[Aries]

Hi,

I'm cooking up some electroplating experiments but I'm having trouble understanding the basics. My goal is to produce a fairly large quantity of Copper (II) acetate solution from household vinegar and electricity. A common DUI recipe calls for hydrogen peroxide which is a costly reagent. A previous set of experiments suggest the following method is feasible as I obtained what I presumed to be Copper (II) acetate crystals upon evaporation of the solution with the characteristic reflective dark blue colour.

Given the conditions in the schematic, I'm trying to answer the following questions:

1) What chemical reactions occur? Specifically, what are the reactions at the positive and negative terminals as well as the overall reaction?

2) What are the upper and lower limits for power dissipation of the circuit? In other words, how much power is needed to provide the activation energy to drive the unfavorable reaction? Likewise, what is the effect of excess power? (side reactions?). In this particular setup I'm able to independently control supply voltage and current adding some degree of control.

3) Other comments on the setup? I have a semester of college chemistry so please keep that in mind as I'm trying to understand the limits of the experiment as I have constructed it. It would be helpful if you could talk about the most important variables that I'm not currently controlling for.

This all started when I was talking to my roommate about the magic of chemistry and my dream to electroplate large items. He decided to go buy 5L of vinegar so now I have to deliver on my big talk! Haha. I'll be performing the experiment in the next few days so I'll post some photos of the electrolyte before I go on to plate stuff. Thanks for the help.

[Hunter2]

The power supplier is very small. I would exchange the 1N4148 to 1N4007.
The anode has to be copper, the cathode carbon in the cell.

[Borek]

1. For a single electron electrode reaction to produce one mole of substance you need to run 1 mole of electrons through the electrode - that means 96500 coulombs (so called Faraday's constant), or 96500 seconds of 1 ampere current (google Faraday's laws of electrolysis). Compare that with the output of your power supply.
2. As long as the voltage applied is too low no reactions take place, what happens later depends on the voltage applied. It can be predicted using standard reduction potentials and the Nernst equation.
3. In your setup I would expect copper to be dissolved on anode and deposited back on the cathode, so no net change in the amount of copper present in the solution. To dissolve copper you need to reduce something else on the cathode, won't be easy (for the reasons signaled in 2.)

[Aries]

1. For a single electron electrode reaction to produce one mole of substance you need to run 1 mole of electrons through the electrode - that means 96500 coulombs (so called Faraday's constant), or 96500 seconds of 1 ampere current (google Faraday's laws of electrolysis). Compare that with the output of your power supply.
2. As long as the voltage applied is too low no reactions take place, what happens later depends on the voltage applied. It can be predicted using standard reduction potentials and the Nernst equation.
3. In your setup I would expect copper to be dissolved on anode and deposited back on the cathode, so no net change in the amount of copper present in the solution. To dissolve copper you need to reduce something else on the cathode, won't be easy (for the reasons signaled in 2.)

Thanks, I think I understand what is going on. Your reply seems to prove my null hypothesis which is my original experiment dissolved the copper by some other means. I asked my teacher about the mechanics and he gave me this hint:



I put this together and came up with a new experiment:



a) Balanced reaction:
2Cu + 4CH3COOH + O2  2Cu2+ + 4CH3COO- + 2H2O

b) Number of ions in starting solution:
1L CH3COOH (5% w/v - 0.83 M) contains 0.83 mol CH3COO- and 0.83 mol H+

c) Maximum mass of copper to be dissolved with excess oxygen:
? g Cu = 0.83 mol CH3COOH x (2 mol Cu / 4 mol CH3COOH) (64.5 g Cu / 1 mol Cu) = 26.8 g Cu

d) Concentration of copper ions in solution upon completion of reaction with no evaporation:
? M Cu2+ = 0.83 mol CH3COO- x (2 mol Cu2+ / 4 mol CH3COO-) (1 / 1 L) = 0.42 M Cu2+

Questions:

1) Are the previous calculations correct?

2) If the previous calculations are correct, how do I determine how much time it will take to run to completion? If I start with 1L of vinegar and excess copper, I'm assuming the rate limiting factor is the diffusion of oxygen into solution?

3) What else can we say about the experiment? For example, since I'm using nominal 5% vinegar in an uncontrolled environment, what are the greatest sources of error that can be easily controlled for?

[Borek]

b) Number of ions in starting solution:
1L CH3COOH (5% w/v - 0.83 M) contains 0.83 mol CH3COO- and 0.83 mol H+

This is a bit more complicated - acetic acid is a weak acid and is not fully dissociated, so it is not like solution contains these amounts of ions. But yes, this is more or less what is in total available for the reaction.

1) Are the previous calculations correct?

Reasonably. Note that quite common procedure when doing synthesis is to use excess of one reagents (typically the cheaper one), to be sure the other one reacted to the end.

2) If the previous calculations are correct, how do I determine how much time it will take to run to completion?

Bad news: by trial and error, and it will be probably painfully slow. I would look for a more concentrated acetic acid (no idea where you live, but here I can buy 10% white vinegar without problems, and in some countries so called "essence of vinegar" is available as a food additive, which is something like 70%) and to aerate the solution, even using simple aquarium air pump.