Question

Calculate the time required to deposit 2.4 g of Cu, when 2.03 A of current is passed through \( \text{CuSO}_4 \) solution. 
(At. mass of Cu = 63.5 g mol\(^{-1}\))

Hint:

Faraday's laws quantitatively link the mass of substances altered at an electrode to the amount of electricity passed through the substance.

Answer 1

Apply Faraday’s first law of electrolysis, which is expressed as: \[ m = \frac{Z I t}{F} \] Definitions: 
- \( m \) = mass of the deposit (2.4 g), 
- \( I \) = electric current (2.03 A), 
- \( F \) = Faraday’s constant (96500 C/mol), 
- \( Z \) = electrochemical equivalent, calculated as \( \frac{M}{nF} \) (for Cu, \( M = 63.5 \), \( n = 2 \)). 
First, calculate the electrochemical equivalent \( (Z) \) for copper: \[ Z = \frac{M}{nF} = \frac{63.5}{2 \times 96500} = \frac{63.5}{193000} \approx 3.296 \times 10^{-4} \, \text{g/C} \] To find the time \( (t) \) required for the deposition: \[ t = \frac{m}{Z I} \] Plugging in the known values: \[ t = \frac{2.4}{(3.296 \times 10^{-4}) \times 2.03} \] \[ t = \frac{2.4}{6.686 \times 10^{-4}} \approx 3593.34 \, \text{seconds} \] 
Therefore, approximately 3593.34 seconds are needed to deposit 2.4 g of copper.