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Faraday Laws of Electrolysis

Key Concepts

First Law
The quantity of a substance produced by electrolysis is proprotional to the quantity of electricty used.
Second Law
For a given quantity of electricity the quantity of substance produced is proportional to its weight.
The quantity of electricity or charge contained in a current running for a specified time can be calculated:
Q = I x t
Q = quantity of electricity or charge in coulombs (C)
I = current in amps (A)
t = time (seconds)

The Faraday constant, F, is the quantity of electricity carried by one mole of electrons.

F = Avogadro's Number x charge on electron in coulombs
F = 6.022 x 10²³ mol^-1 x 1.602192 x 10^-19 C
F = 96,484 C mol^-1
This is usually rounded off to 96,500 C mol^-1 for calculations in chemistry.

The quantity of electricity required to deposit an amount of metal can be calculated:
Q = n(e) x F
Q = quantity of electricity in coulombs (C)
n(e) = moles of electrons
F = Faraday constant = 96, 500 C mol^-1
Electrical Energy, E, can be calculated:
E = Q x V
E = electrical energy in joules (J)
Q = quantity of electricity in coulombs (C)
V = voltage (or EMF) in volts (V)
1 kilowatt-hour, kWH, is a unit of electrical energy.
1 kWH = 3.6 x 10⁶ J

Examples

Q = I x t
Calculate the quantity of electricity, Q, obtained when a current of 25 amps runs for 1 minute.
Q = ? C
I = 25 A
t = 1 minute = 60 seconds
Q = 25 x 60 = 1,500 C
I = Q ÷ t
Calculate the current needed to provide 30,000 coulombs of electricity in 5 minutes.
Q = 30,000 C
I = ? A
t = 5 minutes = 5 x 60 = 300 seconds
I = Q ÷ t = 30,000 ÷ 300 = 100 amps
t = Q ÷ I
Calculate the time required to produce 12,000 C of electricity using a current of 10 amps.
Q = 12,000 C
I = 10 A
t = ?
t = Q ÷ I = 12,000 ÷ 10 = 1,200 seconds = 1,200 ÷ 60 = 20 minutes
Q = n(e) x F
Calculate the quantity of electricity obtained from 2 moles of electrons
Q = n x F
Q = ?
n = 2 mol
F = 96,500 C mol^-1
Q = 2 x 96,500 = 193,000 C
n(e) = Q ÷ F
Calculate the moles of electrons obtained from 250 C of electricity
n(e) = ? mol
Q = 250 C
F = 96,500 C mol^-1
n(e) = 250 ÷ 96,500 = 2.59 x 10^-3 mol
Calculate the time required to deposit 56g of silver from a silver nitrate solution using a current of 4.5A.
1. Calculate the moles of electrons required for the reaction:
  Ag⁺ + e → Ag(s)
  moles of Ag(s) deposited, n(Ag) = moles of electrons required, n(e)
  moles of Ag = n(Ag) = mass ÷ MM
  mass Ag deposited = 56g
  MM = 107.9 g mol^-1 (from Periodic Table)
  n (Ag) 56 ÷ 107.9 = 0.519 mol = n(e)
2. Calculate the quantity of electricity required: Q = n(e) x F
  Q = ? C
  n(e) = 0.519 mol
  F = 96,500 C mol^-1
  Q = 0.519 x 96,500 = 50,083.5 C
3. Calculate the time required: t = Q ÷ I
  Q = 50,083.5 C
  I = 4.5 A
  t = 50,083.5 ÷ 4.5 = 11,129.67 seconds
  t = 11,129.67 ÷ 60 = 185.5 minutes
  t = 185.5 ÷ 60 = 3.1 hours
What mass of copper could be deposited from a copper (II) sulphate solution using a current of 0.50 A over 10 seconds?
1. Calculate the quantity of electricity: Q = I x t
  I = 0.50 A
  t = 10 seconds
  Q = 0.50 x 10 = 5.0 C
2. Calculate the moles of electrons: n(e) = Q ÷ F
  Q = 5.0 C
  F = 96,500 C mol^-1
  n(e) = 5.0 ÷ 96,500 = 5.18 x 10^-5 mol
3. Calculate mass of copper: mass = n x MM
  Cu²⁺ + 2e → Cu(s)
  1 mole of copper is deposited from 2 moles electrons
  n(Cu) = ½n(e) = ½ x 5.18 x 10^-5 = 2.59 x 10^-5 mol
  MM = 63.55 g mol^-1 (from Periodic Table)
  mass (Cu) = (2.59 x 10^-5) x 63.55 = 1.65 x 10^-3 g = 1.65 mg
An EMF of 4.5 V produces 1 kg of sodium metal by the electrolysis of Na⁺.
Calculate the minimum number of kilowatt-hours of electricity needed to produce the sodium metal.
1. Calculate the moles of electrons, n(e), required
  Write the equation for the electrolysis of Na⁺:
  Na⁺ + e → Na(s)
  moles of Na(s) = moles of electrons used n(e)
  n(e) = n(Na) = mass ÷ MM
  mass = 1kg = 1,000g
  MM = 22.99 g mol^-1 (from Periodic Table)
  n(e) = 1,000 ÷ 22.99 = 43.497 mol
2. Calculate the quantity of electricity required: Q = n(e) x F
  n(e) = 43.497 mol
  F = 96,500 C mol^-1
  Q = n(e) x F = 43.497 x 96,500 = 4.2 x 10⁶ C
3. Calculate the electrical energy: E = Q x V
  Q = 4.2 x 10⁶
  V = 4.5 V
  E = 4.2 x 10⁶ x 4.5 = 1.89 x 10⁷ J
4. Convert Electrical Energy to kilowatt-hours: kilowatt-hours = E ÷ 3.6 x 10⁶
  E = (1.89 x 10⁷) ÷ (3.6 x 10⁶) = 5.25 kWH