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9.11Thermodynamics and Electrochemistry

Electrolysis and Faraday’s Law — AP CHEMISTRY Revision Notes & Topical Questions

5 min read 9 parts Lesson 15 of 15 1 quiz

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9.11 Electrolysis and Faraday's Law — AP CHEMISTRY notes: Calculate the amount of substance produced in electrolysis using charge, current and time.Charge — Q = It, with Q in coulombs, I in amperes and t in secondsWhat the symbols mean: Q = charge (C) · I = current (A) · t = time…

  • Link to constants
  • Moles of electrons
  • Non-metals
  • Amount produced
  • Faraday constant
  • Half-equation

Section 1 of 9Charge and moles of electrons

Charge
  • Q = It, with Q in coulombs, I in amperes and t in seconds
    What the symbols mean: Q = charge (C) · I = current (A) · t = time (s)
Faraday
  • Faraday one faraday is the charge on one mole of electrons, 96 485 C mol-1
Link to constants
  • F = NAe
    • (Avogadro constant × electron charge)
Moles of electrons
  • Moles of electrons n(e−) = Q/F

Section 2 of 9Electrons needed per mole of product

Half-equationElectrons per moleCharge per mole
Na+ + e− → Na11 F
Mg2+ + 2e− → Mg22 F
2F− → F2 + 2e−22 F
4OH− → O2 + 2H2O + 4e−44 F

Section 3 of 9Method to find mass

1
Write the half-equation at the electrode.
2
Calculate Q = It (convert minutes to seconds).
3
Divide by F to get moles of electrons.
4
Use the half-equation ratio to get moles of product.
5
Multiply by molar mass.

Section 4 of 9Pattern: magnesium

Data
  • 2.20 A for 15 min
Charge
  • Q = 2.20 × 900 = 1980 C
Moles
  • 1980/(2 × 96 485) = 0.0103 mol Mg, so 0.25 g

Section 5 of 9Pattern: 2.18 A for 65 min

Charge
  • 2.18 × 3900 = 8502 C
Al
  • 8502/(3 × 96 485) = 0.0294 mol, so 0.79 g
Ag
  • 8502/96 485 = 0.0881 mol, so 9.51 g
Cu
  • 8502/(2 × 96 485) = 0.0441 mol, so 2.80 g

Section 6 of 9Products of electrolysis

Metals
  • Metals formed at the cathode
Non-metals
  • Non-metals formed at the anode
Amount produced
  • Amount produced proportional to the charge passed

Section 7 of 9More from the specification

More from the specification
Electrolytic cells versus voltaic cells
  • Electrolytic cells versus voltaic cells non-spontaneous electrochemical cells drive endergonic electrolysis in electrolytic cells using external electrical power supplies.
Concentration
  • Moles ÷ volume in dm³, or g/dm³.
Electrode potential
  • A measure of how easily a species is reduced, relative to a reference half-cell.
Nernst equation
  • Relates cell potential to standard potential and Q.
Electrolytic cell
  • A cell using electricity to drive a non-spontaneous reaction.
Reaction quotient
  • The ratio of product to reactant concentrations at any moment.
Electrochemical cell
  • A device linking redox reactions and electricity.
Standard conditions
  • 298 K and 101 kPa.
Voltaic cell
  • A cell that generates electricity from a spontaneous redox reaction.
Half-cell
  • One electrode and its solution, in which a redox equilibrium is set up.
Oxidation
  • Gain of oxygen, loss of electrons or an increase in oxidation number.
Reduction
  • Loss of oxygen, gain of electrons or a decrease in oxidation number.
Equilibrium
  • The state where forward and reverse transfer rates are equal.

Section 8 of 9Key terms

TermMeaning
Faraday constant96 485 C per mole of electrons.
ChargeQ = It, measured in coulombs.
ElectrolysisUse of electrical energy to drive a non-spontaneous reaction.
Half-equationEquation showing electron gain or loss at one electrode.

Section 9 of 9Exam technique

1
Using minutes instead of seconds.
2
Forgetting to divide by the number of electrons per ion.
3
Using Faraday's constant as charge per ion instead of per mole.
4
Placing the metal at the anode.

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