Determine how long it takes to cover both sides of the plate

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Homework Help Overview

The discussion revolves around determining the time required to electroplate a square copper plate with a specified thickness using a given current in an electroplating bath. The problem involves concepts from electrochemistry and material science, specifically focusing on electrochemical equivalents and mass calculations.

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  • Mixed

Approaches and Questions Raised

  • Participants explore calculations related to volume, mass, and time for electroplating, questioning the accuracy of initial values and units used. There is a focus on understanding the relationships between area, thickness, and the resulting mass of copper deposited.

Discussion Status

Several participants have provided corrections and clarifications regarding the calculations, particularly concerning unit conversions and the interpretation of the problem setup. There appears to be a productive exchange of ideas, with some participants confirming the final answer while others continue to question earlier assumptions.

Contextual Notes

Participants are addressing potential discrepancies in the initial calculations and the provided answer, as well as clarifying the correct use of units in the context of the problem. There is an ongoing discussion about the density of copper and its impact on the final result.

DevonZA
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A square plate has a side of 100mm. It is connected to the cathode in a copper electroplating bath and a current of 4A is passed through the bath. Neglecting the thickness of the plate, determine how long it takes to cover both sides of the plate with copper to thickness of 0.1mm

Density of copper to be 9000kg/m^3
The electrochemical equivalent of copper to be 0.330mg/C

Answer given = 227 min 16 sec

Volume = area x height
= 20 000x10^-7 X 0.23x10^-3
= 4x10^-7

Mass = volume x density
= 4x10^-7 X 9000kg/m^3
= 3.6x10^-3kg = 3600mg

m = zIt
t = m/zI
= 3600/(0.330)(4)
= 43636.36364s = 727 min and 16 sec

I am quite confident in my answer but it is way off from the supplied answer - unless that was a typo.
Any input would be appreciated.

Thanks,
Devon
 
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DevonZA said:
A square plate has a side of 100mm. It is connected to the cathode in a copper electroplating bath and a current of 4A is passed through the bath. Neglecting the thickness of the plate, determine how long it takes to cover both sides of the plate with copper to thickness of 0.1mm

Density of copper to be 9000kg/m^3
The electrochemical equivalent of copper to be 0.330mg/C

Answer given = 227 min 16 sec

Volume = area x height
= 20 000x10^-7 X 0.23x10^-3
= 4x10^-7

Where are the red numbers from?
 
DevonZA said:
Volume = area x height
= 20 000x10^-7 X 0.23x10^-3
Where do those numbers come from and what are their units?

Edit: ehild beat me.
 
ehild said:
Where are the red numbers from?

square plate has 100mm side, therefore area = 100mmx100mm=10000mm^2 X 2 sides = 20000mm^2 = 20000x10^-7
copper thickness = 0.1mm each side = 0.2mm = 0.2x10^-3

Sorry should have been 0.2x10^-3 not 0.23x10^-3
 
1mm^2 = 10-7? An area is a number?
If that is supposed to be square meters, the number is wrong.

I can confirm the given answer.
 
Last edited:
DevonZA said:
square plate has 100mm side, therefore area = 100mmx100mm=10000mm^2 X 2 sides = 20000mm^2 = 20000x10^-7
copper thickness = 0.1mm each side = 0.2mm = 0.2x10^-3

Sorry should have been 0.2x10^-3 not 0.23x10^-3
You counted the two sides twice. And 1 mm2=10-6 m2.
 
mfb said:
1mm^2 = 10-7? An area is a number?
If that is supposed to be square meters, the number is wrong.

I can confirm the given answer.

What is 20000mm^2 in m^2?
 
ehild said:
You counted the two sides twice. And 1 mm2=10-6 m2.

How did I count them twice? 1 side = 100mmx100mm = 10 000mm2 so 2 sides = 20 000mm2
 
DevonZA said:
What is 20000mm^2 in m^2?
1mm = 10-3m, square both sides, or see post 6.

To get closer to the given answer, you might also want to use a more precise density estimate for copper, but that just changes the result by about one minute.
 
  • #10
DevonZA said:
How did I count them twice? 1 side = 100mmx100mm = 10 000mm2 so 2 sides = 20 000mm2
After that, you said that 0.1 mm layer on both sides, so it is 0.2 mm on both sides.
 
  • #11
mfb said:
1mm = 10-3m, square both sides, or see post 6.

To get closer to the given answer, you might also want to use a more precise density estimate for copper, but that just changes the result by about one minute.

Alright so 20 000mm2 = 20 000 x10-6?
If I use that number I get an answer of 7272 min
The density was given as 9000kg/m3
 
  • #12
ehild said:
After that, you said that 0.1 mm layer on both sides, so it is 0.2 mm on both sides.

I see what you mean, thank you.

volume = 20 000x10-7 X 0.1x10-3
= 2x10-7

mass = 2x10-7 X 9000kg/m3
= 1.8x103kg = 1800mg

t = m/zI
= 1800/(0.33)(4)
= 1363.636364s = 22 min 43 seconds
 
  • #13
Now you used the wrong 10-7 again. Apart from that it is right.
 
  • #14
volume = 20 000x10-6 X 0.1x10-3
= 2x10-6

mass = 2x10-6 X 9000kg/m3
= 0.018kg = 18000mg

t = m/zI
= 18000/(0.33)(4)
= 13636.36364s = 227 min 16 seconds

My goodness finally. Thank you very much mfb and ehild for your help!
 

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