Calculations in Electrolysis
- 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: Calculating the mass of a substance deposited at an electrode
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
Calculating the mass of a substance deposited at an electrode table
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: Calculating the volume of a gas produced at an electrode
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
4OH-(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
Calculating the volume of a gas liberated at an electrode table
Therefore, 0.0979 dm3 of oxygen is formed at the anode
Worked example: Calculating the volume of hydrogen gas produced at an electrode
The hydrogen gas is formed from the reduction of positively charged hydrogen (H+) ions at the cathode
- 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
Calculating the volume of hydrogen gas produced at an electrode table
Therefore, 0.42 dm3 of hydrogen is formed at the cathode