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Overview

🌳 Concept tree

Lesson 3→Electrolysis, Faraday's Laws and Industrial Applications

├── The electrolytic cell

├── Example→electrolysis of copper(II) chloride solution

├── First law

├── Second law

├── The faraday

├── The working formula

├── Worked examples

├── Electroplating

├── Extraction of aluminium

├── Electrorefining of copper

├── Galvanic cells versus electrolytic cells

└── Scientists and the value of electrochemistry

Lesson 3: Electrolysis, Faraday’s Laws and Industrial Applications · 1 / 14

The electrolytic cell

An electrolytic cell is a cell in which electrical energy from an external source is converted into chemical energy by driving a non-spontaneous redox reaction.
It consists of a container with an electrolyte (a solution of a salt, acid or base, or a molten salt) and two electrodes (of the same material such as graphite or platinum, or of different materials such as carbon, copper, zinc).

The electrode joined to the positive pole of the external source is the anode — it attracts anions and oxidation occurs there.
The electrode joined to the negative pole is the cathode — it attracts cations and reduction occurs there.

Electrolytes conduct by movement of ions (ionic conductors), unlike metals, which conduct by movement of electrons (electronic conductors).
Liquid electrolytes are of two kinds solutions of acids, bases or salts, and molten salts.
In this course we study the electrolysis of molten salts and of solutions of salts of metals that come after hydrogen in the series.

Lesson 3: Electrolysis, Faraday’s Laws and Industrial Applications · 2 / 14

Example: electrolysis of copper(II) chloride solution

When the voltage applied is slightly above the reverse potential of the cell, current flows.
Cu²⁺ ions move to the cathode and Cl⁻ ions to the anode:

Anode (+), oxidation2Cl⁻(aq) → Cl₂(g) + 2e⁻ (E°ox = −1.36 V)

Cathode (−), reductionCu²⁺(aq) + 2e⁻ → Cu(s) (E°red = +0.34 V)

TotalCu²⁺(aq) + 2Cl⁻(aq) → Cu(s) + Cl₂(g)

Cell potential= −1.36 + 0.34 = −1.02 V.

The negative sign means the reaction would not occur by itself in a galvanic cell; it happens only because an external source of more than 1.02 V supplies the energy.
Chlorine gas bubbles off at the anode and copper is deposited on the cathode.
Electrolysis is therefore the chemical decomposition of a substance caused by passing an electric current through its molten state or solution.

(Extra, for general understanding) if the electrolyte is the solution of a salt of an active metal (like NaCl solution), water molecules compete with the metal ions at the cathode, and hydrogen — not the active metal — is produced there.
For molten salts there is no water, so the metal itself is formed.
In the course the electrodes behave as shown above only for metals after hydrogen.

Lesson 3: Electrolysis, Faraday’s Laws and Industrial Applications · 3 / 14

First law

The mass of a substance liberated, deposited or dissolved at an electrode is directly proportional to the quantity of electricity that passes through the electrolyte.

Doubling the charge passed doubles the mass formed at the same electrode.
To verify it experimentally, pass different quantities of electricity through the same solution, weigh the metal deposited each time (or the metal lost from the anode) and show that the ratio of the masses equals the ratio of the quantities of electricity.
The quantity of electricity:
Q (coulomb) = I (ampere) × t (second) (1 C = 1 A × 1 s)

Lesson 3: Electrolysis, Faraday’s Laws and Industrial Applications · 4 / 14

Second law

When the same quantity of electricity passes through several electrolytes connected in series, the masses of the different substances liberated are proportional to their equivalent masses.
m₁ / m₂ = (equivalent mass of 1) / (equivalent mass of 2)
The gram equivalent mass of a substance is the mass that gains or loses one mole of electrons during the reaction:
equivalent mass = gram atomic mass / number of charges on the ion (valency, z)
To verify it, connect cells of aluminium chloride (or nitrate), copper(II) sulphate and silver nitrate in series, pass one current through all, and compare the masses at the cathodes.
The masses come in the ratio of their equivalent masses: Al : Cu : Ag = 9 : 31.75 : 108 (27/3, 63.5/2 and 108/1).

Lesson 3: Electrolysis, Faraday’s Laws and Industrial Applications · 5 / 14

The faraday

It was found experimentally that 1 coulomb deposits 0.001118 g of silver.
To deposit one equivalent mass of silver (107.88 g) you need

107.88 / 0.001118 ≈ 96 500 C

By the second law, the same quantity of electricity liberates one equivalent mass of any substance. This quantity is the faraday:
1 F = 96 500 C (the charge carried by one mole of electrons).
General statement passing 1 faraday through an electrolyte dissolves, liberates or deposits one gram equivalent mass of a substance at any electrode.
Because the gram atomic mass = equivalent mass × z, the quantity of electricity needed to deposit one mole (one gram atom) of an element is z faradays: 1 F for Ag (Ag⁺ + e⁻ → Ag), 2 F for Cu (Cu²⁺ + 2e⁻ → Cu), 3 F for Al.

Lesson 3: Electrolysis, Faraday’s Laws and Industrial Applications · 6 / 14

The working formula

mass (g) = (I × t × equivalent mass) / 96 500 = (Q × M) / (z × 96 500)
or mass = (number of faradays) × equivalent mass, number of faradays = Q / 96 500.
The charge passed through the cell is Q = I × t, with Q in coulombs when I is in amperes and t in seconds.
One faraday is 96,500 C, so liberating one mole of an element of valency z requires z faradays.
The equivalent mass of an element is its molar mass divided by its valency z.

Lesson 3: Electrolysis, Faraday’s Laws and Industrial Applications · 7 / 14

Worked examples

Example 1 — mass of metal and gas.→A charge of 12 000 C is passed through molten gold(III) chloride (Au = 197, Cl = 35.5). Find the mass of gold and chlorine.

Au³⁺ + 3e⁻ → Au, equivalent mass of Au = 197/3 = 65.67 g.
2Cl⁻ → Cl₂ + 2e⁻ ; equivalent mass of Cl = 35.5 g.

mass Au = 12 000 × 65.67 / 96 500 = 8.17 g

mass Cl₂ = 12 000 × 35.5 / 96 500 = 4.41 g

(They are produced by the same charge, so the ratio AuCl = 65.67 : 35.5 as the second law predicts.)

Example 2 — charge required.How much electricity is needed to deposit 8.4 g of iron from iron(III) chloride solution (Fe = 56)?

Fe³⁺ + 3e⁻ → Fe, equivalent mass = 56/3 = 18.67 g.
18.67 g needs 96 500 C, so 8.4 g needs (8.4 × 96 500)/18.67 = 43 420 C (≈ 0.45 F).
Example 3 — using current and time. A current of 5.0 A flows for 30 minutes through copper(II) sulphate solution (Cu = 63.5). Mass of copper?

Q = 5.0 × 30 × 60 = 9000 C.
equivalent mass of Cu = 63.5/2 = 31.75 g.
mass = 9000 × 31.75 / 96 500 = 2.96 g.
Example 4 — time required. For how long must a current of 4.0 A flow through silver nitrate solution to deposit 10.8 g of silver (Ag = 108)?

10.8 g = 0.1 mole of Ag, which needs 0.1 F = 9650 C.
t = Q/I = 9650/4.0 = 2412.5 s ≈ 40.2 min.
Example 5 — series cells. The same current flows through AgNO₃ and CuSO₄ cells in series; 2.16 g of silver is deposited. What mass of copper is deposited? Ag : Cu = 108 : 31.75, so Cu = 2.16 × 31.75/108 = 0.635 g.
Example 6 — volume of gas. What volume of chlorine at STP (molar volume 22.4 L) is liberated by 19 300 C? 19 300 C = 0.2 F. 1 F liberates 1 mole of Cl atoms = ½ mol Cl₂ = 11.2 L, so 0.2 F gives 2.24 L.

Lesson 3: Electrolysis, Faraday’s Laws and Industrial Applications · 8 / 14

Electroplating

Electroplating

is forming a thin layer of one metal on the surface of another to give it a shiny attractive appearance or to protect it from corrosion.
Car fittings made of steel are chromium-plated; taps and mixers are plated with chromium or gold; cheap metals are given the value and look of silver or gold.
To silver-plate a (copper) spoon or jug:

Clean the object thoroughly (remove grease and oxide).
Dip it in a solution containing silver ions (e.g. silver nitrate).
Place a bar of pure silver in the same solution.
Connect the object to the negative pole of a d.c. source (it becomes the cathode) and the silver bar to the positive pole (the anode).

Comparison aspect	Main idea	Keywords
Anode	Ag(s) → Ag⁺(aq) + e⁻ (the silver bar dissolves)	silverdissolves
Cathode	Ag⁺(aq) + e⁻ → Ag(s) (silver deposits on the object) The concentration of Ag⁺ in the bath therefore stays constant. The thickness of the layer depends on current and time (Faraday’s laws)	silverdepositsobjectconcentration

Lesson 3: Electrolysis, Faraday’s Laws and Industrial Applications · 9 / 14

Extraction of aluminium

Aluminium is produced by electrolysis of bauxite (Al₂O₃) dissolved in molten cryolite (Na₃AlF₆) with a little fluorspar (CaF₂).
Pure Al₂O₃ melts at about 2045 °C, which is far too high; the cryolite mixture melts at about 950 °C and conducts electricity.
(Newer cells use a mixture of Al, Na and Ca fluorides, which gives a melt with a low melting point and a lower density, so the denser molten aluminium sinks to the bottom more easily.)
The cell is a steel tank lined with graphite (this is the cathode); the anodes are graphite (carbon) rods dipping in the melt.

Comparison aspect	Main idea	Keywords
Cathode (−)	2Al³⁺ + 6e⁻ → 2Al(l)	2Al³
Anode (+)	3O²⁻ → 3/2 O₂(g) + 6e⁻
Total	2Al³⁺ + 3O²⁻ → 2Al + 3/2 O₂ (i.e. 2Al₂O₃ → 4Al + 3O₂)	2Al³2Al₂O₃
•	The oxygen evolved reacts with the hot carbon anodes forming CO and CO₂ so the anodes burn away and must be replaced: C + O₂ → CO₂ 2C + O₂ → 2CO	oxygenevolvedreactscarbon
•	Molten aluminium collects at the bottom and is tapped from a special outlet	Moltenaluminiumcollectsbottom

Lesson 3: Electrolysis, Faraday’s Laws and Industrial Applications · 10 / 14

Electrorefining of copper

Industrial copper (about 99 % pure) still contains iron, zinc, silver and gold, which lower its electrical conductivity, so for electrical wires it is refined to about 99.95 %:

Comparison aspect	Main idea	Keywords
Anode	thick slabs of impure copper	thickslabsimpurecopper
Cathode	thin sheets or wires of pure copper	thinsheetswirespure
Electrolyte	copper(II) sulphate solution (acidified)	coppersulphatesolutionacidified
Voltage	only slightly above the copper cell potential	slightlyabovecoppercell
Anode	Cu(s) → Cu²⁺(aq) + 2e⁻
Cathode	Cu²⁺(aq) + 2e⁻ → Cu(s)
•	Impurities more active than copper (Zn Fe) also dissolve at the anode (Zn → Zn²⁺ + 2e⁻ Fe → Fe²⁺ + 2e⁻) but stay in solution because their reduction potentials are lower than copper’s so they do not deposit at the cathode	Impuritiesactivecopperdissolve
•	Impurities less active than copper (Ag Au) are not oxidised at that voltage they fall from the anode as anode mud at the bottom of the cell and are collected — a valuable by-product (gold and silver can be recovered)	Impuritieslessactivecopper

How to obtain pure gold from a copper wire containing gold impurities: make the impure wire the anode in a copper sulphate cell; copper dissolves and re-deposits on the cathode, and the gold falls as anode mud.

Lesson 3: Electrolysis, Faraday’s Laws and Industrial Applications · 11 / 14

Galvanic cells versus electrolytic cells

Point	Galvanic cell	Electrolytic cell
Energy change	chemical → electrical	electrical → chemical
Reaction	spontaneous	non-spontaneous (forced)
Anode	negative pole, oxidation	positive pole, oxidation
Cathode	positive pole, reduction	negative pole, reduction
Source of current	the cell itself	external d.c. source
Containers	two half-cells + salt bridge	usually one container
Cell emf	positive	negative (for the reaction as a galvanic cell)
Examples	Daniell cell dry cells car battery during discharge	electroplating aluminium extraction copper refining battery being charged

Common to both→oxidation occurs at the anode and reduction at the cathode; electrons move in the external wire from anode to cathode; ions carry the current in the solution.

Lesson 3: Electrolysis, Faraday’s Laws and Industrial Applications · 12 / 14

Scientists and the value of electrochemistry

Electrochemistry developed through the work of many scientists — Volta and Galvani (first cells), Daniell (the Daniell cell), Faraday (laws of electrolysis) and others.
Its applications — batteries for phones, cars and spacecraft; protection against corrosion; plating; and the production of metals like aluminium — are part of everyday life and of the progress of human civilisation.
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Lesson 3: Electrolysis, Faraday’s Laws and Industrial Applications · 13 / 14

Part 14

📖

🔍 In-Depth Explanations, Deductions & Multi-Book Insights:

Chemistry Thanaweya Amma (2026/2027)

Lesson 3 – Electrolysis, Faraday's Laws and Industrial Applications
Detailed Academic Lesson Overview:
Electrolytic cells consume electrical energy from external sources to force non-spontaneous redox reactions.
Michael Faraday established the quantitative relationships between electric charge and mass of substance liberated.
Industrial applications include electroplating, aluminum extraction from bauxite, and copper electro-refining.

🌳 Hierarchy Classification & Conceptual Roadmap

Electrolytic Cells and Faraday's Quantitative Laws

├── 1. Faraday's Laws of Electrolysis

Comparison aspect	Main idea	Keywords
│ ├── First Law	Mass of substance liberated (m) ∝ Electric quantity Q (Q = I · t)	MasssubstanceliberatedElectric
│ ├── Second Law	Masses liberated by same Q ∝ Their chemical equivalent masses	Massesliberatedsamechemical
•	│ └── Chemical Equivalent Mass = Molar Mass / Valency (z)	ChemicalEquivalentMassMolar

├── 2. The Faraday Constant (F)

│ ├── 1 Faraday = 96,500 Coulombs

│ ├── Law of Mass LiberationMass = (I · t · Equivalent Mass) / 96,500

│ └── To liberate 1 mole of any elementQ = z × 1 Faraday (where z is valence)

└── 3. Industrial Electrolysis Applications

Comparison aspect	Main idea	Keywords
├── Electroplating	Object to be plated is the Cathode (-) pure plating metal is Anode (+)	ObjectplatedCathodepure
├── Extraction of Aluminum	Hall-Héroult process electrolyzing Bauxite (Al2O3) dissolved in Cryolite and Fluorspar	HallHéroultprocesselectrolyzing
└── Refining of Blister Copper (99% pure)	Impure copper is Anode pure copper is Cathode recovers Gold and Silver as anode slime	ImpureAnodeCathoderecovers
•	📋 Master Teacher's Blackboard Rules & Exam Secrets	MasterTeacherBlackboardRules
RULE 1	Electric charge Q in Coulombs = Current intensity (Amperes) × Time (Seconds)	ElectricchargeCoulombsCurrent
RULE 2	To precipitate 1 mole of an element of valence z exactly z Faradays (z × 96 500 C) are required	precipitatemoleelementvalence
RULE 3	In copper refining impurities more active than copper (Fe Zn) oxidize into solution but DO NOT deposit at the cathode. Precious impurities less active than copper (Ag Au) do not oxidize and fall as anode mud beneath the anode	copperrefiningimpuritiesactive
📊 Detailed Comparison Matrix	Electrochemical Refining Parameters for Copper Metallurgy	ElectrochemicalRefiningParametersCopper

Electrode / Medium	Material Composition	Electrochemical Reaction Occurring
Anode (+)	Impure Blister Copper (99% Cu + Fe Au)	Cu → Cu²⁺ + 2e⁻ Fe → Fe²⁺ + 2e⁻ Zn → Zn²⁺ + 2e⁻
Cathode (-)	Ultra-pure thin Copper sheet (99.99%)	Cu²⁺ + 2e⁻ → Cu(s) (Deposits pure Cu)
Electrolyte	Aqueous Copper(II) sulfate (CuSO₄) + H₂SO₄	Maintains constant Cu²⁺ concentration in bath
Anode Mud	Precious unreactive metals (Gold Au Silver Ag)	Precipitate as solid gravity slime beneath anode

Lesson 3: Electrolysis, Faraday’s Laws and Industrial Applications · 14 / 14

Curriculum

Chemistry

Chapter 1 - Transition Elements and Iron Metallurgy

Chapter 2 - Chemical Analysis

Chapter 3 - Chemical Equilibrium

Chapter 4 - Electrochemistry

Chapter 5 - Organic Chemistry

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Lesson 3: Electrolysis, Faraday’s Laws and Industrial Applications