Electrolytic Cells and Quantitative Electrolysis

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

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:

External power source (battery, DC supply)
Electrolyte (molten or aqueous)
Two electrodes (inert or active)
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:

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

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

Charge:

$Q = I \times t$

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

Stoichiometry of Electrolysis

Step-by-step:

Calculate charge: Q = I × t
Calculate moles e⁻: mol e⁻ = Q/F
Use stoichiometry: Relate e⁻ to substance
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:

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

Where n = electrons per molecule:

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

Volume at STP:

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

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)

$Q = I \times t$

$Q = 3.00 \text{ A} \times 7200 \text{ s}$

$Q = 21,600 \text{ C}$

Step 2: Calculate moles of electrons

$\text{mol e}^- = \frac{Q}{F}$

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

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

Step 3: Use stoichiometry

From equation: Cu²⁺ + 2e⁻ → Cu

2 mol e⁻ → 1 mol Cu

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

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

Step 4: Calculate mass

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

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

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

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:

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

Calculate moles electrons:

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

(a) Moles Na produced

Cathode: Na⁺ + e⁻ → Na

1 mol e⁻ → 1 mol Na

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

Answer (a): 0.0933 mol Na

(b) Volume Cl₂ at STP

Anode: 2Cl⁻ → Cl₂ + 2e⁻

2 mol e⁻ → 1 mol Cl₂

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

Volume at STP:

$V = \text{mol} \times 22.4 \text{ L/mol}$

$V = 0.0467 \times 22.4$

$V = 1.05 \text{ L}$

Answer (b): 1.05 L Cl₂

Verification:

Overall reaction: 2NaCl → 2Na + Cl₂

2 mol Na : 1 mol Cl₂

$\frac{0.0933}{0.0467} = 2.00$ ✓

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

$\text{mol H}_2 = \frac{10.0 \text{ L}}{22.4 \text{ L/mol}}$

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

Step 2: Moles electrons

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

2 mol e⁻ → 1 mol H₂

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

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

Step 3: Charge needed

$Q = \text{mol e}^- \times F$

$Q = 0.892 \times 96,485$

$Q = 86,049 \text{ C}$

Step 4: Calculate current

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

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

$I = 15.9 \text{ A}$

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:

$10.0 \text{ L} \to 0.446 \text{ mol H}_2 \to 0.892 \text{ mol e}^- \to 86,049 \text{ C} \to 15.9 \text{ A}$

What happens at anode?

Anode: 2Cl⁻ → Cl₂ + 2e⁻

Same 0.892 mol e⁻:

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

Volume Cl₂ = 0.446 × 22.4 = 10.0 L

Equal volumes H₂ and Cl₂ produced!

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

🎴

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