Electrochemistry & Applications of ThermodynamicsElectrochemistryContent level: Core 26 min

Faraday's Law and Electrolysis Calculations

What you'll be able to do: Relate current and time to the mass of substance produced or consumed at an electrode using Faraday's law.

Best after: Electrolytic Cells and Driven Reactions

Introduction

Electrolysis stoichiometry is ordinary stoichiometry with one extra conversion at the front: electric charge becomes moles of electrons. Learn that single bridge and every plating problem becomes a road map you already know how to follow.

These are recommended, not required. You can start this lesson at any time.

Learning objectives

  • Convert current and time into moles of electrons
  • Use a half-reaction to relate electrons to moles of product
  • Calculate the mass or gas volume produced by electrolysis
  • Solve for current or time when the mass produced is known

Lesson

Electrons as a reagent

In an electrolysis, the balanced half-reaction says exactly how many electrons are consumed per ion converted. Ag + e → Ag needs one, Cu²⁺ + 2 e → Cu needs two, and Al³⁺ + 3 e → Al needs three. Once electrons are treated as a reactant with a coefficient, standard stoichiometry takes over.

Write the half-reaction first. Every Faraday problem hinges on its electron coefficient.

From current to charge

An ampere is a coulomb per second, so charge is simply q = It. Time must be in seconds, which means converting minutes or hours before multiplying. This unit conversion accounts for a large share of wrong answers on this topic.

q = I t

From charge to moles of electrons

One mole of electrons carries 96485 C, a value equal to the elementary charge times Avogadro number. Dividing total charge by F gives moles of electrons.

mol e = q / F = I t / 96485

The full road map

Amperes and seconds give coulombs; coulombs divided by F give moles of electrons; the half-reaction ratio gives moles of substance; molar mass gives grams, or PV = nRT gives a gas volume. Any single unknown along the chain can be solved by running the same road map backward.

I, t → q → mol e → mol substance → mass or volume

Charge matters

For a fixed quantity of charge, the moles deposited are inversely proportional to the ionic charge. The same current for the same time deposits three times as many moles of silver as of aluminum. In cells wired in series the current and time are shared, so this comparison is exact.

mol substance = (mol e) / (electrons per formula unit)

Efficiency in real cells

Industrial cells rarely convert every electron into the desired product, since side reactions such as hydrogen evolution consume some charge. Current efficiency is the fraction that does useful work, so the actual mass equals the theoretical mass times that efficiency.

actual mass = theoretical mass x current efficiency

Key ideas

Rule
Charge

q = It, with I in amperes and t in seconds.

Definition
Faraday constant

F = 96485 C per mole of electrons.

Rule
Moles of electrons

mol e = It/F.

Rule
Electron ratio

Taken from the balanced half-reaction, for example 2 e per Cu.

Rule
Series cells

Cells in series pass the same charge, so their electron counts are equal.

Equation
Charge

q = I t

  • I = current in amperes
  • t = time in seconds
Equation
Faraday relation

mol e = I t / F

  • F = 96485 C/mol e
Equation
Mass produced

mass = (I t / F) × (1 / z) x M

  • z = electrons required per formula unit
  • M = molar mass in g/mol

Worked examples

Worked example 1

A current of 2.00 A is passed through a AgNO solution for 30.0 minutes. What mass of silver plates onto the cathode? M(Ag) = 107.87 g/mol.

Try it first: Write the cathode half-reaction and note the electron coefficient.

    0 of 4 steps revealed.

    Worked example 2

    How long must a 5.00 A current run to deposit 10.0 g of copper from Cu²⁺? M(Cu) = 63.55 g/mol.

    Try it first: Run the road map backward from mass to time.

      0 of 4 steps revealed.

      Worked example 3

      Molten alumina is electrolyzed at 1.00 × 10 A for 1.00 hour. What mass of aluminum is produced at 100 percent efficiency? M(Al) = 26.98 g/mol.

      Try it first: Aluminum is 3+, which changes the electron ratio.

        0 of 4 steps revealed.

        Common mistakes

        Using minutes or hours directly in q = It.

        Why it's wrong: An ampere is a coulomb per second, so only seconds give the correct charge.

        Check instead: Convert time to seconds first.

        Skipping the electron-to-substance ratio.

        Why it's wrong: Moles of electrons equal moles of product only when the half-reaction needs one electron.

        Check instead: Divide by the electron coefficient from the half-reaction.

        Multiplying instead of dividing by F.

        Why it's wrong: F converts coulombs to moles, so it belongs in the denominator.

        Check instead: Check that the coulombs cancel in the unit analysis.

        Assuming equal masses deposit for equal charge.

        Why it's wrong: Different ionic charges and molar masses give different masses per mole of electrons.

        Check instead: Compare moles of electrons first, then convert each metal separately.

        Ignoring current efficiency in an industrial problem.

        Why it's wrong: Side reactions consume charge, so the real yield is lower than the calculation.

        Check instead: Multiply the theoretical mass by the stated efficiency.

        Practice this skill

        No practice questions are available for this topic yet. You can still practice the whole unit.

        What you should now know

        Faraday's law makes electrons a stoichiometric reagent. Charge in coulombs is current times time, q = It, with current in amperes and time in seconds. Dividing by the Faraday constant, 96485 C per mole of electrons, converts charge into moles of electrons. From there the balanced half-reaction supplies the mole ratio between electrons and the substance being deposited or evolved, and molar mass converts to grams or the ideal gas law converts to a gas volume. The complete chain is current and time, then coulombs, then moles of electrons, then moles of substance, then mass or volume, and every step can be run in reverse when the unknown is the current or the plating time. Because the electron-to-substance ratio depends on ionic charge, the same quantity of charge deposits half as many moles of Cu²⁺ as of Ag, and one third as many of Al³⁺. When electrolytic cells are connected in series they pass identical charge, which is the basis of the classic comparison problem.

        • Charge is current times time, with time in seconds
        • Moles of electrons equal charge divided by 96485 C/mol
        • The half-reaction gives the electrons-per-formula-unit ratio
        • The road map runs I and t to coulombs to moles of electrons to moles of substance to mass
        • Higher ionic charge means fewer moles deposited for the same charge

        Sources and further reading

        • Chemistry 2e, Section 17.7: Electrolysis
          OpenStax · Rice University · Chapter 17.7
          View source

          Access for free at openstax.org License

        • CODATA recommended value of the Faraday constant
          CODATA / NIST · NIST · Fundamental constants
          View source

          NIST Reference on Constants, Units, and Uncertainty