Law of electrolysis

Laws of electrolysis

Faraday put forward two laws expressing the results of experiments in electrolysis. These laws assert that the amount of an element liberated during electrolysis depends on:

• The time of passing the steady current
• The magnitude of the steady current
• The charge on the ion of the element

Faraday's first law of electrolysis

Faraday's First Law states that:

"The mass of a substance liberated at or absorbed from an electrode during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte."

If symbols are used, the first law can be stated mathematically as:

$m = ZIt$

Where:

• $m$ = mass of the substance liberated
• $t$ = time in seconds
• $I$ = current in Amperes

To replace the symbol of proportionality by an equal sign, we must introduce a proportionality constant, $Z$:

$m = ZQ$

Where:

• $Q$ = quantity of electricity ($Q = It$)
• $Z$ = electrochemical equivalent of an element

The electrochemical equivalent of a substance is the mass of the substance that can be liberated by the passage of one coulomb. Its SI unit is g/c (grams per coulomb).

Every element has its own value of electrochemical equivalent, which can be calculated using the formula:

$Z = \frac{A_r \times F}{V}$

Where:

• $A_r$ = relative atomic mass of an element
• $V$ = valency of the element
• $F$ = Faraday's constant (96500 C)

Note: Faraday's constant is the quantity of electric charge carried by electrons (96500 C per mole).

Faraday's second law of electrolysis

Faraday's Second Law states that:

"When the same quantity of electricity is passed through solutions of different electrolytes, the amounts of elements deposited are proportional to the chemical equivalent of the element."

The chemical equivalent of an element is obtained by dividing its relative atomic mass by the charge of its ion.

Mathematically

$\text{Mass of substance deposited} = \text{Chemical Equivalent of Substance} \times \text{Quantity of Electricity}$

Example

In an experiment, 1930 C liberated 0.64g of copper. When the same quantity of electricity was passed through a solution containing silver ions (Ag$^{+}$), what amount of Ag was deposited?

Given:

• Mass of copper ($M_{Cu}$) = 0.64g
• Quantity of electricity ($Q$) = 1930 C

Now, to calculate the mass of silver ($M_{Ag}$) deposited, we use the chemical equivalent of copper (Cu) and silver (Ag):

$M_{Ag} = \frac{M_{Cu} \times E_g}{E_c}$

Where:

• $E_g$ = Chemical equivalent of Ag
• $E_c$ = Chemical equivalent of Cu

Substitute values and solve:

$M_{Ag}$ = 21.6g

The mass of silver deposited is 21.6g. This result is larger than the mass of copper because silver has a smaller chemical equivalent than copper.