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Q. Determine the time required to obtain a Cu deposit of thickness 1 μm when electrodeposition...

Q. Determine the time required to obtain a Cu deposit of thickness 1 μm when electrodeposition is done at 4 A and 6 A, respectively.

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Answer #1

Cu atomic weight of 63.5 and its valance is 2, a Faraday of electricity will deposit
63.5/2 = 31.75 grams of copper
But 1 amp is 60/96485 Faradays, so it will deposit 60/96485 x 31.75 = 0.0197 grams of copper
The density of copper is about 8.96 g/cm​3, or 8960 g/dm3 so 1 amp current deposits about 0.0197/8960 = 2.19 x 10-6​dm3
So, if the area plated is a square decimeter, the thickness plated is 0.219 micrometer

Thickness of plating in microns = 0.219 * A/sq/dm * minutes of plating time * current in ampere

​(i) 4 A current
​ 1 = 0.219*4*plating time in minutes
Plating time = 1.141 min

(ii) 6 A current
1 = 0.219*6*plating time in minutes
​Plating time = 0.79 min

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