What amount of current, in amperes, would be required to deposit 19.7 g of Au metal in 1.00 hour from a solution of Au(NO3)3?
You can deposit 196.97/3 = approx 66 g Au with 96,485 coulombs of electricity. So how much will be required to deposit 19.7g
96,485 x (19.7/66) = approx 29,000 coulombs.
coulombs = amperes x seconds.
Substitute coulombs and seconds and solve for amperes.
Note those numbers I used are approximate. You need to redo all of them.
To determine the amount of current required, we can use Faraday's Law of Electrolysis. The formula is as follows:
1. Amount of Substance (in moles) = Mass of Substance (in grams) / Molar Mass of Substance
For gold (Au), the molar mass is 196.967 g/mol. Let's calculate the number of moles of Au:
Amount of Substance (in moles) = 19.7 g / 196.967 g/mol
Next, we need to use Faraday's Law to relate the amount of substance to the amount of electricity passed through the solution:
2. Amount of Electricity (in coulombs) = Amount of Substance (in moles) x Faraday's Constant
The Faraday's constant is approximately 96485 C/mol.
Let's calculate the amount of electricity required:
Amount of Electricity (in coulombs) = (19.7 g / 196.967 g/mol) x 96485 C/mol
Now, since we know the time in hours, we need to convert it to seconds:
1 hour = 3600 seconds
Finally, we can calculate the current in amperes using the formula:
3. Current (in amperes) = Amount of Electricity (in coulombs) / Time (in seconds)
Let's substitute the values into the equation to find the current:
Current (in amperes) = [(19.7 g / 196.967 g/mol) x 96485 C/mol] / (1 hour x 3600 seconds/hour)
By evaluating this equation, we can determine the current required to deposit 19.7 g of Au metal in 1.00 hour from a solution of Au(NO3)3.