Faraday's Law & Avogadro (CIE A Level Chemistry)

• The amount of substance that is formed at an electrode during electrolysis is proportional to:
  • The amount of time where a constant current to passes
  • The amount of charge, in coulombs, that passes through the electrolyte (strength of electric current)
  • The relationship between the current and time is:

Q = I x t

• • • Q = charge (coulombs, C)
    • I = current (amperes, A)
    • t = time, (seconds, s)
• The amount or the quantity of electricity can also be expressed by the faraday (F) unit
  • One faraday is the amount of electric charge carried by 1 mole of electrons or 1 mole of singly charged ions
  • 1 faraday is 96 500 C mol^-1
  • Thus, the relationship between the Faraday constant and the Avogadro constant (L) is:

F = L x e

• • • F = Faraday’s constant (96 500 C mol^-1)
    • L = Avogadro’s constant (6.02 x 10²³ mol^-1)
    • e = charge on an electron

• The Avogadro’s constant (L) is the number of entities in one mole
  • L = 6.02 x 10²³ mol^-1
  • For example, four moles of water contains 2.41 x 10²⁴ (6.02 x 10²³ x 4) molecules of H₂O

• The value of L (6.02 x 10²³ mol^-1) can be experimentally determined by electrolysis using the following equation:

L = 

Finding L experimentally

• The charge on one mole of electrons is found by using a simple electrolysis experiment using copper electrodes

Apparatus set-up for finding the value of L experimentally

• Method
  • The pure copper anode and pure copper cathode are weighed
  • A variable resistor is kept at a constant current of about 0.17 A
  • An electric current is then passed through for a certain time interval (e.g. 40 minutes)
  • The anode and cathode are then removed, washed with distilled water, dried with propanone, and then reweighed

• Results
  • The cathode has increased in mass as copper is deposited
  • The anode has decreased in mass as the copper goes into solution as copper ions
  • Often, it is the decreased mass of the anode which is used in the calculation, as the solid copper formed at the cathode does not always stick to the cathode properly
  • Let’s say the amount of copper deposited in this experiment was 0.13 g

• Calculation:
  • The amount of charge passed can be calculated as follows:

Q = I x t

= 0.17 x (60 x 40)

= 408 C

• To deposit 0.13 g of copper (2.0 x 10^-3 mol), 408 C of electricity was needed
• The amount of electricity needed to deposit 1 mole of copper can therefore be calculated using simple proportion using the relative atomic mass of Cu

Calculating the amount of charge required to deposit one mole of copper table

Charge (C)	Amount of Cu deposited (mol)	Amount of Cu deposited (g)
408		0.13
	1	63.5

• Therefore, 199 292 C of electricity is needed to deposit 1 mole of Cu
• The half-equation shows that 2 mol of electrons are needed to deposit one mol of copper:

Cu²⁺ (aq) +  2e^- →   Cu (s)

• So, the charge on 1 mol of electrons is:
  • Q = 
• Given that the charge on one electron is 1.60 x 10^-19 C, then L equals:
  • L =
  • L =  = 6.23 x 10²³ mol^-1

• The experimentally determined value for L of 6.23 x 10²³ mol^-1 is very close to the theoretical value of 6.02 x 10²³ mol^-1