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Electrolytic Cells and Quantitative Electrolysis

Explore electrolysis, compare galvanic vs electrolytic cells, and use Faraday's laws for quantitative calculations.

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Electrolytic Cells and Quantitative Electrolysis

Galvanic vs Electrolytic Cells

Galvanic (Voltaic) Cell:

  • Spontaneous redox reaction
  • Chemical → Electrical energy
  • E°_cell > 0
  • ΔG < 0
  • Example: Batteries

Electrolytic Cell:

  • Non-spontaneous redox forced by external voltage
  • Electrical → Chemical energy
  • E°_cell < 0 (reversed)
  • ΔG > 0 (reversed)
  • Examples: Electroplating, aluminum production, charging batteries

Electrolysis

Electrolysis: Using electric current to drive non-spontaneous redox

Requirements:

  1. External power source (battery, DC supply)
  2. Electrolyte (molten or aqueous)
  3. Two electrodes (inert or active)
  4. Complete circuit

Applications:

  • Metal purification (copper, aluminum)
  • Electroplating (chrome, gold)
  • Producing chemicals (Cl₂, NaOH, H₂)
  • Charging batteries

Electrode Identification in Electrolytic Cells

Anode:

  • Connected to positive terminal
  • Oxidation occurs
  • Electrons leave
  • Opposite of galvanic!

Cathode:

  • Connected to negative terminal
  • Reduction occurs
  • Electrons enter
  • Opposite of galvanic!

Mnemonic still works: AN OX, RED CAT

  • But voltage source reverses polarity

Electrolysis of Molten Salts

Molten NaCl example:

Cathode (−): Na⁺ + e⁻ → Na(l) Anode (+): 2Cl⁻ → Cl₂(g) + 2e⁻

Overall: 2NaCl(l) → 2Na(l) + Cl₂(g)

Simple: Only one species to reduce (Na⁺) and one to oxidize (Cl⁻)

Industrial: Produces Na metal and Cl₂ gas

Electrolysis of Aqueous Solutions

More complex: Water can be oxidized or reduced!

Water reduction: 2H₂O + 2e⁻ → H₂(g) + 2OH⁻ (E° = -0.83 V) Water oxidation: 2H₂O → O₂(g) + 4H⁺ + 4e⁻ (E° = +1.23 V)

Competition at cathode:

  • Most easily reduced species wins
  • Higher (more positive) E°

Competition at anode:

  • Most easily oxidized species wins
  • Lower (more negative) E°

General rules:

Cathode (reduction):

  • Active metals (Na⁺, K⁺, Mg²⁺): H₂O reduced instead → H₂(g)
  • Less active (Cu²⁺, Ag⁺): Metal ion reduced → metal
  • Very negative E°: water wins

Anode (oxidation):

  • Active anions (Cl⁻, Br⁻, I⁻): Anion oxidized → X₂(g)
  • Oxyanions (NO₃⁻, SO₄²⁻): H₂O oxidized → O₂(g)
  • Inert electrode required

Electrolysis Examples

Aqueous NaCl:

Cathode: 2H₂O + 2e⁻ → H₂ + 2OH⁻ (Na⁺ too active) Anode: 2Cl⁻ → Cl₂ + 2e⁻

Overall: 2H₂O + 2Cl⁻ → H₂ + Cl₂ + 2OH⁻

Products: H₂, Cl₂, NaOH (chlor-alkali process)


Aqueous CuSO₄:

Cathode: Cu²⁺ + 2e⁻ → Cu (copper deposits) Anode: 2H₂O → O₂ + 4H⁺ + 4e⁻ (SO₄²⁻ not oxidized)

Overall: 2Cu²⁺ + 2H₂O → 2Cu + O₂ + 4H⁺

Application: Copper purification

Faraday's Laws of Electrolysis

First Law: Mass deposited ∝ charge passed

Second Law: For same charge, mass ∝ molar mass / electrons

Quantitative relationship:

moles e−=charge (C)F\text{moles e}^- = \frac{\text{charge (C)}}{F}

Where F = 96,485 C/mol (Faraday constant)

Charge:

Q=I×tQ = I \times t

  • Q = charge (coulombs, C)
  • I = current (amperes, A)
  • t = time (seconds, s)

Stoichiometry of Electrolysis

Step-by-step:

  1. Calculate charge: Q = I × t
  2. Calculate moles e⁻: mol e⁻ = Q/F
  3. Use stoichiometry: Relate e⁻ to substance
  4. Calculate amount: mol, mass, or volume

Example: Cu²⁺ + 2e⁻ → Cu

  • 2 moles e⁻ → 1 mole Cu
  • If 10 mol e⁻: 5 mol Cu deposited

Electroplating

Electroplating: Coating object with thin metal layer

Setup:

  • Anode: Pure metal (dissolves)
  • Cathode: Object to plate (metal deposits)
  • Electrolyte: Metal ion solution

Example: Chrome plating

Cathode: Cr³⁺ + 3e⁻ → Cr (chrome layer forms) Anode: Cr → Cr³⁺ + 3e⁻ (chrome dissolves)

Thickness controlled by:

  • Current
  • Time
  • Current density (A/cm²)

Metal Purification

Copper purification:

Anode: Impure Cu → Cu²⁺ + 2e⁻ Cathode: Cu²⁺ + 2e⁻ → Pure Cu

Impurities:

  • More active metals (Fe, Zn): Stay dissolved
  • Less active metals (Ag, Au): Fall to bottom (anode sludge)
  • Recovered for value!

Result: 99.99% pure copper

Calculating Products

Gas production:

moles gas=moles e−n\text{moles gas} = \frac{\text{moles e}^-}{n}

Where n = electrons per molecule:

  • H₂: 2e⁻ → 1 H₂
  • O₂: 4e⁻ → 1 O₂
  • Cl₂: 2e⁻ → 1 Cl₂

Volume at STP:

V=moles×22.4 L/molV = \text{moles} \times 22.4 \text{ L/mol}

Battery Charging

Charging = Electrolysis!

Discharging (galvanic):

  • Spontaneous
  • Chemical → Electrical

Charging (electrolytic):

  • Apply reverse voltage
  • Electrical → Chemical
  • Restores reactants

Lead-acid battery:

Discharge: Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O Charge: Reverse reaction forced

Overpotential

Overpotential: Extra voltage needed beyond theoretical

Reasons:

  • Activation energy for electrode reactions
  • Resistance in electrolyte
  • Concentration polarization

Practical: Need E_applied > |E°_cell| to start electrolysis

Example: Water electrolysis E° = -1.23 V, but need ~2 V in practice

📚 Practice Problems

1Problem 1easy

❓ Question:

How many grams of Cu will be deposited at the cathode when a 3.00 A current passes through a CuSO₄ solution for 2.00 hours? Cu²⁺ + 2e⁻ → Cu. F = 96,485 C/mol, Cu = 63.5 g/mol.

💡 Show Solution

Given:

  • Current I = 3.00 A
  • Time t = 2.00 hours = 2.00 × 3600 = 7200 s
  • Cathode reaction: Cu²⁺ + 2e⁻ → Cu
  • F = 96,485 C/mol
  • Molar mass Cu = 63.5 g/mol

Step 1: Calculate charge (Q)

Q=I×tQ = I \times t

Q=3.00 A×7200 sQ = 3.00 \text{ A} \times 7200 \text{ s}

Q=21,600 CQ = 21,600 \text{ C}


Step 2: Calculate moles of electrons

mol e−=QF\text{mol e}^- = \frac{Q}{F}

mol e−=21,60096,485\text{mol e}^- = \frac{21,600}{96,485}

mol e−=0.224 mol e−\text{mol e}^- = 0.224 \text{ mol e}^-


Step 3: Use stoichiometry

From equation: Cu²⁺ + 2e⁻ → Cu

2 mol e⁻ → 1 mol Cu

mol Cu=0.224 mol e−2\text{mol Cu} = \frac{0.224 \text{ mol e}^-}{2}

mol Cu=0.112 mol\text{mol Cu} = 0.112 \text{ mol}


Step 4: Calculate mass

mass=mol×molar mass\text{mass} = \text{mol} \times \text{molar mass}

mass=0.112×63.5\text{mass} = 0.112 \times 63.5

mass=7.11 g\text{mass} = 7.11 \text{ g}

Answer: 7.11 g Cu deposited


Summary:

3.00 A × 2.00 hr → 21,600 C → 0.224 mol e⁻ → 0.112 mol Cu → 7.11 g Cu

Check: Makes sense - about 1/9 mole Cu in 2 hours at 3 A

2Problem 2medium

❓ Question:

A current of 5.00 A is passed through molten NaCl for 30.0 minutes. Calculate: (a) moles of Na produced, (b) volume of Cl₂ gas at STP. Na⁺ + e⁻ → Na, 2Cl⁻ → Cl₂ + 2e⁻.

💡 Show Solution

Given:

  • Current I = 5.00 A
  • Time t = 30.0 min = 30.0 × 60 = 1800 s
  • Cathode: Na⁺ + e⁻ → Na
  • Anode: 2Cl⁻ → Cl₂ + 2e⁻
  • F = 96,485 C/mol

Calculate charge:

Q=I×t=5.00×1800=9000 CQ = I \times t = 5.00 \times 1800 = 9000 \text{ C}

Calculate moles electrons:

mol e−=QF=900096,485=0.0933 mol e−\text{mol e}^- = \frac{Q}{F} = \frac{9000}{96,485} = 0.0933 \text{ mol e}^-


(a) Moles Na produced

Cathode: Na⁺ + e⁻ → Na

1 mol e⁻ → 1 mol Na

mol Na=0.0933 mol\text{mol Na} = 0.0933 \text{ mol}

Answer (a): 0.0933 mol Na


(b) Volume Cl₂ at STP

Anode: 2Cl⁻ → Cl₂ + 2e⁻

2 mol e⁻ → 1 mol Cl₂

mol Cl2=0.09332=0.0467 mol\text{mol Cl}_2 = \frac{0.0933}{2} = 0.0467 \text{ mol}

Volume at STP:

V=mol×22.4 L/molV = \text{mol} \times 22.4 \text{ L/mol}

V=0.0467×22.4V = 0.0467 \times 22.4

V=1.05 LV = 1.05 \text{ L}

Answer (b): 1.05 L Cl₂


Verification:

Overall reaction: 2NaCl → 2Na + Cl₂

2 mol Na : 1 mol Cl₂

0.09330.0467=2.00\frac{0.0933}{0.0467} = 2.00 ✓

Ratio checks!


Summary:

  • 9000 C → 0.0933 mol e⁻
  • Cathode: 0.0933 mol e⁻ → 0.0933 mol Na
  • Anode: 0.0933 mol e⁻ → 0.0467 mol Cl₂ → 1.05 L at STP

Same electrons flow through both electrodes!

3Problem 3hard

❓ Question:

Electrolysis of aqueous NaCl produces H₂ and Cl₂. What current is needed to produce 10.0 L of H₂ gas at STP in 1.50 hours? 2H₂O + 2e⁻ → H₂ + 2OH⁻.

💡 Show Solution

Given:

  • Electrolysis of aqueous NaCl
  • Volume H₂ = 10.0 L at STP
  • Time t = 1.50 hours = 1.50 × 3600 = 5400 s
  • Cathode: 2H₂O + 2e⁻ → H₂ + 2OH⁻
  • Find: Current I

Step 1: Moles H₂

At STP: 1 mol gas = 22.4 L

mol H2=10.0 L22.4 L/mol\text{mol H}_2 = \frac{10.0 \text{ L}}{22.4 \text{ L/mol}}

mol H2=0.446 mol\text{mol H}_2 = 0.446 \text{ mol}


Step 2: Moles electrons

From cathode equation: 2H₂O + 2e⁻ → H₂

2 mol e⁻ → 1 mol H₂

mol e−=0.446×2\text{mol e}^- = 0.446 \times 2

mol e−=0.892 mol\text{mol e}^- = 0.892 \text{ mol}


Step 3: Charge needed

Q=mol e−×FQ = \text{mol e}^- \times F

Q=0.892×96,485Q = 0.892 \times 96,485

Q=86,049 CQ = 86,049 \text{ C}


Step 4: Calculate current

I=QtI = \frac{Q}{t}

I=86,049 C5400 sI = \frac{86,049 \text{ C}}{5400 \text{ s}}

I=15.9 AI = 15.9 \text{ A}

Answer: 15.9 A


Verification:

Working backward:

  • 15.9 A × 5400 s = 85,860 C ✓
  • 85,860/96,485 = 0.890 mol e⁻ ✓
  • 0.890/2 = 0.445 mol H₂ ✓
  • 0.445 × 22.4 = 9.97 L ≈ 10.0 L ✓

Summary:

To produce 10.0 L H₂ in 1.50 hr:

10.0 L→0.446 mol H2→0.892 mol e−→86,049 C→15.9 A10.0 \text{ L} \to 0.446 \text{ mol H}_2 \to 0.892 \text{ mol e}^- \to 86,049 \text{ C} \to 15.9 \text{ A}


What happens at anode?

Anode: 2Cl⁻ → Cl₂ + 2e⁻

Same 0.892 mol e⁻:

mol Cl2=0.8922=0.446 mol\text{mol Cl}_2 = \frac{0.892}{2} = 0.446 \text{ mol}

Volume Cl₂ = 0.446 × 22.4 = 10.0 L

Equal volumes H₂ and Cl₂ produced!

Overall: 2H₂O + 2Cl⁻ → H₂ + Cl₂ + 2OH⁻

Explain using:

📋 AP Chemistry — Exam Format Guide

⏱ 3 hours 15 minutes📝 67 questions📊 3 sections
SectionFormatQuestionsTimeWeightCalculator
Multiple ChoiceMCQ6090 min50%✅
Free Response (Long)FRQ369 min30%✅
Free Response (Short)FRQ436 min20%✅

📊 Scoring: 1-5

5
Extremely Qualified
~12%
4
Well Qualified
~16%
3
Qualified
~24%
2
Possibly Qualified
~24%
1
No Recommendation
~24%

💡 Key Test-Day Tips

  • ✓Memorize common polyatomic ions
  • ✓Practice dimensional analysis
  • ✓Know your gas laws

⚠️ Common Mistakes: Electrolytic Cells and Quantitative Electrolysis

Avoid these 3 frequent errors

🌍 Real-World Applications: Electrolytic Cells and Quantitative Electrolysis

See how this math is used in the real world

📝 Worked Example: Stoichiometry — Limiting Reagent

Problem:

22 mol of H2H_2 reacts with 11 mol of O2O_2. How many grams of water are produced? Which is the limiting reagent? (2H2+O2→2H2O2H_2 + O_2 \to 2H_2O)

2Determine the limiting reagent
3Calculate moles of product
4Convert moles to grams

📌 Related Topics in Electrochemistry

❓ Frequently Asked Questions

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Explore electrolysis, compare galvanic vs electrolytic cells, and use Faraday's laws for quantitative calculations.
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