Faraday Laws of Electrolysis

Summary

• 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 10^{6 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

  To calculate the quantity of electricity, Q Enter current  amps The answer will appear below: Enter time  seconds Calculation :   Then click Calculate   Q = C To perform another calculation, Click Reset

• I = Q ÷ t

  Calculate the current needed to 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

  To calculate the current, I Enter quantity of electricity  C The answer will appear below: Enter time  seconds Calculation :   Then click Calculate   I = amps To perform another calculation, Click Reset

• 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 ÷ t = 12,000 ÷ 10 = 1,200 seconds = 1,200 ÷ 60 = 20 minutes

  To calculate the time, t Enter quantity of electricity  C The answer will appear below: Enter current  amps Calculation :   Then click Calculate   t = seconds To perform another calculation, Click Reset

• 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

  To calculate the quantity of electricity, Q Enter moles of electrons  mol The answer will appear below: Then click Calculate   Calculation :   To perform another calculation, Click Reset   Q = 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

  To calculate the moles of electrons, n(e) Enter quantity of electricity  C The answer will appear below: Then click Calculate   Calculation :   To perform another calculation, Click Reset   n(e) = mol

• Calculate the time required to deposit 56g of silver from a silver nitrate solution using a current of 4.5A.
  
  • 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)
  • 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
  • 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?
  • Calculate the quantity of electricity: Q = I x t
    I = 0.50 A
    t = 10 seconds
    Q = 0.50 x 10 = 5.0 C
  • 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
  • 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.
  
  • 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
  • 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
  • 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
  • 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