Faraday’s Laws of Electrolysis
Faraday put forward two laws of electrolysis and he also discovered from his experiment that the quantity of products of electrolysis depends on the following
- Time of the flow of electric current
- The ionic charge of the element deposited
- The amount of current
Faraday’s First Law of Electrolysis
The first law states that the amount of the element deposited during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte (electrolytic solution).
Mathematically
M ∞ Q
Where
Q = quantity of electricity measured in Coulomb
M = mass of the substance deposited
Since the quantity of electricity is given by
Q = current x time in seconds
Faraday’s first law can also be stated as “the mass of element deposited during an electrolysis is proportional to the size of the current and the time taken for the flow of the current.”
M ∞ I x t
Or
M = є I t
Where є is a constant called electrochemical equivalent.
I stands for the amount of current and t stands for the time taken for the flow.
Faraday’s Second Law of Electrolysis
The second law states that when the same quantity of electric current is passed through electrolytes at the same time, the masses of the different substances deposited at the respective electrodes are inversely proportional to the charges on the ions of the elements.
Faraday’s Law of Electrolysis Calculations
- Calculate the amount of gold deposited when a current of 10A is passed through a solution of gold salt for 4 hours. If the same current is used, find the time taken for 12g of gold to be deposited.
(Au = 197, 1faraday = 96500C)
Solution
Au+ + e- → Au
From the ionic equation above, gold requires one mole of an electron to deposit one mole of a gold element. It also means that one faraday is required to deposit one mole of gold.
1F = 1mole of gold or 197g of gold
96500C = 197g of gold
Now, To find the quantity of electricity that will deposit the required mass of gold.
Quantity of electric current = current x time in seconds
= 10 x 4 x 60 x60 = 144000C
96500C = 197g of gold
144000 = 197/96500 x 144000 = 293.97g of gold.
To solve the second question, use the same quantity of electricity.
If 293.97g of gold is required = 144000C
12g of gold will need = 144000/ 293.97 x 12 = 5878.2C
To calculate the time, we are going to use the relation below;
Q = It
Where Q is 5878.2C , I is 10A
5878.2 = 10 x t
time = 5878.2/10 = 587.8 seconds or 9.8 minutes.
- 0.221g of a divalent metal is deposited when a current of 0.45A is passed through a solution of its salt for 25 minutes. Calculate the relative atomic mass of the metal. ( 1F = 96500C)
Solution
Let the symbol of the divalent metal be = X
X2+ + 2e- → X
Since X is divalent metal, it will need two electrons to deposit one mole of
The metal. The first approach is to calculate the quantity of electricity required to deposit 0.221g of the metal
Q = 0.45 x 25 x 60 = 675C
675C = 0.221g of X metal
2x 96500C =
0.221/675 x 139000 =63.2
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