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Answer :
To solve for the current needed to electroplate 5.00 grams of silver in 3600 seconds, we need to use Faraday's laws of electrolysis, which relate the amount of substance deposited at an electrode to the amount of electric charge passed through the solution.
Step-by-step solution:
Calculate the number of moles of silver (Ag) to be electroplated.
The molar mass of silver is approximately 107.87 g/mol. To find the number of moles of silver, use the formula:
[tex]ext{moles of Ag} = \frac{5.00 \, \text{g}}{107.87 \, \text{g/mol}} \approx 0.04637 \, \text{mol}[/tex]
Use the electrochemical equivalent of silver.
The electrochemical equivalent of silver is such that it takes 1 mole of electrons to deposit 1 mole of silver, and the charge of 1 mole of electrons (1 Faraday) is approximately 96485 coulombs.
Calculate the required total charge (Q).
Using Faraday’s law, the total charge required can be calculated as:
[tex]Q = ext{moles of Ag} \times 96500 \, \text{C/mol}[/tex]
[tex]Q = 0.04637 \, \text{mol} \times 96500 \, \text{C/mol} \approx 4472.91 \, \text{C}[/tex]
Calculate the current (I).
The relationship between charge (Q), current (I), and time (t) is given by:
[tex]I = \frac{Q}{t}[/tex]
Where:
- [tex]Q[/tex] is charge in coulombs
- [tex]t[/tex] is time in seconds (3600 seconds in this case)
Therefore, the current needed is:
[tex]I = \frac{4472.91 \, \text{C}}{3600 \, \text{s}} \approx 1.24 \, \text{A}[/tex]
Conclusion:
The current needed to electroplate 5.00 grams of silver in 3600 seconds is approximately 1.24 amperes (A).
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