Verification of cathode current density calculation in an electro-refining cell

[001anuradha]

Dear All,

I would like to verify my calculation of cathode current density in an electro-refining cell.
It is refining cast copper anodes from scrap copper in a CuSO4/H2SO4 bath.
Each tank has a set of impure anodes and copper cathodes forming cells in parallel.

The current passing into a cell is 134A, surface area of one side of cathode is 0.54m^2

My current density calculation is below:
*Each cathode receives copper from anodes on both sides, so according to my understanding the current passing through one side of cathode is I = 134A/2 = 67A
*As this current passes through a single side of cathode, the area,A = 0.54m^2

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From this I get the current density, J=I/A = 67/0.54 = 124 A/m

This value I get is roughly half the typical industry values which are in the range of 250 - 280 A/m^2
Am I doing the calculation wrong, is there a industry specific calculation for "cathode current density"

Thank you all in advance, love the work in this forum which I read from time to time but this is my first time posting!

 

[cloa]

Yes you calculated it wrong. The current is same across both electrodes i.e. 134A- its a series circuit current which is an electrical flow per second can't be lost and that's what you are saying by dividing by 2. What is lost is voltage across the circuit- the losses will vary with conditions at electrode. Its not really a corrosion engineering question. Your question comes out awfully like a homework question- please explain you got the current.

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[001anuradha]

Dear cloa thank you for your reply.

I was torn between posting this on the metallurgy forum or the corrosion engineering forum, decided to post on corrosion forum as I thought members would be more familiar with current densities here. Sorry for posting in the wrong forum.

I am undergoing training at a plant that is carrying out electro-refining of copper. I have been tasked with verifying whether their operating parameters are within the typical industry ranges (I am in a developing country and factories aren't always working at the optimum, I refer "Extractive Metallurgy of Copper" by Davenport, Biswas et al for getting the typical industry values) This is why my question might have sounded like homework.

The circuit is set up such that there are 30 electrolytic tanks in series and each tank has 12 cells in parallel. The total current is about 1608 A, which works out to 1608/12 = 134A going though a single cell. The total voltage is 12.72V, which works out to 12.72/30 = 0.424V across a cell (this is more or less the reading from the voltmeter across a cell).

Please forgive me for the previous post which was confusing. Would it be more understandable if I put it this way: The current going through a cell is 134A, and in a cell, the copper deposits on both sides of the cathode on a total area of 1.08 m^2. So the current density on cathode is 134/1.08 = 124 A/m^2

If this calculation is correct then the factory needs to increase current density to 250 - 280 A/m^2 to run at a more efficient level.

Thanks again for your reply, truly appreciate it.

 

[cloa]

I'd say you really should be focused on the solution chemistry to achieve the desired result. Also what is the voltage loss, the deposition rate and quality of deposit. I honestly have no idea about what would be regular industry practices- not covered in my very broad materials engineering degree- about current density but what they achieve is probably with the optimum control of the solution and voltage.

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[001anuradha]

Yes solution chemistry (copper/ferrous ion conc, pH, grain refining agent conc etc) is one of the factors I intend to check. They have a direct effect on the quality and morphology of the copper deposit. The temperature and the current density (from Faraday's law of electrolysis) have a major bearing on the deposition rate.

Average deposition rate is currently about 260g/hr.

Theoretically the cell voltage should be zero, because deposition at cathode is the reverse of the dissolution at anode. The voltage termed over-voltages in literature (which I believe are the voltage losses for overcoming the various resistances)are within the typical ranges of 0.1 - 0.6 V.

I will try changing the current and see whether I can improve the deposition rate.

Thanks again!