🎯⭐ 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

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

Electrode Conventions in Electrolytic Cells

PropertyGalvanic CellElectrolytic Cell
AnodeOxidation ✓Oxidation ✓
CathodeReduction ✓Reduction ✓
Anode sign− (negative)+ (positive)
Cathode sign+ (positive)− (negative)
Spontaneous?YesNo

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

Energy Considerations

For Electrolysis

ΔG>0(non-spontaneous)\Delta G > 0 \quad \text{(non-spontaneous)} Ecell<0(negative cell potential)E_{\text{cell}} < 0 \quad \text{(negative cell potential)}

The external power source must supply at least Ecell|E_{\text{cell}}| volts to drive the reaction.

In Practice: Overpotential

The actual voltage required is usually higher than the theoretical minimum due to overpotential — extra voltage needed to overcome kinetic barriers at the electrode surfaces.

Vapplied=Ecell+overpotentialV_{\text{applied}} = |E_{\text{cell}}| + \text{overpotential}

Example: Electrolysis of Water

2H2O(l)2H2(g)+O2(g)2\text{H}_2\text{O}(l) \rightarrow 2\text{H}_2(g) + \text{O}_2(g)

  • E°=1.23E° = -1.23 V (non-spontaneous)
  • Minimum applied voltage: 1.23 V
  • Typical actual voltage: ~1.8 - 2.0 V (due to overpotential)

Electrolysis Concept Quiz 🎯

Electrolytic Cell Basics 🔽

Electrolysis Energy 🧮

  1. The electrolysis of water has E°=1.23E° = -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

FeatureGalvanic CellElectrolytic Cell
Spontaneous?Yes (ΔG<0\Delta G < 0)No (ΔG>0\Delta G > 0)
EcellE_{\text{cell}}PositiveNegative
Energy conversionChemical → ElectricalElectrical → Chemical
External power?No (produces power)Yes (requires power)
AnodeOxidation (−)Oxidation (+)
CathodeReduction (+)Reduction (−)
Electron flowAnode → CathodeAnode → Cathode
Salt bridgeUsually presentOften not needed
ExampleBattery, fuel cellElectroplating, 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

What CHANGES

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

Recharging: Galvanic → Electrolytic

When you recharge a battery, you convert it from a galvanic cell to an electrolytic cell:

Discharging (Galvanic Mode)

Pb(s)+PbO2(s)+2H2SO4(aq)2PbSO4(s)+2H2O(l)\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)

  • Spontaneous: E>0E > 0
  • Produces electrical energy

Charging (Electrolytic Mode)

2PbSO4(s)+2H2O(l)Pb(s)+PbO2(s)+2H2SO4(aq)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)

  • Non-spontaneous: E<0E < 0
  • Consumes electrical energy (from the charger)
  • The reaction is driven in reverse

Key Point

The anode and cathode swap when switching between galvanic and electrolytic modes! The electrode that was the anode during discharge becomes the cathode during charging.

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

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

At the anode (oxidation): 2Cl(l)Cl2(g)+2e2\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

SaltCathode ProductAnode Product
NaClNa(l)Cl₂(g)
MgCl₂Mg(l)Cl₂(g)
Al₂O₃Al(l)O₂(g)
CaBr₂Ca(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.

At the Cathode (Which Gets Reduced?)

Compare the metal ion vs. water:

2H2O(l)+2eH2(g)+2OH(aq)E°=0.83 V2\text{H}_2\text{O}(l) + 2e^- \rightarrow \text{H}_2(g) + 2\text{OH}^-(aq) \quad E° = -0.83 \text{ V}

  • If the metal has E°>0.83E° > -0.83 V (e.g., Cu²⁺, Ag⁺): metal is deposited
  • If the metal has E°<0.83E° < -0.83 V (e.g., Na⁺, K⁺, Al³⁺): H₂ gas forms

At the Anode (Which Gets Oxidized?)

Compare the anion vs. water:

2H2O(l)O2(g)+4H+(aq)+4eE°=+1.23 V2\text{H}_2\text{O}(l) \rightarrow \text{O}_2(g) + 4\text{H}^+(aq) + 4e^- \quad E° = +1.23 \text{ V}

  • Simple anions (Cl⁻, Br⁻, I⁻): anion is oxidized (due to overpotential effects)
  • Oxyanions (SO₄²⁻, NO₃⁻) or F⁻: water is oxidized → O₂ forms

Summary Rules for Aqueous Electrolysis

CathodeAnode
Less active metals (Cu, Ag, Au) depositSimple halides (Cl⁻, Br⁻, I⁻) oxidize
Active metals (Na, K, Al) → H₂ formsOxyanions → O₂ forms

Electrolysis Product Quiz 🎯

Predicting Electrolysis Products 🔽

Product Identification 🧮

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

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

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

SymbolMeaningUnits
IICurrentAmperes (A) = C/s
ttTimeSeconds (s)
nnElectrons per ion in the half-reaction
FFFaraday's constant96,48596{,}485 C/mol ee^-
ItItTotal chargeCoulombs (C)

Step-by-Step Problem Solving

  1. Calculate total charge: q=Itq = It (coulombs)
  2. Find moles of electrons: mol e=q/F=It/F\text{mol } e^- = q/F = It/F
  3. Use stoichiometry: relate moles of electrons to moles of substance using nn
  4. Convert to mass if needed: m=mol×Mm = \text{mol} \times M

Important: What Is n?

nn = number of electrons in the balanced half-reaction

Half-Reactionnn
Ag++eAg\text{Ag}^+ + e^- \rightarrow \text{Ag}1
Cu2++2eCu\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}2
Al3++3eAl\text{Al}^{3+} + 3e^- \rightarrow \text{Al}3
2ClCl2+2e2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^-2

Worked Example

How many grams of Cu are deposited by passing a current of 2.002.00 A through CuSO4\text{CuSO}_4 solution for 1.001.00 hour?

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

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

Step 2: Moles of electrons mol e=720096,485=0.07462 mol\text{mol } e^- = \frac{7200}{96{,}485} = 0.07462 \text{ mol}

Step 3: Moles of Cu mol Cu=mol en=0.074622=0.03731 mol\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 gm = (0.03731)(63.55) = 2.37 \text{ g}

Alternative One-Step Formula

m=ItMnF=(2.00)(3600)(63.55)(2)(96,485)=2.37 gm = \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=96,485F = 96{,}485 C/mol, MAg=107.87M_{\text{Ag}} = 107.87 g/mol

  1. A current of 5.005.00 A flows for 10001000 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? (Ag++eAg\text{Ag}^+ + e^- \rightarrow \text{Ag}, n=1n = 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

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

  1. At the anode: plating metal dissolves → Ag(s)Ag+(aq)+e\text{Ag}(s) \rightarrow \text{Ag}^+(aq) + e^-
  2. Ag⁺ ions migrate through solution
  3. At the cathode: metal ions deposit → Ag+(aq)+eAg(s)\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 (II): higher current → faster deposition
  • Time (tt): longer time → thicker coating
  • Faraday's law: m=ItM/(nF)m = ItM/(nF)

Common Plating Metals

MetalApplication
ChromeCar bumpers, faucets
SilverJewelry, silverware
GoldElectronics, jewelry
NickelCorrosion protection
ZincGalvanization of steel

Major Industrial Processes

1. Hall-Héroult Process (Aluminum Production)

2Al2O3(l)4Al(l)+3O2(g)2\text{Al}_2\text{O}_3(l) \rightarrow 4\text{Al}(l) + 3\text{O}_2(g)

  • Al₂O₃ is dissolved in molten cryolite (Na3AlF6\text{Na}_3\text{AlF}_6) to lower the melting point
  • Enormous current (100,000+ A!)
  • Carbon anodes are consumed: C+O2CO2+e\text{C} + \text{O}^{2-} \rightarrow \text{CO}_2 + e^-
  • Produces ~65 million tonnes of Al per year worldwide

2. Chlor-Alkali Process

2NaCl(aq)+2H2O(l)Cl2(g)+H2(g)+2NaOH(aq)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=2.0I = 2.0 A for 2020 minutes. Ag++eAg\text{Ag}^+ + e^- \rightarrow \text{Ag}, n=1n = 1, MAg=107.87M_{\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

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

  1. ✅ Convert time to seconds (1 min=60 s1 \text{ min} = 60 \text{ s}, 1 hr=3600 s1 \text{ hr} = 3600 \text{ s})
  2. ✅ Calculate charge: q=Itq = It
  3. ✅ Find mol electrons: mol e=q/F\text{mol } e^- = q/F
  4. ✅ Write the half-reaction to find nn
  5. ✅ Find mol substance: mol=mol e/n\text{mol} = \text{mol } e^-/n
  6. ✅ Convert to mass or volume if needed

For Product Prediction

SystemCathode ProductAnode Product
Molten saltMetalNonmetal (Cl₂, O₂, Br₂)
Aqueous, active metalH₂Depends on anion
Aqueous, less active metalMetal depositsDepends on anion
Aqueous, halide anionHalogen (Cl₂, Br₂, I₂)
Aqueous, oxyanion/F⁻O₂

The One-Step Mass Formula

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

This combines all steps into one equation.

Mixed Electrolysis Problems 🎯

Calculation Workshop 🧮

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

  2. Using the charge from (1), how many grams of Ni deposit from Ni²⁺? (n=2n = 2, MNi=58.69M_{\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

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

GalvanicElectrolytic
ΔG\Delta G<0< 0>0> 0
EcellE_{\text{cell}}>0> 0<0< 0
EnergyChemical → ElectricalElectrical → Chemical
Anode+
Cathode+

Predicting Aqueous Electrolysis Products

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

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

Faraday's Law

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

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

Industrial Applications

ProcessInputProduct
Hall-HéroultAl₂O₃ in cryoliteAl metal
Chlor-alkaliNaCl(aq)Cl₂, H₂, NaOH
ElectrorefiningImpure Cu99.99% pure Cu
ElectroplatingMetal ion solutionMetal-coated object

Comprehensive AP Review 🎯

Integration Problems 🧮

  1. How many grams of Al can be produced from Al³⁺ (n=3n = 3, M=26.98M = 26.98 g/mol) using I=100I = 100 A for 1.001.00 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