To solve this problem, we will use Faraday's laws of electrolysis to determine the mass of copper (Cu) deposited during the electrochemical reaction.
Firstly, we need to convert the time from minutes to seconds, because the unit of electric current (Ampere) is coulombs per second. We have:
15 minutes x 60 seconds/minute = 900 seconds
Now, we will calculate the total charge \( Q \) that passes through the solution, which is the product of the current \( I \) and the time \( t \). With an electric current of 12 A for 900 seconds, we get:
\( Q = I \times t \)
\( Q = 12 \, \text{A} \times 900 \, \text{s} \)
\( Q = 10800 \, \text{C} \)
Next, we need to calculate the electrochemical equivalent of copper. This constant tells us how much copper will be deposited per coulomb of charge passed. The electrochemical equivalent \( z \) can be determined using the molar mass \( M \) of copper and Faraday's constant \( F \):
\( z = \frac{M}{n \times F} \)
Where:
- \( M \) is the molar mass of copper (63.5 g/mol),
- \( n \) is the valency of copper, which is 2 for copper in CuSO4,
- \( F \) is Faraday's constant (96500 C/mol),
\( z = \frac{63.5 \, \text{g/mol}}{2 \times 96500 \, \text{C/mol}} \)
Now, calculate the electrochemical equivalent of copper:
\( z = \frac{63.5 \, \text{g/mol}}{193000 \, \text{C/mol}} \)
Now let's calculate this value:
\( z \approx \frac{63.5}{193000} \, \text{g/C} \)
\( z \approx 0.000329 \, \text{g/C} \)
Finally, we will calculate the mass \( m \) of copper deposited, which is the product of the electrochemical equivalent \( z \) and the total charge \( Q \):
\( m = z \times Q \)
\( m = 0.000329 \, \text{g/C} \times 10800 \, \text{C} \)
\( m \approx 3.5532 \, \text{g} \)
So, when rounding to one decimal place, the mass of copper deposited is approximately 3.6 grams. Therefore, the answer to the question is 3.6 grams.
