Electrochemistry How to calculate copper deposition on cathodes per hour?

[Knowledgethirst1]

Friends, I know it is a gold refining forum. I am sorry to disturb some people for this. But, I have seen lot of occasion where people gave very good piece of advice about the matter which was completley different in nature from what people like to disscuss on this forum.
So, please help me too.

I have a tank of 5 ft x 2 ft 6 inches and height-3ft
The concentration of copper sulphate is 250 grams of copper per litre
25 lead anodes thickness 1 mm and size 2ft x2ft
25 pure copper anodes thickness 1mm size 2ft x 2ft
Space between electrodes is one inch
Voltage 1.5 volt
Amperes- 30 amperes per square feet
Rectifier used: 5000 amperes 12 volts
My question is how much copper will deposit on cathodes per hour?
What is the formula to calculate the deposition of copper on cathodes?

 

[goldsilverpro]

At 100% cathode efficiency (the highest it could possibly be), 1.185 g of copper will deposit per amp-hour (per amp - per hour). 5000A would deposit about 13.05 pounds per hour, max.

 

[Knowledgethirst1]

Thanks a lot for your answer. So, this 1.185 grams of copper will deposit on one cathode individually or this will be the weight on all the cathodes?

 

[goldsilverpro]

Thanks a lot for your answer. So, this 1.185 grams of copper will deposit on one cathode individually or this will be the weight on all the cathodes?

It's based on amperage and time and, for 1 amp for 1 hour, 1.185g will be the total weight deposited on all of the cathodes. Total copper deposited in grams = amps x number of hours x 1.185. This is based on 100% cathode efficiency, which rarely is possible. In reality, you will usually get at least a little less than 100%.

This only holds true for copper(II), which is the state of copper in your copper sulfate solution. If the copper were in a cyanide solution, you would have copper(I) and the rate of deposition would be double - 2.370 g/A-hr

 

[Knowledgethirst1]

Thanks again. You are very co-operative.