Electrolysis: Calculations (Cambridge (CIE) A Level Chemistry): Revision Note
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 |
| 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
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
Charge (C) | Amount of O2 liberated (mol) | Amount of O2 liberated (dm3) |
---|---|---|
386 000 | 1 | 24 |
1575 |
| 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 (dm3) |
---|---|---|
193 000 | 1 | 24 |
3413 |
| 1.76 x 10-2 x 24 = 0.42 |
Therefore, 0.42 dm3 of hydrogen is formed at the cathode
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