Electrolysis and Faraday's Laws

Introduction

Every aluminum can you drink from, every iPhone chip, and every gold-plated connector in your laptop exists because of electrolysis! This forced electrochemical process uses electrical energy to drive non-spontaneous reactions. From refining metals to producing chemicals, electrolysis is the backbone of modern industry.

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Interactive: Electrolysis Demonstration

Watch electrolysis of water - see hydrogen and oxygen forming at electrodes:

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What is Electrolysis?

Electrolysis is the process of using electrical energy to drive a non-spontaneous chemical reaction.

Galvanic vs Electrolytic Cells

Feature	Galvanic Cell	Electrolytic Cell
Energy conversion	Chemical → Electrical	Electrical → Chemical
ΔG	Negative	Positive
E°cell	Positive	Negative
Spontaneity	Spontaneous	Non-spontaneous
Anode	Negative (-)	Positive (+)
Cathode	Positive (+)	Negative (-)
Process at anode	Oxidation	Oxidation
Process at cathode	Reduction	Reduction
Example	Battery	Charging battery, electroplating

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Basic Principles of Electrolysis

Requirements

1. Electrolyte: Molten salt or aqueous solution containing ions
2. Electrodes: Usually inert (Pt, graphite) or reactive (metal being plated)
3. External power source: Battery or DC supply
4. Closed circuit: Complete path for electron flow

Electrode Processes

At Anode (+):

• Oxidation occurs
• Anions migrate toward anode
• Electrons are removed
• Example: $2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^-$

At Cathode (-):

• Reduction occurs
• Cations migrate toward cathode
• Electrons are supplied
• Example: $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$

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Electrolysis of Molten Salts

Example: Molten NaCl

At Cathode (-):

$$\text{Na}^+ + e^- \rightarrow \text{Na}_{(l)}$$

At Anode (+):

$$2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^-$$

Overall:

$$\boxed{2\text{NaCl}_{(l)} \xrightarrow{\text{electrolysis}} 2\text{Na}_{(l)} + \text{Cl}_2}$$

This is how sodium metal is produced industrially!

Example: Molten Al₂O₃ (Hall-Héroult Process)

At Cathode (-):

$$\text{Al}^{3+} + 3e^- \rightarrow \text{Al}_{(l)}$$

At Anode (+):

$$2\text{O}^{2-} \rightarrow \text{O}_2 + 4e^-$$ $$\text{C} + \text{O}_2 \rightarrow \text{CO}_2$$

(carbon anode burns)

Overall:

$$\boxed{2\text{Al}_2\text{O}_3 \xrightarrow{\text{electrolysis}} 4\text{Al} + 3\text{O}_2}$$

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Electrolysis of Aqueous Solutions

Competing Reactions

In aqueous solutions, both the salt ions and water can undergo electrolysis!

At Cathode (reduction):

• Metal cations: $\text{M}^{n+} + ne^- \rightarrow \text{M}$
• Water: $2\text{H}_2\text{O} + 2e^- \rightarrow \text{H}_2 + 2\text{OH}^-$ (E° = -0.83 V)
• H⁺ ions: $2\text{H}^+ + 2e^- \rightarrow \text{H}_2$ (E° = 0.00 V)

At Anode (oxidation):

• Anions: $2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^-$, $2\text{Br}^- \rightarrow \text{Br}_2 + 2e^-$
• Water: $2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4e^-$ (E° = 1.23 V)
• OH⁻ ions: $4\text{OH}^- \rightarrow \text{O}_2 + 2\text{H}_2\text{O} + 4e^-$ (E° = 0.40 V)

Preferential Discharge Rules

At Cathode: Species with higher reduction potential (less negative) is reduced first.

Order: Metal ions (high E°) > H⁺ > Active metals (low E°) > Water

At Anode: Species with lower oxidation potential (less positive) is oxidized first.

Order: Anions (Cl⁻, Br⁻, I⁻) > OH⁻/H₂O > Oxyanions (SO₄²⁻, NO₃⁻)

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Examples of Aqueous Electrolysis

Example 1: Electrolysis of CuSO₄ (aq) with Pt Electrodes

At Cathode (-):

• Choices: Cu²⁺ + 2e⁻ → Cu (E° = +0.34 V) OR 2H₂O + 2e⁻ → H₂ + 2OH⁻ (E° = -0.83 V)
• Cu²⁺ wins (higher E°)

$$\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$$

At Anode (+):

• Choices: SO₄²⁻ oxidation (very difficult) OR 2H₂O → O₂ + 4H⁺ + 4e⁻
• Water wins

$$2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4e^-$$

Overall:

$$\boxed{2\text{CuSO}_4 + 2\text{H}_2\text{O} \rightarrow 2\text{Cu} + 2\text{H}_2\text{SO}_4 + \text{O}_2}$$

Observation: Copper deposits on cathode, O₂ gas at anode, solution becomes acidic

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Example 2: Electrolysis of NaCl (aq) with Pt Electrodes

At Cathode (-):

• Choices: Na⁺ + e⁻ → Na (E° = -2.71 V) OR 2H₂O + 2e⁻ → H₂ + 2OH⁻ (E° = -0.83 V)
• Water wins (much higher E°)

$$2\text{H}_2\text{O} + 2e^- \rightarrow \text{H}_2 + 2\text{OH}^-$$

At Anode (+):

• Choices: 2Cl⁻ → Cl₂ + 2e⁻ (E° = 1.36 V) OR 2H₂O → O₂ + 4H⁺ + 4e⁻ (E° = 1.23 V)
• In concentrated NaCl, Cl⁻ wins (due to overpotential)

$$2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^-$$

Overall:

$$\boxed{2\text{NaCl} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2 + \text{Cl}_2}$$

This is the Chlor-alkali process!

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Example 3: Electrolysis of H₂SO₄ (aq) with Pt Electrodes

At Cathode (-):

$$2\text{H}^+ + 2e^- \rightarrow \text{H}_2$$

At Anode (+):

$$2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4e^-$$

Overall:

$$\boxed{2\text{H}_2\text{O} \xrightarrow{\text{electrolysis}} 2\text{H}_2 + \text{O}_2}$$

This is how hydrogen fuel is produced!

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Faraday’s Laws of Electrolysis

Michael Faraday (1834) discovered quantitative relationships between electricity and chemical change.

Faraday’s First Law

The mass of substance deposited/liberated at an electrode is directly proportional to the quantity of electricity passed.

$$\boxed{m \propto Q}$$ $$\boxed{m = Z \times Q = Z \times I \times t}$$

where:

• $m$ = mass deposited (grams)
• $Z$ = electrochemical equivalent (g/C)
• $Q$ = charge in coulombs (C)
• $I$ = current in amperes (A)
• $t$ = time in seconds (s)

Electrochemical Equivalent (Z)

$$\boxed{Z = \frac{M}{nF}}$$

where:

• $M$ = molar mass (g/mol)
• $n$ = number of electrons transferred
• $F$ = Faraday’s constant = 96,500 C/mol ≈ 96,485 C/mol

1 Faraday (F) = Charge on 1 mole of electrons = 96,500 C

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Faraday’s Second Law

When the same quantity of electricity is passed through different electrolytes, the masses of substances deposited are proportional to their equivalent weights.

$$\boxed{\frac{m_1}{m_2} = \frac{E_1}{E_2} = \frac{M_1/n_1}{M_2/n_2}}$$

where $E$ = equivalent weight = $\frac{\text{Molar mass}}{n}$

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Quantitative Calculations

Formula Summary

Mass deposited:

$$\boxed{m = \frac{M \times I \times t}{nF}}$$ $$\boxed{m = \frac{M \times Q}{nF}}$$

Moles deposited:

$$\boxed{\text{Moles} = \frac{Q}{nF} = \frac{It}{nF}}$$

Volume of gas liberated (at STP):

$$\boxed{V = \frac{22.4 \times I \times t}{nF} \text{ liters}}$$

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Example 1: Copper Deposition

Problem: How much copper is deposited when a current of 2 A flows through CuSO₄ solution for 1 hour?

Given: M(Cu) = 63.5 g/mol, F = 96,500 C/mol

Solution:

Step 1: Calculate charge

$$Q = I \times t = 2 \times 3600 = 7200 \text{ C}$$

Step 2: For Cu²⁺ + 2e⁻ → Cu, n = 2

Step 3: Calculate mass

$$m = \frac{M \times Q}{nF} = \frac{63.5 \times 7200}{2 \times 96500}$$ $$m = \frac{457,200}{193,000} = \boxed{2.37 \text{ g}}$$

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Example 2: Hydrogen Gas Volume

Problem: What volume of H₂ (at STP) is liberated when 5 A current is passed through acidified water for 30 minutes?

Solution:

$$t = 30 \times 60 = 1800 \text{ s}$$ $$Q = 5 \times 1800 = 9000 \text{ C}$$

For 2H⁺ + 2e⁻ → H₂, n = 2

$$\text{Moles of H}_2 = \frac{Q}{nF} = \frac{9000}{2 \times 96500} = 0.0466 \text{ mol}$$ $$V = 0.0466 \times 22.4 = \boxed{1.044 \text{ L}}$$

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Example 3: Faraday’s Second Law Application

Problem: The same current deposits 0.54 g of Ag and x g of Cu from their salt solutions. Find x.

Given: M(Ag) = 108, M(Cu) = 63.5, n(Ag) = 1, n(Cu) = 2

Solution:

$$\frac{m_{\text{Cu}}}{m_{\text{Ag}}} = \frac{M_{\text{Cu}}/n_{\text{Cu}}}{M_{\text{Ag}}/n_{\text{Ag}}}$$ $$\frac{x}{0.54} = \frac{63.5/2}{108/1} = \frac{31.75}{108}$$ $$x = 0.54 \times \frac{31.75}{108} = \boxed{0.159 \text{ g}}$$

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Industrial Applications of Electrolysis

1. Electroplating

Coating a metal object with a thin layer of another metal

Example: Chromium Plating

• Cathode: Object to be plated (e.g., car bumper)
• Anode: Pure chromium
• Electrolyte: Chromic acid (H₂CrO₄)

At cathode: $\text{Cr}^{3+} + 3e^- \rightarrow \text{Cr}$

Purpose: Decoration, corrosion protection

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2. Electrorefining (Purification of Metals)

Example: Copper Refining

• Anode: Impure copper
• Cathode: Pure copper (thin sheet)
• Electrolyte: CuSO₄ + H₂SO₄

At anode: $\text{Cu} \rightarrow \text{Cu}^{2+} + 2e^-$ (impurities fall as anode mud)

At cathode: $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$ (pure copper deposits)

Result: 99.99% pure copper for electronics!

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3. Manufacture of Chemicals

Chlor-Alkali Industry (Nelson-Solvay Cell):

Electrolysis of brine (NaCl solution):

$$2\text{NaCl} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2 + \text{Cl}_2$$

Products:

• Chlorine (Cl₂): PVC plastics, water treatment
• Sodium hydroxide (NaOH): Paper, soap, chemicals
• Hydrogen (H₂): Fuel, ammonia synthesis

Global production: 70+ million tons/year!

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4. Extraction of Metals

Aluminum (Hall-Héroult Process):

• Electrolysis of Al₂O₃ dissolved in molten cryolite (Na₃AlF₆)
• Temperature: 950-1000°C
• Uses 13-15 kWh per kg of Al!

Sodium (Downs Cell):

• Electrolysis of molten NaCl
• Operating temperature: 600°C
• Produces Na and Cl₂

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5. Anodization

Electrolytic passivation to increase oxide layer thickness

Example: Aluminum anodization

• Anode: Aluminum object
• Electrolyte: Sulfuric acid
• Forms thick, protective Al₂O₃ layer (5-25 μm)

Applications:

• iPhone cases, cookware, architectural panels
• Can be dyed for colored finishes!

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Practice Problems

Level 1: JEE Main

Q1. Calculate the mass of copper deposited when 0.5 A current flows through CuSO₄ for 2 hours. Given: M(Cu) = 63.5, F = 96,500 C/mol

Q2. How long will it take to deposit 2.7 g of Al from Al₂(SO₄)₃ using a 3 A current? Given: M(Al) = 27

Q3. What volume of O₂ (at STP) is liberated at the anode when 1 F of electricity is passed through dilute H₂SO₄?

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Level 2: JEE Main/Advanced

Q4. The same quantity of electricity that deposits 0.583 g of Ag from AgNO₃ will deposit how much Cu from CuSO₄? Given: M(Ag) = 108, M(Cu) = 63.5

Q5. A current of 1.5 A is passed through CuSO₄ solution for 40 minutes using platinum electrodes.

• (a) Mass of Cu deposited
• (b) Volume of O₂ released at STP

Q6. During electrolysis of molten NaCl, 2.3 g of Na is deposited. Find the volume of Cl₂ gas (STP) liberated at the anode. Given: M(Na) = 23

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Level 3: JEE Advanced

Q7. Three electrolytic cells containing AgNO₃, CuSO₄, and AlCl₃ are connected in series. The same current deposits 1.08 g of Ag. Calculate:

• (a) Mass of Cu deposited
• (b) Mass of Al deposited

Q8. A solution of Ni²⁺ is electrolyzed between Pt electrodes using 0.1 F of electricity. If 60% of this is used for H₂ evolution, what mass of Ni is deposited? Given: M(Ni) = 58.7

Q9. A solution contains Cu²⁺ and Ag⁺ ions. When electrolysis is carried out, all Ag⁺ deposits first, then Cu²⁺. If total charge passed is 1000 C and 1.62 g of Ag is deposited, find:

• (a) Charge used for Ag deposition
• (b) Mass of Cu deposited Given: M(Ag) = 108, M(Cu) = 63.5

Q10. In the electrolysis of acidified water, if 2.24 L of H₂ (STP) is collected, calculate:

• (a) Volume of O₂ collected
• (b) Total charge passed

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Solutions to Practice Problems

A1.

$$Q = 0.5 \times 2 \times 3600 = 3600 \text{ C}$$ $$m = \frac{63.5 \times 3600}{2 \times 96500} = \boxed{1.185 \text{ g}}$$

A2.

$$Q = \frac{m \times nF}{M} = \frac{2.7 \times 3 \times 96500}{27} = 28,950 \text{ C}$$ $$t = \frac{Q}{I} = \frac{28950}{3} = 9650 \text{ s} = \boxed{160.83 \text{ min}}$$

A3. 4OH⁻ → O₂ + 2H₂O + 4e⁻ (n = 4)

$$V = \frac{1 \times 22.4}{4} = \boxed{5.6 \text{ L}}$$

A4. Moles of Ag = 0.583/108 = 0.0054 mol → 0.0054 F used Cu²⁺ + 2e⁻ → Cu: 0.0054 F deposits 0.0027 mol Cu

$$m = 0.0027 \times 63.5 = \boxed{0.172 \text{ g}}$$

A5. Q = 1.5 × 40 × 60 = 3600 C (a) m(Cu) = (63.5 × 3600)/(2 × 96500) = 1.185 g (b) 2H₂O → O₂ + 4H⁺ + 4e⁻: V = (3600 × 22.4)/(4 × 96500) = 0.209 L

A6. Moles of Na = 2.3/23 = 0.1 mol 2Cl⁻ → Cl₂ + 2e⁻: 0.1 mol e⁻ → 0.05 mol Cl₂

$$V = 0.05 \times 22.4 = \boxed{1.12 \text{ L}}$$

A7. Moles of Ag = 1.08/108 = 0.01 mol = 0.01 F (a) Cu: 0.01 F → 0.005 mol → 0.318 g (b) Al: 0.01 F → 0.00333 mol → 0.09 g

A8. 40% for Ni: 0.04 F Ni²⁺ + 2e⁻ → Ni: 0.04/2 = 0.02 mol

$$m = 0.02 \times 58.7 = \boxed{1.174 \text{ g}}$$

A9. (a) Ag⁺ + e⁻ → Ag: 1.62/108 = 0.015 mol → 1448.25 C (b) Remaining charge = 1000 - 1448.25 = -448.25 (impossible!) Error in problem: Need more charge! If 1.62 g Ag deposited, Q used = 1448 C, so total Q must be > 1448 C.

A10. Moles of H₂ = 2.24/22.4 = 0.1 mol (a) 2H₂O → 2H₂ + O₂: V(O₂) = 1.12 L (b) 2H⁺ + 2e⁻ → H₂: Q = 0.1 × 2 × 96500 = 19,300 C

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Common Mistakes to Avoid

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Key Takeaways for JEE

Essential Formulas

$$\boxed{Q = I \times t}$$ $$\boxed{m = \frac{M \times I \times t}{nF}}$$ $$\boxed{\frac{m_1}{m_2} = \frac{E_1}{E_2}}$$ $$\boxed{1 \text{ Faraday (F)} = 96,500 \text{ C}}$$

Must Remember

1. Faraday’s constant: F = 96,500 C/mol
2. Preferential discharge: Higher E° at cathode, lower E° at anode
3. Electrolysis reverses spontaneous cell reactions (needs external voltage)
4. Same current, different cells → Use Faraday’s Second Law
5. n value depends on electrons in half-reaction, not ion charge alone

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Related Topics

Within Electrochemistry

• Oxidation-Reduction — Electron transfer fundamentals
• Electrochemical Cells — Galvanic vs electrolytic
• Electrode Potentials — Predicting discharge order
• Batteries — Rechargeable batteries use electrolysis

Cross-Chapter Connections

• Chemical Kinetics — Rate of electrode reactions
• Thermodynamics — Energy requirements (ΔG > 0)
• Metallurgy — Metal extraction and refining

Physics Connections

• Current Electricity — Current, resistance, power
• Magnetism — Electromagnetic effects in electrolysis

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