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).

In a solution of CuSO₄, 0.5 A current passes for 60 minutes. How much copper is deposited if current efficiency is 50%? (Cu = 63.5 g)

0.296 g

The correct answer is option 2. 0.296 g.

To determine how much copper is deposited when a current of 0.5 A passes through a solution of CuSO4 for 60 minutes (which is 3600 seconds) with a current efficiency of 50%, we can follow these steps:

Given:

Current \( I = 0.5 \) A

Time \( t = 60 \) minutes = \( 60 \times 60 = 3600 \) seconds

\(\text{Charge} = I \times t = 0.5 \, \text{A} \times 3600 \, \text{s} = 1800 \, \text{C}\)

Determine the moles of electrons transferred:

Faraday's constant \( F \) is approximately \( 96485 \) C/mol.

\(\text{Moles of electrons} = \frac{\text{Charge}}{F} = \frac{1800 \, \text{C}}{96485 \, \text{C/mol}} \approx 0.01867 \, \text{mol}\)

Calculate the moles of copper deposited: The reduction half-reaction for copper ion (\( \text{Cu}^{2+} \)) to copper metal (Cu) is:

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

According to the reaction, 2 moles of electrons are required to deposit 1 mole of copper.

\(\text{Moles of Cu} = \frac{\text{Moles of electrons}}{2} = \frac{0.01867 \, \text{mol}}{2} = 0.00934 \, \text{mol}\)

Calculate the mass of copper deposited:

\(\text{Mass of Cu} = \text{Moles of Cu} \times \text{Molar mass of Cu} = 0.00934 \, \text{mol} \times 63.5 \, \text{g/mol} = 0.592 \, \text{g}\)

Consider the current efficiency: Current efficiency is given as 50%, meaning only half of the calculated amount of copper (0.592 g) will actually be deposited.

\(\text{Actual mass of Cu deposited} = 0.592 \, \text{g} \times 0.5 = 0.296 \, \text{g}\)

Therefore, the mass of copper deposited, considering a current efficiency of 50%, is approximately: 0.296 g.