thebillsfan

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I had some confusing question on electroplating. I think I just dont understand it fundamentally. When you put a gold bar in a solution and want to plate an aluminum spoon with that gold bar, both the gold and aluminum are in their ELEMENTAL forms, correct? let's say you run an electrolytic cell so that gold is oxidized and alum is reduced. since aluminum has a more negative reduction potential you need to use an ELECTROLYTIC cell.

That's the set up. now, for this electrolytic cell to work, aluminum needs to be reduced like I said. But for it to be reduced, it needs to be in its cationic form. But I already said earlier that you drop the spoon into solution in its elemental form...so where is the cationic form coming from? It's an acidic solution, so does that have anything to do with it?

I just took a FL with 3 passages on electrochemistry...
 

sleepy425

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I had some confusing question on electroplating. I think I just dont understand it fundamentally. When you put a gold bar in a solution and want to plate an aluminum spoon with that gold bar, both the gold and aluminum are in their ELEMENTAL forms, correct? let's say you run an electrolytic cell so that gold is oxidized and alum is reduced. since aluminum has a more negative reduction potential you need to use an ELECTROLYTIC cell.

That's the set up. now, for this electrolytic cell to work, aluminum needs to be reduced like I said. But for it to be reduced, it needs to be in its cationic form. But I already said earlier that you drop the spoon into solution in its elemental form...so where is the cationic form coming from? It's an acidic solution, so does that have anything to do with it?

I just took a FL with 3 passages on electrochemistry...
In the electroplating technique, the thing you want to coat (like aluminum or w/e) and the electroplating metal (like gold) are placed in the same solution compartment. The gold bar is connected to the anode (the positive terminal of your external voltage source) and the aluminum object is connected to the negative terminal of the voltage source. The gold is first oxidized at the anode to give you gold cations in solution. This cationic gold is subsequently reduced at the cathode to regenerate solid gold, which deposits onto the aluminum at the cathode. So the aluminum does not participate in the reaction (since it would require an extremely high potential to reduce elemental aluminum). Instead, the metal at the anode is first oxidized to go into solution, and then reduced at the cathode to deposit it onto the other metal.
 
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thebillsfan

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In the electroplating technique, the thing you want to coat (like aluminum or w/e) and the electroplating metal (like gold) are placed in the same solution compartment. The gold bar is connected to the anode (the positive terminal of your external voltage source) and the aluminum object is connected to the negative terminal of the voltage source. The gold is first oxidized at the anode to give you gold cations in solution. This cationic gold is subsequently reduced at the cathode to regenerate solid gold, which deposits onto the aluminum at the cathode. So the aluminum does not participate in the reaction (since it would require an extremely high potential to reduce elemental aluminum). Instead, the metal at the anode is first oxidized to go into solution, and then reduced at the cathode to deposit it onto the other metal.
if this is the case then why do we use the reduction potential of aluminum in determing the EMF?
 

sleepy425

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if this is the case then why do we use the reduction potential of aluminum in determing the EMF?
You don't use the reduction potential of aluminum to determine the EMF. The EMF for an electroplating reaction is 0, since there is no net redox. The metal ions in your electroplating bath (like gold ions) are reduced, and the solid metal at the anode is oxidized to replenish the ions in solution. However, since the EMF is 0, you need a battery to drive the electroplating, since you need a positive EMF for the reaction to proceed spontaneously.
 
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thebillsfan

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You don't use the reduction potential of aluminum to determine the EMF. The EMF for an electroplating reaction is 0, since there is no net redox. The metal ions in your electroplating bath (like gold ions) are reduced, and the solid metal at the anode is oxidized to replenish the ions in solution. However, since the EMF is 0, you need a battery to drive the electroplating, since you need a positive EMF for the reaction to proceed spontaneously.
fair enough, but then whats causing the aluminum from getting oxidized (it actually has a higher ox potential i believe) and also, do you use the aluminum reduction potential to calculate the minimum voltage required to run the rxn? how do you figure that out?

related question...how do you figure out the amount of time it takes to plate X grams of gold? do you have to have a current given? kaplan gave the question without a current and then in the explanation proceeded to say "since 1 amp is" ...
 

sleepy425

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fair enough, but then whats causing the aluminum from getting oxidized (it actually has a higher ox potential i believe) and also, do you use the aluminum reduction potential to calculate the minimum voltage required to run the rxn? how do you figure that out?
Do you mean why doesn't the aluminum get oxidized and plate the gold bar as well? Well, there are two reasons. First, the way the external voltage source is hooked up will determine which metal gets plated. You attach the metal you want to coat (i.e. the aluminum) to the negative terminal of your battery/voltage source so that the dissolved metal ions (like gold) can get reduced there and coat the surface of the aluminum. The gold is attached to the positive terminal of the battery so that the gold will be oxidized there. In this configuration, since the aluminum is hooked up to the negative terminal, it cannot be oxidized. If you reversed the terminals of the battery, you'd get the opposite plating. The gold would get coated with aluminum.

The second reason is that, typically, the solution that the gold and aluminum are both dipped in contains gold ions already (in other words, you use something like gold nitrate to make the solution) so that there will already be a significant concentration of ions in solution to be reduced. Thus, even if you accidentally switched the battery terminals around, there wouldn't be any aluminum ions in solution to be reduced.

related question...how do you figure out the amount of time it takes to plate X grams of gold? do you have to have a current given? kaplan gave the question without a current and then in the explanation proceeded to say "since 1 amp is" ...
I would have to see that question to know for sure if current is necessary. However, in general, if you have X grams of gold, you can convert that to moles of gold. Now I can't remember what the oxidation state of gold is, but I vaguely remember the most common one is +3. If that's the case, then one of the half reactions is Au---->Au(3+) + 3 e-. So, if I'm right about this oxidation state (even if I'm wrong you'll get the idea from this), that means that for every mole of gold you want to plate, you'll need to move 3 moles of electrons. So you multiply the number of moles you want to plate by 3, and multiply by avogadro's number. That'll give you the number of electrons. Now, one amp is 1 coulomb per second. 1.6*10^-19 coulombs is the charge of one electron, so multiply 1.6*10^-19 by the number of electrons you've calculated and that'll give you the number of coulombs you require to plate that gold. Now divide that number of coulombs by your current to give you the time it'll take in seconds.

Sorry bout the long explanation, but if you just follow the units you can calculate the time it'll take.
 
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thebillsfan

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alright, so you do HAVE to know the current. kaplan gave no current. wow, the whole FL was full of errors. thank you very much for all your help