How many grams of copper will be deposited at the cathode of an electrolytic cell if a current of 680.0 mA is run through a 2.5 M solution of copper sulfate for 20.0 minutes?

Q = coulombs = amperes x seconds

Q = 0.680A x 20m x (60s/m) = ? = about 816
96,485 coulombs will deposit (63.54/2)g Cu. You have 816 coulombs; therefore,
63.54/2 x (816/96,485) = about 0.3g if I punched the right buttons.

To determine the amount of copper deposited at the cathode, we can use Faraday's law of electrolysis. The formula is:

mass of substance = (current (in amperes) x time (in seconds) x molar mass) / (number of electrons x Faraday's constant)

First, we need to convert the given values to the proper units:

current = 680.0 mA = 680.0 / 1000.0 A = 0.680 A
time = 20.0 minutes = 20.0 x 60 seconds = 1200 seconds
molar mass of copper = 63.55 g/mol (from the periodic table)
number of electrons involved in the copper deposition reaction = 2 (from the balanced chemical equation)
Faraday's constant = 96,485 C/mol

Now, we can substitute these values into the formula:

mass of copper = (0.680 A x 1200 s x 63.55 g/mol) / (2 x 96,485 C/mol)

Calculating this, we get:

mass of copper = (972.96 g*s) / (192,970 C)

mass of copper = 0.00504 g

Therefore, approximately 0.00504 grams of copper will be deposited at the cathode of the electrolytic cell.

To find the number of grams of copper deposited at the cathode, we need to use Faraday's law of electrolysis. This law states that the amount of substance deposited at an electrode is directly proportional to the current passing through the cell and the time of electrolysis.

First, let's calculate the charge (Q) passed through the electrolytic cell. We can use the formula:

Q = I * t

where:
Q is the charge passed through the electrolytic cell in coulombs (C)
I is the current in amperes (A)
t is the time in seconds (s)

Converting the given time, which is 20.0 minutes, to seconds:

t = 20.0 minutes * 60 seconds/minute = 1200 seconds

Now, substituting the given current and time into the formula:

Q = 0.680 A * 1200 s = 816 C

Next, we need to determine the number of moles of copper deposited. Since copper has a +2 charge in copper sulfate, we need to divide the charge (Q) by the Faraday constant (F), which is 96,485 C/mol.

n = Q / F

n = 816 C / 96,485 C/mol = 0.0085 mol

Finally, we can find the mass of copper deposited using the molar mass of copper (Cu), which is 63.55 g/mol.

Mass = n * molar mass

Mass = 0.0085 mol * 63.55 g/mol ≈ 0.540 g

Therefore, approximately 0.540 grams of copper will be deposited at the cathode of the electrolytic cell.