What current would you need to apply to a solution of Cu2+ to plate out 40.0 g of copper metal in 24 hours? Assume you have the required voltage and there is sufficient copper in solution for this to take place.
The correct answer and explanation is:
To calculate the current needed to plate out 40.0 g of copper metal (Cu) in 24 hours, we can use Faraday’s law of electrolysis. The equation is: m=M⋅I⋅tn⋅Fm = \frac{M \cdot I \cdot t}{n \cdot F}
Where:
- mm is the mass of the metal deposited (in grams),
- MM is the molar mass of the metal (Cu is 63.55 g/mol),
- II is the current (in amperes, which we are solving for),
- tt is the time (in seconds),
- nn is the number of electrons involved in the reaction (for Cu2+ to Cu, n=2n = 2),
- FF is Faraday’s constant (F=96485 C/molF = 96485 \, \text{C/mol}).
Step 1: Convert time to seconds
24 hours = 24×60×60=86400 seconds24 \times 60 \times 60 = 86400 \, \text{seconds}.
Step 2: Rearrange the equation to solve for current II
I=m⋅n⋅FM⋅tI = \frac{m \cdot n \cdot F}{M \cdot t}
Step 3: Substitute the known values
- m=40.0 gm = 40.0 \, \text{g},
- n=2n = 2,
- M=63.55 g/molM = 63.55 \, \text{g/mol},
- F=96485 C/molF = 96485 \, \text{C/mol},
- t=86400 secondst = 86400 \, \text{seconds}.
I=40.0 g⋅2⋅96485 C/mol63.55 g/mol⋅86400 secondsI = \frac{40.0 \, \text{g} \cdot 2 \cdot 96485 \, \text{C/mol}}{63.55 \, \text{g/mol} \cdot 86400 \, \text{seconds}}
Step 4: Perform the calculation
I=40.0×2×9648563.55×86400≈77188005483040≈1.41 AI = \frac{40.0 \times 2 \times 96485}{63.55 \times 86400} \approx \frac{7718800}{5483040} \approx 1.41 \, \text{A}
Thus, the required current is approximately 1.41 A.
Explanation:
To plate out 40.0 g of copper, the current is calculated based on the electrolysis process where copper ions (Cu2+Cu^{2+}) are reduced to copper metal at the cathode. Faraday’s law provides a relationship between the mass of metal deposited and the current passed through the solution. By considering the molar mass of copper, the number of electrons involved in the reduction, and the time allowed for the plating process, the required current can be determined. In this case, applying a current of approximately 1.41 amperes over 24 hours will deposit 40.0 g of copper from the solution.