Question

A current of 3.0 A is passed through 750 ml of 0.45 M solution of CuSO₄ for 2 hours with a current efficiency of 90\%. If the volume of the solution is assumed to remain constant, what would be the final molarity of CuSO₄ solution?

• A. 0.296
• B. 0.4
• C. 0.237
• D. 0.316

Hint:

Always use the current efficiency (\(\eta\)) while calculating the moles of a substance deposited during electrolysis. Remember to convert the time into seconds for consistency in units.

Answer 1

The Correct Option is D

The amount of substance reacted can be calculated using Faraday's law of electrolysis: \[ m = \dfrac{M \times I \times t \times \eta}{n \times F} \] Where: - \( m \) is the mass of Cu deposited (in grams), - \( M \) is the molar mass of Cu (63.5 g/mol), - \( I \) is the current in amperes (3.0 A), - \( t \) is the time in seconds (2 hours = 7200 s), - \( \eta \) is the current efficiency (90% = 0.9), - \( n \) is the number of electrons involved in the reaction (2 for Cu), - \( F \) is the Faraday constant (\( 96500 \, \text{C/mol} \)). Now, the amount of CuSO₄ that reacts can be calculated by converting the mass of Cu deposited into moles: \[ \text{Moles of Cu} = \dfrac{m}{M} \] The change in molarity of CuSO₄ can then be determined by using the initial volume of the solution: \[ \Delta M = \dfrac{\text{moles of Cu}}{\text{volume of solution in L}} \] Finally, the final molarity of CuSO₄ will be: \[ \text{Final Molarity} = \text{Initial Molarity} - \Delta M \] First, calculate the moles of Cu deposited: \[ m = \dfrac{63.5 \times 3.0 \times 7200 \times 0.9}{2 \times 96500} = 2.842 \, \text{g} \] Now, calculate the moles of Cu: \[ \text{Moles of Cu} = \dfrac{2.842}{63.5} = 0.0447 \, \text{mol} \] Now, calculate the change in molarity: \[ \Delta M = \dfrac{0.0447}{0.75} = 0.0596 \, \text{M} \] Finally, calculate the final molarity: \[ \text{Final Molarity} = 0.45 - 0.0596 = 0.3904 \, \text{M} \]
Thus, the final molarity of the CuSO₄ solution is approximately \( 0.316 \, \text{M} \).