Michael Faraday was the first scientist who described the quantitative aspects of electrolysis. Now Faraday’s laws also flow from what has been discussed earlier. Faraday’s Laws of Electrolysis After his extensive investigations on electrolysis of solutions and melts of electrolytes, Faraday published his results during 1833-34 in the form of the following well known Faraday’s two laws of electrolysis:

(i) First Law: The amount of chemical reaction which occurs at any electrode during electrolysis by a current is proportional to the quantity of electricity passed through the electrolyte (solution or melt).

(ii) Second Law: The amounts of different substances liberated by the same quantity of electricity passing through the electrolytic solution are proportional to their chemical equivalent weights (Atomic Mass of Metal ÷ Number of electrons required to reduce the cation).

When electricity is passed through a solution of AlCl₃, 13.5 g of Al is deposited. What is the number of faradays involved? 

1.5

The correct answer is option 2. 1.5.

To determine the number of faradays involved in depositing 13.5 g of aluminum \((Al)\) from a solution of aluminum chloride \((AlCl_3)\), we need to use the relationship between the amount of substance deposited, the equivalent weight, and Faraday's laws of electrolysis.

Step-by-Step Calculation:

Determine the molar mass of aluminum (Al):

\(\text{Molar mass of Al} = 27 \, \text{g/mol}\)

Calculate the number of moles of aluminum deposited:

\(\text{Moles of Al} = \frac{\text{Mass of Al}}{\text{Molar mass of Al}} = \frac{13.5 \, \text{g}}{27 \, \text{g/mol}} = 0.5\, \text{mol}\)

Write the half-reaction for the reduction of \(Al^{3+}\) to \(Al\):

\(\text{Al}^{3+} + 3e^- \rightarrow \text{Al}\)

This indicates that 3 moles of electrons (3 faradays) are required to deposit 1 mole of aluminum.

Calculate the number of faradays required to deposit 0.5 moles of aluminum:

\(\text{Faradays} = \text{Moles of Al} \times \frac{\text{Faradays per mole of Al}}{\text{Moles of Al}} = 0.5 \, \text{mol} \times 3 \, \text{F/mol} = 1.5 \, \text{F}\)

So, the number of faradays involved in depositing 13.5 g of Al is: 1.5.