Electrolysis & Faraday’s Law (College Board AP Chemistry)

Determine the amount of electricity required for the following reactions:

1) To deposit 1 mol of sodium 

Na⁺ (aq) + e^– → Na (s) 

2) To deposit 1 mol of magnesium

Mg²⁺ (aq) + 2e^– → Mg (s) 

3) To form 1 mol of fluorine gas

2F^– (aq)  → F₂ (g) + 2e^–

4) To form 1 mole of oxygen

4OH^– (aq)  → O₂ (g) + 2H₂O (l) +  4e^–

Answers:

• One Faraday is the amount of charge (96 485 C) carried by 1 mole of electrons

Answer 1:

• One mole of electrons is used
• Therefore, one Faraday of electricity (96,485 C mol^–1) is needed to deposit one mole of sodium.

Answer 2:

• Now, two moles of electrons are used
• Therefore, two Faradays of electricity (2 x 96,485 C mol^–1) are required to deposit one mole of magnesium.

Answer 3:

• Two moles of electrons are released
• Therefore, it requires two Faradays of electricity (2 x 96,485 C mol^–1) to form one mole of fluorine gas.

Answer 4:

• Four moles of electrons are released
• Therefore it requires four Faradays of electricity (4 x 96,485 C mol^–1) to form one mole of oxygen gas.

Calculate the amount of magnesium deposited when a current of 2.20 A flows through the molten bromide for 15 minutes.

Answer:

• The magnesium (Mg²⁺) ion is a positively charged cation that will move towards the cathode.
• Step 1: Write the half-equation at the cathode

Mg²⁺(aq)          +          2e^-       →         Mg(s)

1 mol                           2 mol               1 mol

• Step 2: Calculate the charge transferred during the electrolysis
  • Q = I x t
  • Q = 2.20 x (60 x 15) = 1980 C 
• Step 3: Determine the number of coulombs required to deposit one mole of magnesium at the cathode

• • For every one mole of electrons, the number of coulombs needed is 96,485 C mol^-1
  • In this case, there are two moles of electrons required
  • So, the number of coulombs needed is = 2 x 96,485 = 192,970 C mol^-1

• Step 4: Calculate the mass of magnesium deposited by simple proportion using the relative atomic mass of Mg

Charge (C)	Amount of Mg deposited (mol)	Amount of Mg deposited (g)
192,970	1	24.30
1980	= 0.0103	0.0103 x 24.30 = 0.25

• Therefore, 0.25 g of magnesium is deposited at the cathode

A current of 2.18 A is delivered to an electrolytic cell for 65 min. 

How many grams of the following will be obtained

1) Al from AlCl₃

2) Ag from AgNO₃

3) Cu from CuCl₂

Answers:

To work out mass:

1. Write half equation
2. find mol of solid by =  (where n = mol e^– per mole of substance in balanced equation)
3. find mass of solid = mol x M = x g

Answer 1:

• Au³⁺ + 3e^– → Al
• mol Au = = 0.02937 mol
• g Au = 0.02937 x 196.97 g mol^–1 = 5.79 g

Answer 2:

• Ag⁺ + 1e^– → Ag
• mol Ag = = 0.08812 mol
• g Ag = 0.08812 x 107.87 g mol^–1 = 9.51 g

Answer 3:

• Cu²⁺ + 2e^– → Cu
• mol Cu =  = 0.04406 mol
• g Cu = 0.04406 x 63.55 g mol^–1 = 2.80 g