a spoon)
Anode: a piece of the plating metal (e.g., silver)
Electrolyte: a solution of the plating metal ions (e.g., $AgNO_{3}$ )

🔑 Key Rule: The object you want to coat is ALWAYS the cathode (where metal deposits). The plating metal is the anode (where it dissolves).

How It Works

At the anode: plating metal dissolves → $\text{Ag}(s) \rightarrow \text{Ag}^+(aq) + e^-$
$Ag^{+}$ ions migrate through solution
At the cathode: metal ions deposit → $\text{Ag}^+(aq) + e^- \rightarrow \text{Ag}(s)$

The object at the cathode gets coated with a layer of silver!

Controlling Thickness

The thickness of the coating depends on:

Current ( $I$ ): higher current → faster deposition
Time ( $t$ ): longer time → thicker coating
Faraday's law: $m = ItM/(nF)$

Common Plating Metals

Metal	Application
Chrome	Car bumpers, faucets
Silver	Jewelry, silverware
Gold	Electronics, jewelry
Nickel	Corrosion protection
Zinc	Galvanization of steel

📌 Major Industrial Processes

1. Hall-Héroult Process (Aluminum Production)

$2\text{Al}_2\text{O}_3(l) \rightarrow 4\text{Al}(l) + 3\text{O}_2(g)$

$Al_{2}O_{3}$ is dissolved in molten cryolite ( $\text{Na}_3\text{AlF}_6$ ) to lower the melting point
Enormous current (100,000+ A!)
Carbon anodes are consumed: $\text{C} + \text{O}^{2-} \rightarrow \text{CO}_2 + e^-$
Produces ~65 million tonnes of Al per year worldwide

💡 Why Electrolysis? Aluminum is too reactive to reduce with carbon alone. The Hall-Héroult process was a breakthrough that made aluminum affordable.

2. Chlor-Alkali Process

$2\text{NaCl}(aq) + 2\text{H}_2\text{O}(l) \rightarrow \text{Cl}_2(g) + \text{H}_2(g) + 2\text{NaOH}(aq)$

Produces three valuable products: chlorine, hydrogen, and sodium hydroxide
Membrane cell separates products
Uses aqueous NaCl (brine)

3. Electrorefining of Copper

Impure Cu = anode; pure Cu = cathode
$Cu^{2+}$ from impure anode deposits as pure Cu on cathode
Impurities fall to the bottom ("anode mud") — contains Ag, Au, Pt!
Produces 99.99% pure copper for electrical wiring

⚠️ Don’t Confuse: In electrorefining, both electrodes are copper! The impure Cu dissolves at the anode, and pure Cu deposits at the cathode. Impurities that don’t dissolve collect as valuable "anode mud."

Applications Quiz 🎯

Electroplating Calculations 🧮

A piece of jewelry is silver-plated using $I = 2.0$ A for $20$ minutes. $\text{Ag}^+ + e^- \rightarrow \text{Ag}$ , $n = 1$ , $M_{\text{Ag}} = 107.87$ g/mol

1) Total charge in coulombs?

2) Moles of Ag deposited? (to 3 significant figures)

3) Mass of Ag deposited in grams? (to 3 significant figures)

Industrial Electrolysis 🔽

Exit Quiz — Applications ✅

Part 6: Problem-Solving Workshop

🛠️ Problem-Solving Workshop — Electrolysis and Faraday

Part 6 of 7 — Practice and Integration

Practice Makes Perfect

This workshop features multi-step problems that mirror the AP Chemistry exam format. Each problem requires you to combine concepts from previous parts and show your work clearly.

🔑 Why this matters: The AP Chemistry exam rewards students who can apply concepts to unfamiliar problems — structured practice is the best preparation.

What You'll Master in Part 6

Working through complete multi-step problems from start to finish
Building problem-solving strategies you can apply on the AP exam
Identifying which concepts to apply and in what order

🛠️ Problem-Solving Checklist

For Faraday's Law Problems

⚠️ First Step Always: Convert time to seconds before calculating!

✅ Convert time to seconds ( $1 \text{ min} = 60 \text{ s}$ , $1 \text{ hr} = 3600 \text{ s}$ )
✅ Calculate charge: $q = It$
✅ Find mol electrons: $\text{mol } e^- = q/F$
✅ Write the half-reaction to find $n$
✅ Find mol substance: $\text{mol} = \text{mol } e^-/n$
✅ Convert to mass or volume if needed

For Product Prediction

System	Cathode Product	Anode Product
Molten salt	Metal	Nonmetal ( $Cl_{2}$ , $O_{2}$ , $Br_{2}$ )
Aqueous, active metal	$H_{2}$	Depends on anion
Aqueous, less active metal	Metal deposits	Depends on anion
Aqueous, halide anion	—	Halogen ( $Cl_{2}$ , $Br_{2}$ , $I_{2}$ )
Aqueous, $oxyanion/F^{-}$	—	$O_{2}$

🔑 Quick Rule: Active metals (Na, K, Ca, Al) can’t be deposited from aqueous solution—you get $H_{2}$ instead. Use molten salts for these metals.

The One-Step Mass Formula

$\boxed{m = \frac{ItM}{nF}}$

This combines all steps into one equation.

Mixed Electrolysis Problems 🎯

Calculation Workshop 🧮

1) $I = 4.00$ A, $t = 50.0$ min. Total charge in coulombs?

2) Using the charge from (1), how many grams of Ni deposit from $Ni^{2+}$ ? ( $n = 2$ , $M_{\text{Ni}} = 58.69$ g/mol) (to 3 significant figures)

3) In the electrolysis of molten $CaCl_{2}$ , what forms at the cathode? (type "Ca" or "Cl2")

Problem Solving Strategies 🔽

Exit Quiz — Problem-Solving Workshop ✅

Part 7: Synthesis & AP Review

🎯 Synthesis & AP Review — Electrolytic Cells and Faraday

Part 7 of 7 — Complete Mastery

Bringing It All Together

This comprehensive review connects every concept from Parts 1–6 with AP-style problems. The questions are designed to mirror what you'll see on the actual exam — multi-step, multi-concept, and requiring clear written explanations.

🔑 Why this matters: AP Chemistry exam questions rarely test one concept in isolation — success requires connecting ideas across topics.

What You'll Master in Part 7

Solving AP-style questions that integrate multiple concepts from this unit
Writing clear, concise explanations using proper chemistry terminology
Identifying and avoiding common AP exam traps and mistakes

📋 Master Summary

Galvanic vs. Electrolytic

	Galvanic	Electrolytic
$\Delta G$	$< 0$	$> 0$
$E_{\text{cell}}$	$> 0$	$< 0$
Energy	Chemical → Electrical	Electrical → Chemical
Anode	−	+
Cathode	+	−

💡 Memory Trick: In both cell types, the anode is where oxidation occurs (AN OX) and the cathode is where reduction occurs (RED CAT). Only the signs flip!

Predicting Aqueous Electrolysis Products

Cathode: Metal deposits if $E°_{\text{metal}} > -0.83$ V; otherwise $H_{2}$

Anode: Halide → halogen; $oxyanion/F^{-}$ → $O_{2}$

🔑 AP Quick Check: If the metal is above $H_{2}$ in the activity series (active metals like Na, K, Al), then $H_{2}$ forms at the cathode instead of the metal.

Faraday's Law

$\boxed{m = \frac{ItM}{nF}}$

$q = It \quad \text{mol } e^- = \frac{q}{F} \quad \text{mol substance} = \frac{\text{mol } e^-}{n}$

Industrial Applications

Process	Input	Product
Hall-Héroult	$Al_{2}O_{3}$ in cryolite	Al metal
Chlor-alkali	NaCl(aq)	$Cl_{2}$ , $H_{2}$ , NaOH
Electrorefining	Impure Cu	99.99% pure Cu
Electroplating	Metal ion solution	Metal-coated object

Comprehensive AP Review 🎯

Integration Problems 🧮

1) How many grams of Al can be produced from $Al^{3+}$ ( $n = 3$ , $M = 26.98$ g/mol) using $I = 100$ A for $1.00$ hour?

2) In the electrolysis of aqueous NaI, what gas forms at the cathode? (type "H2" or "O2" or "Na")

3) In the electrolysis of aqueous NaI, what forms at the anode? (type "I2" or "O2" or "Na")

Round all answers to 3 significant figures.

Final Concept Review 🔽

Final Exit Quiz — Electrolysis Mastery ✅

← Back to Standard Lesson

Ion	$E°$ (V)	Ion	$E°$ (V)
$Au^{3+}$	+1.50	$Zn^{2+}$	−0.76
$Ag^{+}$	+0.80	$Al^{3+}$	−1.66
$Cu^{2+}$	+0.34	$Na^{+}$	−2.71
$Ni^{2+}$	−0.26	$K^{+}$	−2.93

Half-reaction	$E°$ (V)
$2Cl^{-}$ → $Cl_{2}$ + $2e^{-}$	+1.36
$2Br^{-}$ → $Br_{2}$ + $2e^{-}$	+1.07
$2I^{-}$ → $I_{2}$ + $2e^{-}$	+0.54
$2H_{2}O$ → $O_{2}$ + $4H^{+}$ + $4e^{-}$	+1.23