3.25x10-3 kg of gold is deposited onto the negative electrode of an electrolytic cell in a period of 2.78 hours, what is the current through this cell in amperes? Assume that the gold ions carry one elementary unit of charge
I worked this below. You should be able to find it. If not I can try and locate it. It is filed under a chemistry banner, not physics.
To find the current through the electrolytic cell, we need to use Faraday's law of electrolysis, which states that the amount of substance deposited or released at an electrode is directly proportional to the amount of electric charge passed through the cell.
The formula for Faraday's law is:
Q = n × F
where
Q = electric charge (in coulombs),
n = number of moles of substance,
F = Faraday's constant (96485 C/mol).
First, let's calculate the number of moles of gold deposited:
n = mass / molar mass
Given:
mass of gold deposited = 3.25 × 10^(-3) kg
molar mass of gold = 197.0 g/mol (or 0.197 kg/mol)
So, the number of moles of gold is:
n = 3.25 × 10^(-3) kg / 0.197 kg/mol
Next, we calculate the electric charge passed through the cell:
Q = n × F
The gold ions carry one elementary unit of charge, so the number of moles of gold is equal to the number of elementary charges. Therefore, the electric charge passed through the cell is:
Q = n × F = n × (96485 C/mol)
Now, we need to convert the time to seconds:
time = 2.78 hours = 2.78 × 3600 seconds
Finally, we can calculate the current using the formula:
Current (I) = Q / time
Substituting the values, we get:
Current (I) = Q / time = (n × F) / time
By plugging in the values for n, F, and time, we can calculate the current in amperes.