Letter 8007

What part does voltage play in electroplating? 

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I am doing a science project about electroplating and the effects of using different voltages in the process. I have tried to research what exactly voltage does in the process but the only results I've found is that the voltage determines the ions travel.

Can someone expand on this or if it is wrong explain what part voltage plays in electroplating?

Scott Ross
- Los Osos, CA, USA

Electricity cannot be seen, so it can be rather abstract. Therefore, for ease of visualization, terms and units that relate to the flow and quantity of electricity are often compared to terms used to describe the flow of water.

The corresponding term to voltage is water pressure.

If you lived in a high-rise apartment alongside a lake and decided to hang a hose out your window to draw up some lake water for your flowers & plants, you would recognize that the water is not going to run up that hose by itself, rather that you'll need a pump, and that how much pressure that pump has to put out will depend on what floor of the high rise you live on.

You need voltage in order to overcome all of the resistances involved in plating, which include the resistance of the copper wires that the electrons flow through, the electrical resistance of the plating solution which the ions flow through, and other resistances. I can answer more deeply what those other resistances are, but that may be beyond the scope of your project.

 
Ted Mooney, P.E. - finishing.com Inc. - Brick, NJ

I guess I should restate my question, I am testing different voltages 1.5 volts, 3 volts and 20 volts. What relationship does the voltage have to the amount of copper plated onto a zinc plate? Is there a way to have too much voltage to where it would cause a weaker bond? or is there a limit to where there wouldn't be any better results after a certain voltage is reached? If you could explain those resistances in depth that would be great.

Thanks for your help!

Scott Ross
- Los Osos, CA, USA

Faraday's Law says you will ideally deposit one gram equivalent weight of metal for every 96,500 ampere-seconds.

I = E / R

So, as a first cut, when you double the voltage from 1.5 volts to 3 volts you more or less double the current so you more or less plate twice as fast. But at 20 volts you will surely deplete the solution in the area of the cathode almost instantly. But the electrons you are making available have to go somewhere--with the result that the electrons will reduce the hydrogen in the water to nascent hydrogen gas. This formation of copious hydrogen gas and atoms of metal being yanked out of solution instantly instead of slowly growing as crystals is called "burning".

 
Ted Mooney, P.E. - finishing.com Inc. - Brick, NJ

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