What are numerical based exampels on Faraday's law of electrolysis?

Answer 1

1. 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

2. 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

3. 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

4. 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

5. 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

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

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: 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

1. What mass of copper could be deposited from a copper (II) sulphate solution using a current of 0.50 A over 10 seconds?
2. Calculate the quantity of electricity: Q = I x t

   I = 0.50 A

   t = 10 seconds

   Q = 0.50 x 10 = 5.0 C

3. 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

4. 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

1. 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.

2. 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

3. 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

4. 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

5. 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