[slambert]

I am stuck with the last question I am working on for chem II and was curious if anyone could help me with that:

How many grams of copper are deposited on the cathode of an electrolytic cell if an electric current of 2.00 A is run through a solution of CuSO₄ for 19.0 minutes.

I have searched to figure it out and keep coming up with answers that do not make sense.
Thank you

[Borek]

Faraday's law. Show your work, even if you think it doesn't make sense.

[slambert]

Okay I understand Faraday's law to a point. So what does the electric current and minutes have to do with the number of grams?

[ARGOS++]

Dear Slambert;

Because the Current in your experiment is constant, so the Product of current and time gives you Q in the Law of Faraday:
 

Q = Current * Time = I *t   = Total electrical Charge during Electrolysis.

"Faraday's Law of Electrolysis”

I hope it may be of help to you.

Good Luck!
                    ARGOS⁺⁺

[slambert]

okay so this is what I came up with, can anyone tell me if im way off or if its right?

m=QM/zF
C=1A*1s
2*(19*60)
C=2(1140)=2280

m=(2280C)(63.54g/mol)/(2)(96,500C/mol^-1)
m=144871.2/193000
m=.75g

[Borek]

m=.75g

Seems OK.