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 10^{23} 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

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

Extract the data from the question:
Q = ? C
n(e^{-}) = 2 mol
F = 96,500 C mol^{-1}

Write the equation:
Q = n(e^{-}) x F

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

Extract the data from the question:
n(e^{-}) = ? mol
Q = 250 C
F = 96,500 C mol^{-1}

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

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.

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

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

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

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

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

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)}

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?

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 moles of copper using the balanced reduction half reaction equation:
Cu^{2+} + 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

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.

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
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^{-}) 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:
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