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).

The density of Cu is 8.94 g/cm³. What is the quantity of electricity needed to plate an area 10 cm x 10 cm to a thickness of 10^-2 cm using CuSO₄ solution?

\((1F = 96500 C)\)

27172 C

The correct answer is option 2. 27172 C.

To calculate the quantity of electricity needed to plate a copper \((Cu)\) layer of a given thickness and area using \(CuSO_4\) solution, we'll follow these steps:

Determine the volume of copper to be plated:

\(\text{Volume} = \text{Area} \times \text{Thickness}\)

Given:

\(\text{Area} = 10 \, \text{cm} \times 10 \, \text{cm} = 100 \, \text{cm}^2\)

\(\text{Thickness} = 10^{-2} \, \text{cm}\)

\(\text{Volume} = 100 \, \text{cm}^2 \times 10^{-2} \, \text{cm} = 1 \, \text{cm}^3\)

Calculate the mass of copper to be plated:

\(\text{Density of Cu} = 8.94 \, \text{g/cm}^3\)

\(\text{Mass} = \text{Volume} \times \text{Density} = 1 \, \text{cm}^3 \times 8.94 \, \text{g/cm}^3 = 8.94 \, \text{g}\)

Determine the moles of copper to be plated:

\(\text{Molar mass of Cu} = 63.55 \, \text{g/mol}\)

\(\text{Moles of Cu} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{8.94 \, \text{g}}{63.55 \, \text{g/mol}} = 0.1407 \, \text{mol}\)

Calculate the charge required to plate the copper:

The reduction reaction for copper is:

\(\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}\)

This means 2 moles of electrons (2 faradays) are needed to deposit 1 mole of copper.

\(\text{Faradays required} = \text{Moles of Cu} \times 2 = 0.1407 \, \text{mol} \times 2 = 0.2814 \, \text{faradays}\)

1 faraday is equivalent to 96485 coulombs (C).

\(\text{Charge (C)} = \text{Faradays} \times 96500 \, \text{C/faraday} = 0.2814 \times 96500 \, \text{C} \approx 27172 \, \text{C}\)

So, the quantity of electricity needed to plate the copper is: 27172 C.