Electrolysis Calculator

Calculate mass, charge, time, or current using Faraday's laws of electrolysis

Faraday's Laws of Electrolysis

Calculate mass of product, charge, time, or current required for electrolysis.

m = (Q × M) / (n × F)

Q = I × t  |  F = 96,485 C/mol

Calculate:

Common Substances:

Example Problems:

Understanding Faraday's Laws

First Law:

The mass of substance deposited/liberated at an electrode is directly proportional to the quantity of electric charge passed through the electrolyte.

Key Relationships:

  • Q = I × t - Charge equals current times time
  • F = 96,485 C/mol - Faraday's constant (charge per mole of electrons)
  • m = (Q × M) / (n × F) - Mass from charge, molar mass, and electron transfer

What is Electrolysis?

Electrolysis is a chemical process that uses electric current to drive a non-spontaneous redox reaction. It converts electrical energy into chemical energy, splitting compounds into their elements or producing new substances at the electrodes.

Key components of an electrolytic cell:

  • Anode: Positive electrode where oxidation occurs (electrons leave)
  • Cathode: Negative electrode where reduction occurs (electrons enter)
  • Electrolyte: Ionic solution or molten salt that conducts electricity
  • External power source: Battery or DC power supply that drives the reaction

Key Concept:

Electrolysis is the opposite of a galvanic (voltaic) cell. While galvanic cells produce electricity from spontaneous reactions, electrolytic cells use electricity to force non-spontaneous reactions to occur.

Faraday's Laws of Electrolysis

Faraday's First Law

The mass of substance deposited or liberated at an electrode during electrolysis is directly proportional to the quantity of electric charge passed through the electrolyte.

Mathematical Expression:

m ∝ Q

m = Z × Q

where Z = electrochemical equivalent (g/C)

Faraday's Second Law

When the same quantity of electric charge passes through different electrolytes, the masses of substances deposited are proportional to their equivalent weights (molar mass divided by the number of electrons transferred).

Mathematical Expression:

m₁/m₂ = (M₁/n₁)/(M₂/n₂)

where M = molar mass, n = electrons transferred

Combined Formula

General Equation:

m = (Q × M) / (n × F)

Where:

  • m = mass deposited/liberated (g)
  • Q = total electric charge (C) = I × t
  • M = molar mass (g/mol)
  • n = number of electrons transferred per ion/molecule
  • F = Faraday's constant = 96,485 C/mol

Also remember:

Q = I × t

Charge (C) = Current (A) × Time (s)

Worked Example: Copper Electroplating

Problem:

A current of 5.0 A is passed through a copper(II) sulfate solution for 1 hour. How many grams of copper will be deposited at the cathode?

Reaction: Cu²⁺ + 2e⁻ → Cu

Given Information:

  • Current (I) = 5.0 A
  • Time (t) = 1 hour = 3600 seconds
  • Molar mass of Cu (M) = 63.55 g/mol
  • Electrons transferred (n) = 2 (Cu²⁺ → Cu)
  • Faraday's constant (F) = 96,485 C/mol

Solution:

Step 1: Calculate total charge (Q)

Q = I × t

Q = 5.0 A × 3600 s = 18,000 C

Step 2: Apply Faraday's law

m = (Q × M) / (n × F)

m = (18,000 C × 63.55 g/mol) / (2 × 96,485 C/mol)

Step 3: Calculate mass

m = 1,143,900 / 192,970

m = 5.93 g

Answer:

5.93 grams of copper will be deposited at the cathode after 1 hour.

This corresponds to 5.93/63.55 = 0.0933 moles of copper, which required 0.1866 moles of electrons (2 electrons per copper atom).

Real-World Applications

1. Electroplating

Coating metal objects with a thin layer of another metal (gold, silver, chromium, nickel) for protection against corrosion or for decorative purposes. Used in jewelry, automotive parts, and electronics manufacturing.

2. Industrial Chemical Production

Large-scale production of chemicals like chlorine and sodium hydroxide (chlor-alkali process), aluminum (Hall-Héroult process), and hydrogen/oxygen gases. Essential for chemical industry and manufacturing.

3. Metal Purification

Refining metals like copper to 99.99% purity through electrorefining. The impure metal is used as the anode, and pure metal is deposited on the cathode. Critical for electronics requiring high-purity materials.

4. Water Splitting (Electrolysis of Water)

Production of hydrogen gas (clean fuel) and oxygen from water using renewable electricity. Important for hydrogen economy, fuel cells, and sustainable energy storage systems.

5. Electroforming

Manufacturing metal parts by electrodeposition on a mold or mandrel, which is later removed. Used to create intricate shapes, mesh screens, waveguides, and precision metal parts that would be difficult to machine.

6. Anodizing

Creating protective oxide layers on aluminum and other metals through electrolysis. The metal is made the anode, and oxygen ions form a durable, corrosion-resistant coating. Used in aerospace, architecture, and consumer products.

Common Electrolysis Reactions

SubstanceCathode Reaction (Reduction)n (electrons)M (g/mol)
CopperCu²⁺ + 2e⁻ → Cu263.55
SilverAg⁺ + e⁻ → Ag1107.87
AluminumAl³⁺ + 3e⁻ → Al326.98
ZincZn²⁺ + 2e⁻ → Zn265.38
Hydrogen2H₂O + 2e⁻ → H₂ + 2OH⁻22.016
Chlorine2Cl⁻ → Cl₂ + 2e⁻ (anode)270.90
Oxygen2H₂O → O₂ + 4H⁺ + 4e⁻ (anode)432.00
GoldAu³⁺ + 3e⁻ → Au3196.97

Problem-Solving Strategy

Step 1: Identify the Reaction

  • Write the balanced half-reaction at the electrode
  • Determine the number of electrons (n) transferred per ion/molecule
  • Find the molar mass (M) of the substance being deposited/liberated

Step 2: Calculate Total Charge

  • Use Q = I × t if current and time are given
  • Ensure time is in seconds (1 hour = 3600 s, 1 min = 60 s)
  • Current should be in amperes (A)
  • Result will be in coulombs (C)

Step 3: Apply Faraday's Law

  • Use m = (Q × M) / (n × F) to find mass
  • Faraday's constant F = 96,485 C/mol
  • Rearrange the formula if solving for Q, I, or t instead
  • Keep track of units throughout the calculation

Step 4: Check Your Answer

  • Does the mass make sense given the current and time?
  • Are the units correct? (mass in g, current in A, time in s)
  • Is the number of significant figures appropriate?
  • Can you calculate moles and verify using stoichiometry?

Common Mistakes to Avoid

❌ Forgetting to Convert Time to Seconds

Time must be in seconds when using Q = I × t. Don't use minutes or hours directly.

Correct: 2 hours = 2 × 3600 = 7200 seconds

❌ Using Wrong Number of Electrons

Always check the balanced half-reaction. Cu²⁺ requires 2 electrons, not 1. Al³⁺ requires 3.

Correct: Write and balance the half-reaction first

❌ Confusing Anode and Cathode

In electrolysis, reduction occurs at the cathode (negative), oxidation at the anode (positive).

Remember: "RED CAT" (Reduction at Cathode) and "AN OX" (Anode Oxidation)

❌ Using Atomic Mass Instead of Molecular Mass for Diatomic Gases

For Clâ‚‚, use 70.90 g/mol (not 35.45). For Hâ‚‚, use 2.016 g/mol (not 1.008).

Correct: Use the molar mass of the actual product formed

❌ Mixing Up F and f

F (capital) is Faraday's constant = 96,485 C/mol. Don't confuse with frequency (f).

Correct: Always use 96,485 C/mol for Faraday's constant

Quick Reference Guide

Key Formulas

m = (Q × M) / (n × F)

Q = I × t

F = 96,485 C/mol

m = mass (g)
Q = charge (C)
I = current (A)
t = time (s)
M = molar mass (g/mol)
n = electrons transferred

Unit Conversions

Time:

1 hour = 3600 s

1 minute = 60 s

Current:

1 A = 1000 mA

Charge:

1 C = 1 A·s

Common n Values

Cu²⁺ → Cu: n = 2

Ag⁺ → Ag: n = 1

Al³⁺ → Al: n = 3

2H₂O → H₂: n = 2

2H₂O → O₂: n = 4

2Cl⁻ → Cl₂: n = 2

Electrode Mnemonics

RED CAT: Reduction at Cathode

AN OX: Anode Oxidation

In electrolysis:
Cathode = negative terminal
Anode = positive terminal

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