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:

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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