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Electrolysis

Electrolysis uses electrical energy to drive non-spontaneous redox reactions. Faraday's laws relate charge passed to mass deposited—essential for electroplating, water splitting, and metal refining.

Concept Fundamentals
Mass
Charge
Time
Current
Calculate Electrolysis ParametersMass, time, or current from Faraday's laws

Why This Chemistry Calculation Matters

Why: Electrolysis powers electroplating, chlor-alkali industry, aluminum production, and hydrogen generation. Faraday's laws give quantitative predictions.

How: Q = I×t; moles e⁻ = Q/F; mass = (M×Q)/(n×F). Choose the unknown and solve.

  • Silver plating: Ag⁺ + e⁻ → Ag (n=1). Copper: Cu²⁺ + 2e⁻ → Cu (n=2).
  • Water splitting: 2H₂O → 2H₂ + O₂ requires ~1.23 V minimum.
  • F = 96,485 C/mol. One Faraday deposits 1 mol of monovalent ions.
  • Current efficiency < 100% in practice due to side reactions.

Electrolysis Examples

⚡ Silver Electroplating

Deposit 5g of silver using 2A current

🔩 Copper Refining

Calculate time to deposit 100g copper at 5A

🏭 Aluminum Production

Mass of aluminum from 1000A for 1 hour

⚗️ Hydrogen Production

Calculate hydrogen gas from water electrolysis

✨ Gold Electroplating

Thin gold layer - 0.1g at 0.5A

🔧 Zinc Galvanizing

Zinc coating calculation

🧪 Chlorine Production

Industrial chlorine from brine

⚙️ Nickel Plating

Nickel electroplating process

🔩 Chrome Plating

Chromium electroplating

🔋 Lead Battery

Lead deposition in battery

💨 Oxygen Production

Oxygen from water electrolysis

🥫 Tin Plating

Tin electroplating

⚒️ Iron Refining

Iron electrorefining

🔋 Cadmium Battery

Cadmium in Ni-Cd battery

🔌 Copper Wire Production

Copper deposition for wire

Calculate Electrolysis Parameters

Molar mass of the substance
Number of electrons per ion

⚠️For educational and informational purposes only. Verify with a qualified professional.

🔬 Chemistry Facts

m = (M × I × t) / (n × F) — Faraday's law for mass.

— IUPAC

🔋

1 F = 96,485 C deposits 1 mol e⁻ worth of product.

— Electrochemistry

Electroplating: control thickness via current and time.

— Industry

📐

Water electrolysis: 2 mol e⁻ per mol H₂; 4 mol e⁻ per mol O₂.

— Faraday

Faraday's Laws of Electrolysis

Faraday's laws describe the quantitative relationship between the amount of substance deposited or dissolved during electrolysis and the quantity of electric charge passed through the electrolyte.

m = (M × I × t) / (n × F)

m = mass (g), M = molar mass (g/mol), I = current (A), t = time (s), n = electrons, F = 96,485 C/mol

Common Electrochemical Substances

SubstanceFormulaMolar Mass (g/mol)ElectronsHalf-ReactionCommon Use
Silver ext{Ag}107.871Ag⁺ + e⁻ → AgElectroplating, jewelry
Copper ext{Cu}63.552Cu²⁺ + 2e⁻ → CuElectroplating, electronics
Zinc ext{Zn}65.382Zn²⁺ + 2e⁻ → ZnGalvanizing, batteries
Aluminum ext{Al}26.983Al³⁺ + 3e⁻ → AlMetal production, aerospace
Nickel ext{Ni}58.692Ni²⁺ + 2e⁻ → NiElectroplating, batteries
Chromium ext{Cr}523Cr³⁺ + 3e⁻ → CrChrome plating
Gold ext{Au}196.971Au⁺ + e⁻ → AuJewelry, electronics
HydrogenH_{2}2.01622H⁺ + 2e⁻ → H₂Fuel production
OxygenO_{2}3242H₂O → O₂ + 4H⁺ + 4e⁻Water electrolysis
ChlorineCl_{2}70.922Cl⁻ → Cl₂ + 2e⁻Chlor-alkali industry
Lead ext{Pb}207.22Pb²⁺ + 2e⁻ → PbBatteries, refining
Tin ext{Sn}118.712Sn²⁺ + 2e⁻ → SnTin plating
Iron ext{Fe}55.852Fe²⁺ + 2e⁻ → FeSteel production
Cadmium ext{Cd}112.412Cd²⁺ + 2e⁻ → CdBatteries, plating

Key Concepts

Faraday's Constant

F = 96,485 C/mol represents the charge of one mole of electrons. This fundamental constant connects electrical charge to chemical amount.

Charge Calculation

Total charge Q = I × t (current × time). This determines how many electrons flow through the circuit.

Stoichiometry

The number of electrons (n) determines the stoichiometry. More electrons per ion means less mass deposited per mole of electrons.

How Does Electrolysis Work?

Electrolysis is the process of using electrical energy to drive a non-spontaneous chemical reaction. When current flows through an electrolyte, ions migrate to electrodes where they gain or lose electrons.

⚡ Step-by-Step Process

1. Charge Flow

Q = I × t

Total charge in Coulombs

Example: 2A × 3600s = 7200 C

2. Moles of Electrons

moles e⁻ = Q / F

7200 C / 96485 C/mol

= 0.0746 mol electrons

3. Moles of Substance

moles = moles e⁻ / n

For Cu²⁺ (n=2):

0.0746 / 2 = 0.0373 mol Cu

4. Mass Deposited

m = moles × M

0.0373 × 63.55

= 2.37 g Cu

When to Use This Calculator

Electrolysis calculations are essential for industrial processes, electroplating operations, metal refining, and electrochemical research.

Electroplating

Calculate coating thickness, deposition time, and current requirements for metal plating operations.

  • Silver/gold jewelry plating
  • Chrome automotive parts
  • Zinc galvanizing
🏭

Metal Refining

Determine production rates and energy requirements for electrolytic metal extraction and purification.

  • Aluminum production
  • Copper electrorefining
  • Zinc extraction
⚗️

Water Electrolysis

Calculate hydrogen and oxygen production rates for fuel generation and industrial gas production.

  • Hydrogen fuel production
  • Oxygen generation
  • Chlor-alkali process

Practical Electrolysis Examples

Example: Silver Electroplating

Given:

  • Current: 2.0 A
  • Time: 1 hour (3600 s)
  • Silver: Ag⁺ + e⁻ → Ag (n=1)
  • Molar mass: 107.87 g/mol

Solution:

Q = 2.0 × 3600 = 7200 C

moles e⁻ = 7200 / 96485 = 0.0746 mol

moles Ag = 0.0746 / 1 = 0.0746 mol

Mass = 0.0746 × 107.87 = 8.05 g

Example: Aluminum Production

Given:

  • Current: 1000 A
  • Time: 1 hour (3600 s)
  • Aluminum: Al³⁺ + 3e⁻ → Al (n=3)
  • Molar mass: 26.98 g/mol

Solution:

Q = 1000 × 3600 = 3,600,000 C

moles e⁻ = 3,600,000 / 96485 = 37.3 mol

moles Al = 37.3 / 3 = 12.4 mol

Mass = 12.4 × 26.98 = 335 g

Limitations and Considerations

⚠️ Real-World Factors

  • • Current efficiency may be less than 100%
  • • Side reactions can consume charge
  • • Overpotential increases energy requirements
  • • Concentration changes affect conductivity
  • • Temperature affects reaction rates

✓ Assumptions Made

  • • 100% current efficiency
  • • No side reactions
  • • Constant current throughout process
  • • Ideal electrode behavior
  • • Uniform current distribution
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