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Faraday's Laws of Electrolysis Tutorial

Key Concepts

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Worked Examples: Q = n(e-)F calculations

Question 1. Calculate the quantity of electricity obtained from 2 moles of electrons

Solution:

(Based on the StoPGoPS approach to problem solving.)

  1. What is the question asking you to do?

    Calculate the quantity of electricity
    Q = ? C

  2. What data (information) have you been given in the question?

    Extract the data from the question:
    moles of electrons = n(e-) = 2 mol
    Faraday constant = F = 96,500 C mol-1 (data sheet)

  3. What is the relationship between what you know and what you need to find out?

    Write the equation:
    Q = n(e-) × F

  4. Substitute the values into the equation and solve for Q:
    Q = 2 × 96,500 = 193,000 C
  5. Is your answer plausible?

    Use your calculated value of Q and the Faraday constant F to calculate moles of electrons and compare that to the value given in the question.
    Q = n(e-)F
    193,000 = n(e-) × 96,500
    n(e) = 193,000 ÷ 96,500 = 2
    Since we were told there were 2 moles of electrons in the question, we are reasonably confident that our value for Q is correct.

  6. State your solution to the problem:

    Q = 193,000 C

Question 2. Calculate the moles of electrons obtained from 250 C of electricity.

Solution:

(Based on the StoPGoPS approach to problem solving.)

  1. What is the question asking you to do?

    Calculate the moles of electrons
    n(e-) = ? mol

  2. What data (information) have you been given in the question?

    Extract the data from the question:
    Q = 250 C
    F = 96,500 C mol-1 (data sheet)

  3. What is the relationship between what you know and what you need to find out?

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

  4. Substitute the values into the equation and solve for n(e-):

    n(e-) = 250 ÷ 96,500 = 2.59 × 10-3 mol

  5. Is your answer plausible?

    Use your calculated value of n(e-) and the Faraday constant F to calculate quantity of charge (Q) required and compare that to the value given in the question.
    Q = n(e-) × F
    Q = 2.59 × 10-3 × 96,500 = 250 C
    Since this value of Q agrees with that given in the question we are reasonably confident that our value for n(e-) is correct.

  6. State your solution to the problem:

    n(e-) = 2.59 × 10-3 mol

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Worked Examples: Calculating amount of substance deposited

Question 1: Calculate the moles of copper metal that can be produced by the electrolysis of molten copper sulfate using 500 C of electricity.

Solution:

(Based on the StoPGoPS approach to problem solving.)

  1. What is the question asking you to do?

    Calculate the moles of copper metal
    n(Cu(s)) = ? mol

  2. What data (information) have you been given in the question?

    Extract the data from the question:
    electrolyte: CuSO4(l)
    Q = 500 C
    F = 96,500 C mol-1 (data sheet)

  3. What is the relationship between what you know and what you need to find out?

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

    Calculate the moles of electrons, n(e-):

    n(e-) = Q ÷ F
        = 500 ÷ 96,500
        = 5.18 × 10-3 mol

  4. Determine the moles of Cu(s) produced using the balanced reduction reaction equation (mole ratio):

    1 mole of electrons produces ½ mole of Cu(s)
    Therefore 5.18 × 10-3 moles of electrons produces ½ × 5.18 × 10-3
    n(Cu(s)) = 2.59 × 10-3 mol

  5. Is your answer plausible?

    Use your calculated value of n(Cu(s)) and the Faraday constant F to calculate quantity of charge (Q) required and compare that to the value given in the question.
    Q = n(e-)F
        n(e-) = 2 × n(Cu) = 2 × 2.59 × 10-3 = 5.18 × 10-3 mol
        F = 96,500
    Q = 5.18 × 10-3 × 96,500 = 500 C
    Since this value for Q is the same as that given in the question, we are reasonably confident that our calculated value for moles of copper deposited is correct.

  6. State your solution to the problem:

    n(Cu(s)) = 2.59 × 10-3 mol

Question 2. Calculate the mass of silver that can be produced by the electrolysis of 1 mol L-1 AgCN(aq) using 800 C of electricity

Solution:

(Based on the StoPGoPS approach to problem solving.)

  1. What is the question asking you to do?

    Calculate the mass of silver deposited
    m(Ag(s)) = ? g

  2. What data (information) have you been given in the question?

    Extract the data from the question:
    electrolyte: AgCN(aq)
    [AgCN(aq)] = 1 mol L-1 (standard solution)
    Q = 800 C
    F = 96,500 C mol-1 (data sheet)

  3. What is the relationship between what you know and what you need to find out?

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

    Calculate the moles of electrons, n(e-):
    n(e-) = Q ÷ F
    n(e-) = 800 ÷ 96,500 = 8.29 × 10-3 mol

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

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

        moles (Ag) = 8.29 × 10-3 mol
        molar mass (Ag) = 107.9 g mol-1 (from periodic table)

    mass (Ag) = 8.29 × 10-3 mol × 107.9 g mol-1
        = 0.894 g

  5. Is your answer plausible?

    Use your calculated value of m(Ag(s)) and the Faraday constant F to calculate quantity of charge (Q) required and compare that to the value given in the question.
    n(e-) = n(Ag) = mass ÷ molar mass = 0.894 ÷ 107.9 = 8.29 × 10-3 mol
    Q = n(e-)F = 8.29 × 10-3 mol × 96,500 = 800 C
    Since this value of Q agrees with that given in the question, we are reasonably confident that our calculated mass of silver is correct.

  6. State your solution to the problem:

    m(Ag(s)) = 0.894 g

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Worked Examples: Q = n(e-)F and Q = It

Question 1. What mass of copper could be deposited from a copper(II) sulphate solution using a current of 0.50 A over 10 seconds?

Solution:

(Based on the StoPGoPS approach to problem solving.)

  1. What is the question asking you to do?

    Calculate the mass of copper deposited
    m(Cu(s)) = ? g

  2. What data (information) have you been given in the question?

    Extract the data from the question:
    electrolyte: copper(II) sulphate solution, CuSO4
    current: I = 0.50 A
    time: t = 10 seconds
    F = 96,500 C mol-1 (data sheet)

  3. What is the relationship between what you know and what you need to find out?

    Calculate the quantity of electricity:
    Q = I x t
        I = 0.50 A
        t = 10 seconds
    Q = 0.50 × 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 × 10-5 mol

    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 (mole ratio)
    moles(Cu) = ½n(e-)
        = ½ × 5.18 × 10-5
        = 2.59 × 10-5 mol

  4. Calculate mass of copper:

    mass = moles × molar mass
        moles (Cu) = 2.59 × 10-5 mol
        molar mass (Cu) = 63.55 g mol-1 (from Periodic Table)
    mass (Cu) = (2.59 × 10-5) × 63.55
        = 1.65 × 10-3 g
        = 1.65 mg

  5. Is your answer plausible?

    Use your calculated value of m(Cu(s)) and the Faraday constant F to calculate quantity of charge (Q(b)) required and compare that to the value of Q(a) = It given in the question.
    Q(a) = It = 0.50 × 10 = 5 C

    Q(b) = n(e-)F
        n(e-) = 2 × n(Cu) = 2 × [m(Cu) ÷ Mr(Cu)] = 2 × [(1.65 × 10-3) ÷ 63.55] = 2 × 2.6 × 10-5 = 5.2 × 10-5 mol
    Q = 5.2 × 10-5 × 96,500 = 5

    Since Q(a) = Q(b) = 5 C, we are reasonably confident that our calculated mass of copper is correct.

  6. State your solution to the problem:

    m(Cu(s)) = 1.65 × 10-3 g (or 1.65 mg)

Question 2. Calculate the time required to deposit 56 g of silver from a silver nitrate solution using a current of 4.5 A.

Solution:

(Based on the StoPGoPS approach to problem solving.)

  1. What is the question asking you to do?

    Calculate time required
    t = ? seconds

  2. What data (information) have you been given in the question?

    Extract the data from the question:
    mass silver = m(Ag(s)) = 56 g
    current = I = 4.5 A
    F = 96,500 C mol-1 (from data sheet)

  3. What is the relationship between what you know and what you need to find out?

    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

    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 (mole ratio)
    therefore 0.519 moles of Ag(s) is deposited by 0.519 moles of electrons
    n(e-) = 0.519 mol

    Calculate the quantity of electricity required:
    Q = n(e-) × F
        n(e-) = 0.519 mol
        F = 96,500 C mol-1
    Q = 0.519 × 96,500 = 50,083.5 C

  4. Calculate the time required:
    Q = I × 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
  5. Is your answer plausible?

    Use your calculated value of time in seconds, the Faraday constant F and the current given in the question to calculate the mass of Ag you could deposit and compare that to the value given in the question.
    Q = It = 4.5 × 11,129.67 = 50083.5 C
    Q = n(e-)F
    so, n(e-) = Q ÷ F = 50083.5 ÷ 96,500 = 0.519 mol
    n(Ag) = n(e-) = 0.519 mol
    m(Ag) = n(Ag) × Mr(Ag) = 0.519 × 107.9 = 56 g
    Since this value for the mass of silver is the same as that given in the question, we are reasonably confident that the time in seconds we have calculated is correct.

  6. State your solution to the problem:

    t = 11,129.67 seconds (or 185.5 minutes or 3.1 hours)

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

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