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:

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:

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

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

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

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

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