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Faraday's Laws of Electrolysis

Key Concepts

  • First Law :
        The mass of a substance produced by electrolysis is proportional to the quantity of electricty used.

        In order to produce more substance electrolytically we must use more electricity.

  • Second Law:
        * The amount of electricity in coulombs required to produce 1 mole of a substance is a simple whole number multiple of 96 500#

        The quantity 96,500 is given the name the Faraday (or Faraday Constant) and the symbol F
        F is equal to the quantity of electricity carried by one mole of electrons:
        F = Avogadro's Number x charge on electron in coulombs = 6.022 x 1023 mol-1 x 1.602192 x 10-19 C
        F = 96,484 C mol-1 (usually rounded up to 96,500 C mol-1 in high school chemistry)

  • Faraday's Laws of Electrolysis Calculations: Q = n(e-) x F

        Q = quantity of electricity measured in coulombs (C)
        n(e-) = moles of electrons used
        F = the Faraday (Faraday constant) = 96,500 C mol-1

    We can calculate the mass of a substance produced during an electrolysis experiment by:

        i) calculating the moles of electrons used: n(e-) = Q/F

        ii) calculating the moles of substance produced using the balanced reduction (or oxidation) half reaction equation

        iii) using the moles of substance to calculate the mass of substance: mass = moles x molar mass

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Examples

1. Calculate the quantity of electricity obtained from 2 moles of electrons
  1. Extract the data from the question:
    Q = ? C
    n(e-) = 2 mol
    F = 96,500 C mol-1

  2. Write the equation:
    Q = n(e-) x F

  3. Substitute the values into the equation and solve:
    Q = 2 x 96,500 = 193,000 C
2. Calculate the moles of electrons obtained from 250 C of electricity
  1. Extract the data from the question:
    n(e-) = ? mol
    Q = 250 C
    F = 96,500 C mol-1

  2. Write the equation:
    Q = n(e-) x F
    Rearrange the equation to find moles of electrons, n(e-):
    n(e-) = Q ÷ F

  3. Substitute the values into the equation and solve:
    n(e-) = 250 ÷ 96,500 = 2.59 x 10-3 mol
3. Calculate the moles of copper metal that can be produced by the electrolysis of molten copper sulfate using 500 C of electricity.

  1. Write the reduction reaction equation for the production of copper metal from molten copper sulfate:
    Cu2+ + 2e- → Cu(s)

  2. Calculate the moles of electrons, n(e-):
    n(e-) = Q/F

    Extract the data from the question:
    Q = 500 C
    n(e-) = ? mol
    F = 96,500 C mol-1

    Substitute the values into the equation:
    n(e-) = 500/96,500 = 5.18 x 10-3 mol

  3. Determine the moles of Cu(s) produced using the balanced reduction reaction equation:
    1 mole of electrons produces ½ mole of Cu(s)
    Therefore 5.18 x 10-3 moles of electrons produces ½ x 5.18 x 10-3 = 2.59 x 10-3 moles Cu(s)
4. Calculate the mass of silver that can be produced by the electrolysis of 1 mol L-1 AgCN(aq) using 800 C of electricity

  1. Write the reduction reaction equation for the production of silver metal from the aqueous solution:
    Ag+ + e- → Ag(s)

  2. Calculate the moles of electrons, n(e-):
    n(e-) = Q/F

    Extract the data from the question:
    Q = 800 C
    n(e-) = ? mol
    F = 96,500 C mol-1

    Substitute the values into the equation:
    n(e-) = 800/96,500 = 8.29 x 10-3 mol

  3. Determine the moles of Ag(s) produced using the balanced reduction reaction equation:
    1 mole of electrons produces 1 mole of Ag(s)
    Therefore 8.29 x 10-3 moles of electrons produces 8.29 x 10-3 moles Ag(s)

  4. Calculate the mass of Ag(s)
    moles (Ag) = mass (Ag) ÷ molar mass (Ag)
    So, mass (Ag) = moles (Ag) x molar mass (Ag)

    moles (Ag) = 8.29 x 10-3 mol
    molar mass (Ag) = 107.9 (from periodic table)
    mass (Ag) = 8.29 x 10-3 mol x 107.9 = 0.894 g

5. 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 moles of copper using the balanced reduction half reaction equation:
    Cu2+ + 2e → Cu(s)
    1 mole of copper is deposited from 2 moles electrons
    moles(Cu) = ½n(e-) = ½ x 5.18 x 10-5 = 2.59 x 10-5 mol

  4. Calculate mass of copper: mass = moles x molar mass
    moles (Cu) = 2.59 x 10-5 mol
    molar mass (Cu) = 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
6. Calculate the time required to deposit 56 g of silver from a silver nitrate solution using a current of 4.5 A.
  1. Calculate the moles of silver deposited:
    moles (Ag) = mass (Ag) ÷ molar mass (Ag)
        mass Ag deposited = 56 g
        molar mass = 107.9 g mol-1 (from Periodic Table)
    moles (Ag) = 56 ÷ 107.9 = 0.519 mol

  2. Calculate the moles of electrons required for the reaction:
    write the reduction reaction equation: Ag+ + e → Ag(s)
    from the equation 1 mole of Ag is deposited by 1 mole of electrons
    therefore 0.519 moles of Ag(s) is deposited by 0.519 moles of electrons
    n(e-) = 0.519 mol

  3. 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

  4. Calculate the time required:
    Q = I x t
    Rearrange the equation to find t: 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


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* More formally we say that for a given quantity of electricity the quantity of substance produced is proportional to its equivalent weight.

#The figure is closer to 96,484 but is usually rounded off to 96,500 for school chemistry calculations

 
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