Electrolysis: Calculations (Cambridge International A-Level Chemistry)

Electrolysis Calculations

• Faraday’s constant can be used to calculate:

  • The mass of a substance deposited at an electrode
  • The volume of gas liberated at an electrode

Calculating the mass of a substance deposited at an electrode

• To calculate the mass of a substance deposited at the electrode, you need to be able to:

  • Write the half-equation at the electrode
  • Determine the number of coulombs needed to form one mole of substance at the specific electrode using Faraday’s constant
  • Calculate the charge transferred during electrolysis
  • Use simple proportion and the relative atomic mass of the substance to find its mass

Worked Example 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: 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 96 500 C mol^-1

In this case, there are two moles of electrons required

So, the number of coulombs needed is:

F = 2 x 96 500

F = 193 000 C mol^-1

• **Step 3: **Calculate the charge transferred during the electrolysis

Q = I x t

Q = 2.20 x (60 x 15) = 1980 C****

• 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)
193 000	1	24.3
1980	$\frac{1980}{193000}$ = 0.0103	0.0103 x 24.3 = 0.25

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

Calculating the volume of gas liberated at an electrode

• To calculate the volume of gas liberated at an electrode, you need to be able to:

  • Write the half-equation at the electrode
  • Determine the number of coulombs needed to form one mole of substance at the specific electrode using Faraday’s constant
  • Calculate the charge transferred during electrolysis
  • Use simple proportion and the relationship 1 mol of gas occupies 24.0 dm³ at room temperature

Worked Example Calculate the volume of oxygen gas produced at room temperature, when a concentrated aqueous solution of sulfuric acid, H₂SO₄, is electrolysed for 35.0 minutes using a current of 0.75 A.

Answer:

• The oxygen gas is formed from the oxidation of negatively charged hydroxide (OH^-) ions at the anode-
• **Step 1: **Write the half-equation at the anode

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

4 mol 1 mol 2 mol 4 mol

• **Step 2: **Determine the number of coulombs required to form one mole of oxygen gas at the anode

For every one mole of electrons, the number of coulombs needed is 96 500 C mol^-1

So, for four moles of electrons, the number of coulombs needed is:

F = 4 x 96 500

F = 386 000 C mol^-1

• **Step 3: **Calculate the charge transferred during the electrolysis

Q = I x t

Q = 0.75 x (60 x 35) = 1575 C

• **Step 4: **Calculate the volume of oxygen liberated by simple proportion using the relationship 1 mol of gas occupies 24.0 dm³ at room temperature

Charge (C)	Amount of O2** liberated (mol)**	Amount of O2** liberated (dm**3)
386 000	1	24
1575	$\frac{1575}{386000}$ = 4.080 x 10-3	4.080 x 10-3 x 24.0 = 0.0979

• Therefore, 0.0979 dm³ of oxygen is formed at the anode

Worked Example Calculating the volume of hydrogen gas produced at room temperature, when a concentrated aqueous solution of sodium sulfate, Na₂SO₄, is electrolysed for 17.5 minutes using a current of 3.25 A.

The hydrogen gas is formed from the reduction of positively charged hydrogen ions, H⁺ at the cathode.

Answer:

• **Step 1: **Write the half-equation at the cathode

2H⁺(aq) + 2e^-→ H₂(g)

2 mols 2 mols 1 mol

• **Step 2: **Determine the number of coulombs required to form one mole of hydrogen gas at the cathode

  • For every one mole of electrons, the number of coulombs needed is:

F = 96 500 C mol^-1

F = 1 x 96 500

F = 96 500 C

• So, for two moles of electrons, the number of coulombs needed is:

F = 2 x 96 500

F = 193 000 C

• **Step 3: **Calculate the charge transferred during the electrolysis

Q = I x t

Q = 3.25 x (60 x 17.5) = 3 413 C

• **Step 4: **Calculate the volume of hydrogen liberated by simple proportion using the relationship 1 mol of gas occupies 24.0 dm³ at room temperature

Charge (C)	Amount of H2** liberated (mol)**	Amount of H2** liberated (dm**3)
193 000	1	24
3413	$\frac{3413}{193000}$ = 1.76 x 10-2	1.76 x 10-2 x 24 = 0.42

• Therefore, 0.42 dm³ of hydrogen is formed at the cathode