How many grams of copper are deposited on the cathode of an electrolytic cell if an electric current of 2.00 A is passed through a solution of CuSO4 for a period of 19.0 min?
coulombs - A x seconds = ?
096,485 coulombs will deposit 63.55/2 g Cu = approx 32 but you need a better number than that.
Then 32? x coulombs/96,485 = g Cu deposited.
what are the coulombs? how do I find them?
and also where can I find that equation?
Well, this question is shocking! Let's calculate the amount of copper deposited with a little spark of humor!
The formula to calculate the amount of substance deposited during electrolysis is:
Amount = (Current x Time) / Faraday's constant
Now, to find out the amount of copper deposited, we need to determine Faraday's constant. It's approximately 96,485 Coulombs per mole.
Given that the electric current is 2.00 A and the time is 19.0 min (which we'll convert to seconds), let's plug in the values:
Amount = (2.00 A x 19.0 min x 60 s/min) / 96485 C/mol
Now, my circuits are a bit rusty, but let me grab my calculator and crunch the numbers for you.
Calculating... Calculating... Voilà!
The amount of copper deposited on the cathode is approximately X grams. Sorry, I can't give you the exact value without knowing the concentration of the copper sulfate solution. You'll have to check that information separately.
Remember, though, this is an estimation, so don't take it too seriously. Have a "copper"-tastic day! 🤡
To determine the number of grams of copper deposited on the cathode, we need to use Faraday's law of electrolysis.
Faraday's law states that the mass of a substance produced or deposited at an electrode is directly proportional to the amount of electric charge passed through the electrolyte solution. The equation is given as:
m = (Q × M) / (z × F)
Where:
- m is the mass of the substance deposited (in grams)
- Q is the electric charge passed through the solution (in Coulombs)
- M is the molar mass of the substance (in grams per mole)
- z is the charge (or number of electrons) transferred per ion
- F is Faraday's constant, which is approximately 96,500 Coulombs per mole of electrons
In this case, we want to find the mass of copper deposited, so M is the molar mass of copper (63.55 g/mol). The charge transferred per ion for copper is 2 (since Cu2+ ions are involved in the reaction).
Now, let's calculate the electric charge passed (Q) using the formula:
Q = I × t
Where:
- I is the electric current (in Amperes)
- t is the time (in seconds)
Given that the electric current (I) is 2.00 A and the time (t) is 19.0 min (which needs to be converted to seconds), we can substitute these values into the equation:
t = 19.0 min × 60 s/min = 1140 s
Now we have all the necessary values to calculate the mass of copper deposited:
m = (Q × M) / (z × F)
Substituting the values:
m = (2.00 A × 1140 s × 63.55 g/mol) / (2 × 96,500 C/mol)
By solving this equation, we can find the mass of copper deposited on the cathode.
See my first line. Not only did I tell you that you needed coulombs but I told you how to calculate it.
coulombs = A x seconds.
amperes in the problem = 2.00 A.
time in the problem = 19.0 min or 19.0 x (60 seconds/min) = ? seconds so
coulombs = 2.00 x 19.0 x 60 = >