🎯⭐ INTERACTIVE LESSON

Electrolytic Cells and Quantitative Electrolysis

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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

In a galvanic cell, a spontaneous reaction produces electricity. In an electrolytic cell, it is the opposite: we use an external power source to force a non-spontaneous reaction to occur. This process is called electrolysis.

🔧 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
$E_{\text{cell}}$

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

Understanding the similarities and differences between galvanic and electrolytic cells is one of the most frequently tested concepts on the AP Chemistry exam. Let's compare them side by side.

⚖️ Complete Comparison

Feature	Galvanic Cell	Electrolytic Cell
Spontaneous?	Yes ( $\Delta G < 0$ )	No ( $\Delta G > 0$ )

Part 3: Electrolysis of Molten Salts

🧪 Electrolysis of Molten Salts and Aqueous Solutions

Part 3 of 7 — Predicting Products

One of the trickiest parts of electrolysis is predicting what forms at each electrode. The products depend on whether you are electrolyzing a molten salt or an aqueous solution.

🔋 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): $\text{Na}^+(l) + e^- \rightarrow \text{Na}(l)$

Part 4: Electrolysis of Aqueous Solutions

⚖️ Faraday's Laws of Electrolysis

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

Faraday's laws connect the amount of substance produced or consumed during electrolysis to the electric current and time. This is one of the most calculation-heavy topics on the AP exam.

📏 Faraday's Laws

The Key Equation

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

Part 5: Faraday\'s Laws of Electrolysis

🏭 Electroplating and Industrial Applications

Part 5 of 7 — Real-World Electrolysis

Electrolysis has enormous industrial importance. From electroplating jewelry to producing aluminum, these applications demonstrate the practical power of electrochemistry.

🔋 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., AgNO₃)

How It Works

At the anode: plating metal dissolves → $\text{Ag}(s) \rightarrow \text{Ag}^+(aq) + e^-$

Part 6: Problem-Solving Workshop

🛠️ Problem-Solving Workshop — Electrolysis and Faraday

Part 6 of 7 — Practice and Integration

This workshop combines all electrolysis concepts: cell comparisons, product prediction, and Faraday's law calculations. These are the exact problem types you will face on the AP exam.

🛠️ Problem-Solving Checklist

For Faraday's Law Problems

✅ Convert time to seconds ( $1 \text{ min} = 60 \text{ s}$ , $1 \text{ hr} = 3600 \text{ s}$ )
✅ Calculate charge:

Part 7: Synthesis & AP Review

🎯 Synthesis & AP Review — Electrolytic Cells and Faraday

Part 7 of 7 — Complete Mastery

This final review integrates everything about electrolytic cells: the comparison with galvanic cells, predicting products, Faraday's law calculations, and industrial applications. Master these and you will own the electrochemistry portion of the AP exam.

📋 Master Summary

Galvanic vs. Electrolytic

	Galvanic	Electrolytic
$\Delta G$	$< 0$	$> 0$

🔑 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!

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

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

🔋 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 ✅

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

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

Salt	Cathode Product	Anode Product
NaCl	Na(l)	Cl₂(g)
MgCl₂	Mg(l)	Cl₂(g)
Al₂O₃	Al(l)	O₂(g)
CaBr₂	Ca(l)	Br₂(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)
$\text{Au}^{3+} + 3e^- \rightarrow \text{Au}$	$+1.50$
$\text{Ag}^+ + e^- \rightarrow \text{Ag}$	$+0.80$
$\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$	$+0.34$
$2\text{H}_2\text{O} + 2e^- \rightarrow \text{H}_2 + 2\text{OH}^-$
$\text{Ni}^{2+} + 2e^- \rightarrow \text{Ni}$	$-0.26$
$\text{Zn}^{2+} + 2e^- \rightarrow \text{Zn}$	$-0.76$
$\text{Al}^{3+} + 3e^- \rightarrow \text{Al}$	$-1.66$
$\text{Na}^+ + e^- \rightarrow \text{Na}$	$-2.71$
$\text{K}^+ + e^- \rightarrow \text{K}$	$-2.93$

🔑 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	Cu²⁺, Ag⁺, Au³⁺, Ni²⁺
$< -0.83$ V	H₂ gas forms	Na⁺, K⁺, Al³⁺

⬆️ 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⁻ → Cl₂, Br⁻ → Br₂, I⁻ → I₂
Oxyanions or F⁻	Water is oxidized → O₂	SO₄²⁻, NO₃⁻, F⁻

⚠️ Why do halides win even though their $E°$ is less favorable? Overpotential! The kinetic barrier for O₂ 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	H₂ forms
3	Anode: Is anion a simple halide?	Halide is oxidized	O₂ 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)
Au³⁺ +1.50 Zn²⁺ −0.76
Ag⁺ +0.80 Al³⁺ −1.66
Cu²⁺ +0.34 Na⁺ −2.71
Ni²⁺ −0.26 K⁺ −2.93

Oxidation potentials (anode):

Half-reaction $E°$ (V)
2Cl⁻ → Cl₂ + 2e⁻ +1.36
2Br⁻ → Br₂ + 2e⁻ +1.07
2I⁻ → I₂ + 2e⁻ +0.54
2H₂O → O₂ + 4H⁺ + 4e⁻ +1.23

Predicting Electrolysis Products 🔽

📋 Reduction potentials:

Ion $E°$ (V) Ion $E°$ (V)
Au³⁺ +1.50 Zn²⁺ −0.76
Ag⁺ +0.80 Al³⁺ −1.66
Cu²⁺ +0.34 Na⁺ −2.71
Ni²⁺ −0.26 K⁺ −2.93

Anode: 2Cl⁻ → Cl₂ (+1.36 V) | 2Br⁻ → Br₂ (+1.07 V) | 2I⁻ → I₂ (+0.54 V) | H₂O → O₂ (+1.23 V)

Rules: Metal $E°$ above −0.83 V → metal deposits. Below → H₂. Simple halides → oxidized. Oxyanions → O₂.

Product Identification 🧮

📋 Reference: Ag⁺ $E°$ = +0.80 V | Cu²⁺ = +0.34 V | Ni²⁺ = −0.26 V | Zn²⁺ = −0.76 V | Water cathode = −0.83 V | Na⁺ = −2.71 V | K⁺ = −2.93 V | Mg²⁺ = −2.37 V

Anode: Halides (Cl⁻, Br⁻, I⁻) → oxidized. Oxyanions (NO₃⁻, SO₄²⁻) → O₂ forms.

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

1) Molten MgCl₂ (cathode product)?

2) Aqueous AgNO₃ (cathode product — is Ag⁺ or H₂O reduced)?

3) Aqueous KI (anode product — is I⁻ or H₂O oxidized)?

Exit Quiz — Electrolysis Products ✅

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

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

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

Half-reaction: $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$ ( $n = 2$ )

Step 1: Total charge $q = It = (2.00)(3600) = 7200 \text{ C}$

Step 2: Moles of electrons $\text{mol } e^- = \frac{7200}{96{,}485} = 0.07462 \text{ mol}$

Step 3: Moles of Cu $\text{mol Cu} = \frac{\text{mol } e^-}{n} = \frac{0.07462}{2} = 0.03731 \text{ mol}$

Step 4: Mass of Cu $m = (0.03731)(63.55) = 2.37 \text{ g}$

Alternative One-Step Formula

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

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 ✅

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)

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

Al₂O₃ 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: $\text{C} + \text{O}^{2-} \rightarrow \text{CO}_2 + e^-$

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
Cu²⁺ 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

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 ✅

✅ 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 (Cl₂, O₂, Br₂)
Aqueous, active metal	H₂	Depends on anion
Aqueous, less active metal	Metal deposits	Depends on anion
Aqueous, halide anion	—	Halogen (Cl₂, Br₂, I₂)
Aqueous, oxyanion/F⁻	—	O₂

The One-Step Mass Formula

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

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 Ni²⁺? ( $n = 2$ , $M_{\text{Ni}} = 58.69$ g/mol) (to 3 significant figures)

3) In the electrolysis of molten CaCl₂, what forms at the cathode? (type "Ca" or "Cl2")

Problem Solving Strategies 🔽

Exit Quiz — Problem-Solving Workshop ✅

Predicting Aqueous Electrolysis Products

Cathode: Metal deposits if $E°_{\text{metal}} > -0.83$ V; otherwise H₂

Anode: Halide → halogen; oxyanion/F⁻ → O₂

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

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

Industrial Applications

Process	Input	Product
Hall-Héroult	Al₂O₃ in cryolite	Al metal
Chlor-alkali	NaCl(aq)	Cl₂, H₂, 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 Al³⁺ ( $n = 3$ , $M = 26.98$ 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 ✅

(2)(96,485)(2.00)(3600)(63.55)

Produces ~65 million tonnes of Al per year worldwide

Electrolytic Cells and Quantitative Electrolysis

⚠️ Overpotential — The Hidden Cost

🧪 Example: Electrolysis of Water

What STAYS THE SAME

What CHANGES

🔋 Recharging: Galvanic → Electrolytic

⬇️ Discharging (Galvanic Mode)

⬆️ Charging (Electrolytic Mode)

🔀 What Swaps During Charging?

Molten Salt Rule

Examples

🧪 Electrolysis of Aqueous Solutions

The Complication: Water Competes!

📋 Key Reduction Potentials to Know

⬇️ At the Cathode — Which Gets Reduced?

⬆️ At the Anode — Which Gets Oxidized?

🗺️ Quick Decision Flowchart

Step-by-Step Problem Solving

Important: What Is n?

🧪 Worked Example

Half-reaction: $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$ ( $n = 2$ )

Alternative One-Step Formula

Controlling Thickness

Common Plating Metals

📌 Major Industrial Processes

1. Hall-Héroult Process (Aluminum Production)

2. Chlor-Alkali Process

3. Electrorefining of Copper

For Product Prediction

The One-Step Mass Formula

Predicting Aqueous Electrolysis Products

Faraday's Law

Industrial Applications

Ion	$E°$ (V)	Ion	$E°$ (V)
Au³⁺	+1.50	Zn²⁺	−0.76
Ag⁺	+0.80	Al³⁺	−1.66
Cu²⁺	+0.34	Na⁺	−2.71
Ni²⁺	−0.26	K⁺	−2.93

Half-reaction	$E°$ (V)
2Cl⁻ → Cl₂ + 2e⁻	+1.36
2Br⁻ → Br₂ + 2e⁻	+1.07
2I⁻ → I₂ + 2e⁻	+0.54
2H₂O → O₂ + 4H⁺ + 4e⁻	+1.23

Electrolytic Cells and Quantitative Electrolysis

Electrolytic Cells and Quantitative Electrolysis - Complete Interactive Lesson

Part 1: Electrolysis Basics

⚡ Electrolysis — Driving Non-Spontaneous Reactions

🔧 How Electrolysis Works

The Key Idea

Requirements

Electrode Conventions in Electrolytic Cells

⚡ Energy Considerations

The Thermodynamic Reality

Part 2: Electrolytic vs Galvanic Cells

🔄 Galvanic vs. Electrolytic Cells

⚖️ Complete Comparison

Part 3: Electrolysis of Molten Salts

🧪 Electrolysis of Molten Salts and Aqueous Solutions

🔋 Electrolysis of Molten Salts

Why Molten?

Simple Case: Molten NaCl

Part 4: Electrolysis of Aqueous Solutions

⚖️ Faraday's Laws of Electrolysis

📏 Faraday's Laws

The Key Equation

Part 5: Faraday\'s Laws of Electrolysis

🏭 Electroplating and Industrial Applications

🔋 Electroplating

Setup

How It Works

Part 6: Problem-Solving Workshop

🛠️ Problem-Solving Workshop — Electrolysis and Faraday

🛠️ Problem-Solving Checklist

For Faraday's Law Problems

Part 7: Synthesis & AP Review

🎯 Synthesis & AP Review — Electrolytic Cells and Faraday

📋 Master Summary

Galvanic vs. Electrolytic

⚠️ Overpotential — The Hidden Cost

🧪 Example: Electrolysis of Water

What STAYS THE SAME

What CHANGES

🔋 Recharging: Galvanic → Electrolytic

⬇️ Discharging (Galvanic Mode)

⬆️ Charging (Electrolytic Mode)

🔀 What Swaps During Charging?

Molten Salt Rule

Examples

🧪 Electrolysis of Aqueous Solutions

The Complication: Water Competes!

📋 Key Reduction Potentials to Know

⬇️ At the Cathode — Which Gets Reduced?

⬆️ At the Anode — Which Gets Oxidized?

🗺️ Quick Decision Flowchart

Step-by-Step Problem Solving

Important: What Is n?

🧪 Worked Example

Half-reaction: Cu2++2e−→Cu\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}Cu2++2e−→Cu (n=2n = 2n=2)

Alternative One-Step Formula

Controlling Thickness

Common Plating Metals

📌 Major Industrial Processes

1. Hall-Héroult Process (Aluminum Production)

2. Chlor-Alkali Process

3. Electrorefining of Copper

For Product Prediction

The One-Step Mass Formula

Predicting Aqueous Electrolysis Products

Faraday's Law

Industrial Applications

Half-reaction: $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$ ( $n = 2$ )