Question

Electrochemistry is the study of the relationship between chemical energy and electrical energy. Many spontaneously occurring chemical reactions liberate electrical energy. In electrolysis, electrical energy is converted directly into chemical energy. The product of an electrolytic reaction depends on the nature of the material being electrolysed and the type of electrode used. Oxidising and reducing species present in the electrolytic cell and their standard electrode potential too, affect the products of electrolysis. Electrolysis plays an important role in most people's daily lives, whether it is for the manufacturing of aluminium, electroplating of metals, or the synthesis of chemical compounds.

Michael Faraday was the first scientist who proposed two laws to explain the quantitative aspects of electrolysis, popularly known as Faraday's laws of electrolysis. Faraday's laws of electrolysis provide a basis for mathematical analysis of the mass deposited at electrodes and the amount of charge passed through them.

Faraday's laws are fundamental in various applications, including electroplating, metal extraction, battery technology and chemical synthesis. These laws also help in environmental monitoring and in various chemistry experiments.

Answer the following questions :

(b) State Faraday's second law of electrolysis.

Hint:

The second law compares different electrolytes. The same charge gives masses proportional to their equivalent weights.

Answer 1

Step 1: Set up the situation:
Take several electrolytic cells joined in series, each holding a different electrolyte.
The same current flows for the same time, so the same quantity of electricity passes through every cell.

Step 2: State the law:
Faraday's second law says: the amounts of different substances liberated by the same quantity of electricity passing through the electrolytic solutions are proportional to their chemical equivalent weights.
The equivalent weight of a substance is its molar mass divided by the number of electrons transferred per ion.

Step 3: Write it as a relation:
\[ \frac{m_1}{m_2} = \frac{E_1}{E_2} \]
Here \( m_1, m_2 \) are the masses deposited and \( E_1, E_2 \) are the equivalent weights.
For example, the same charge deposits 108 g of Ag but only 31.75 g of Cu, which is the ratio of their equivalent weights.

Final Answer:
When the same charge passes through different electrolytes, the masses of the substances liberated are in the ratio of their equivalent weights.