Faraday's Law & Avogadro (Cambridge International A-Level Chemistry)

Faraday's Law

• 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

Worked Example Determine the amount of electricity required for the following reactions:

1. To deposit 1 mol of sodium

Na⁺ (aq) + e^– → Na (s)

1. To deposit 1 mol of magnesium

Mg²⁺ (aq) + 2e^– → Mg (s)

1. To form 1 mol of fluorine gas

2F^– (aq) → F₂ (g) + 2e^–

1. To form 1 mole of oxygen

4OH^– (aq) → O₂ (g) + 2H₂O (l) + 4e^–

Answers:

• One Faraday is the amount of charge (96 500 C) carried by 1 mole of electrons

Answer 1:

• As there is one mole of electrons, one faraday of electricity (96 500 C) is needed to deposit one mole of sodium.

Answer 2:

• Now, there are two moles of electrons, therefore, two faradays of electricity (2 x 96 500 C) are required to deposit one mole of magnesium.

Answer 3:

• Two moles of electrons are released, so it requires two faradays of electricity (2 x 96 500 C) to form one mole of fluorine gas.

Answer 4:

• Four moles of electrons are released, therefore it requires four faradays of electricity (4 x 96 500 C) to form one mole of oxygen gas.

Determining Avogadro's Constant by Electrolysis

• 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 = $\frac{charge on1molofelectrons}{charge on1electron}$

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

Electrolysis of copper(II) sulfate solution

• 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

• To scale our experimental results up to the molar level, we can use a ratio method
• The scaling factor tells us how many times larger our target amount (1 mole) is compared to our experimental result.

• 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 = $\frac{199292}{2}=99646C$

• Given that the charge on one electron is 1.60 x 10^-19 C, then L equals:

  • L = $\frac{\mathrm{charge} \mathrm{on}1\mathrm{mol} \mathrm{of} \mathrm{electrons}}{\mathrm{charge} \mathrm{on}1\mathrm{electron}}$
  • L = $\frac{99646}{1.60\times10^{-19}}$ = 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