The power supply should be able to supply at least 750ma to be safe. While a constant current supply is not an absolute necessity, it does however make the process time consistent and removes some of the variables related to the submerged surface area and distance between electrodes. The power supply used for this experiment was a 30vdc @ 1 ampere unit.
The Process:
Arrange the apparatus so that the beaker is on the hotplate or suspended above the bunsen burner with a ring stand.
Add 250ml of distilled water to the beaker.
Suspend the electrodes in the beaker and bring the water to a boil. Leave the power supply off.
With the power supply turned off, connect the gold electrode to the positive supply terminal, and the stainless steel electrode to the negative. Make sure that there are no short circuits, and that there is no exposed wiring which could cause a high voltage shock. Do not power it up yet.
If you have a magnetic stirrer, put the magnet into the beaker. Make sure the stirrer magnet does not contact either electrode.
Bring water to a boil.
Energize the power source and add 3.6 ml of stock KCl solution to the beaker.
Allow to run several minutes.
Add 2 drops Hydrogen Peroxide to the beaker
Add 4.1 ml of the stock K-Citrate solution to the beaker.
Within minutes, a red tint should start to appear.
When the desired color depth is produced, or no more color change is detected, turn off the power first, then remove the electrodes. Top off the water to restore it to 250ml. You should now have red colloidal gold. Filter, and bottle.

Fig. 2 Final Product
Half Reactions:
Cathode:
6KCl + 6e --> 6K +6Cl
- Chloride ion travels to Anode where it combines with gold
6K + 6H
20 --> 6KOH +3H
2 (which bubbles off as hydrogen gas)
6KOH --> 6K
+ +6OH
- New hydroxyls journey to anode continuing electrolysis of water.
2Au
+++ +6e --> 2Au We don't want this to happen as it merely plates gold onto the cathode
4H
20 +4e --> 2H
2 + 4OH
- This reaction may not happen because of the preference to electrolyze the potassium
Anode:
6Cl
- --> 3Cl
2 + 6e
2Au +3Cl
2 --> 2AuCl
3 Gold is liberated as gold chloride
4OH
- --> O
2 + 2H
2O + 4e Oxygen gas is liberated as hydroxyl ions are oxidized
Secondary Reactions:
AuCl
3 is known to react with the Cl
- of the KCl
- So -
AuCl
3 + KCl +H
2O --> HAuCl
4 +KOH
Reduction Reaction:
4HAuCl
4 + 8H
2O
2 --> 4Au + 16HCl + 2H
20 + 7O
2
Stoichiometry:
50ppm CG in 250ml water is 12.5mg, or 63.5 microMoles of Au
Since it takes 4 atoms of chlorine to oxidize one atom of gold at the anode and produce HAuCl
4, we need 254 microMoles of KCL
To stabilize each Au particle with a citrate ion, we need 63.5 microMoles of K-Citrate
To reduce 63.5 microMoles of AuCl
3, we need 127 microMoles of H
2O
2.
Therefore:
254 microMoles of KCl = .018 gm == 1.8 ml of 1% KCL solution
127 microMoles of H
2O
2 = .0043 gm == .143 ml of 3% solution (about 2 drops)
127 microMoles of K-Citrate = .041 gm == 4.1 ml of 1% solution
Useful Data:
| Atom/Molecule |
Weight
grams/mol |
Electronegativity
|
Electrode Potential
Volts |
Electron Affinity
mJoules/mol |
| Au |
196.96 |
2.54 |
+.8 |
228.8 |
| K |
39.098 |
.82 |
-2.92 |
48.4 |
| O |
15.999 |
3.44 |
+1.23 |
141 |
| H |
1.0079 |
2.2 |
0.0 |
72.8 |
| Cl |
35.453 |
3.16 |
+1.36 |
349 |
| K-Citrate |
324.41 |
  |
  |
  |
| NA-Ascorbate |
198.11 |
  |
  |
  |
H2O2 |
34 |
  |
  |
  |
| OH- |
17 |
  |
-.40 |
  |
Further Observations:
While the spacing between the electrodes does not affect the reactions, it does alter the amount of voltage needed from the power supply to produce a fixed amount of current. The closer the electrodes are, the lower the voltage needed. The number of turns of wire on the electrodes (submerged surface area) also affects the voltage somewhat; more electrode area requires less power supply voltage for the same current. Both of these effects are to be expected. These variables are cancelled out when using a constant current power supply.
Actual voltage is inconsequential to the electrochemisty. What matters is current (amperes). (See Faraday's Laws) As you can see from the half reactions, it takes 3 electrons to liberate 1 atom of metallic gold. Additional electrons are required for the inevitable electrolysis of water.
As the Au+++ ions enter solution, they migrate to and are pulled to the cathode because of the electric field between the electrodes. If the Au ions reach the cathode, they will be reduced back to metallic gold, which is not what we want. (This is the basic electroplating mechanism). By adding the reducing agent (Hydrogen peroxide, and K-Citrate) at the start of electrolysis, the Au ions have a high probability of finding a molecule of the reducing agent before reaching the cathode. When this happens, the ionic charge is neutralized, and free metallic gold appears in the solution instead of on the cathode. However, a small amount of the Au ions may be reduced at the cathode, causing some loss gold particles.
Note that the potassium atoms are essentially trapped at the cathode. As soon as they are reduced to potassium metal, they immediately react with water to again produce potassium hydroxide which immediately ionizes, allowing the potassium cation to be reduced to metal and start the cycle all over again. This in turn produces a steady stream of hydroxyl ions moving towards the anode to be turned into oxygen gas and water again. At this point, I do not know how to determine the ratio of the potassium cycle to the gold cycle. (If this were known, the exact current vs time for the process could be calculated.)
A similar occurence seems to happen with the Cl- ion at the anode. As soon as the Cl- reacts with gold to produce gold chloride it would immediately ionize, again producing Cl- ions attracted to the anode. Perhaps this is why so little of the gold actually plates out onto the cathode.
Sodium salts are normally used for the Turkevich method, and should work the same as the potassium salts used in this demonstration.
Notes:
1) A fixed voltage power supply can be used, but nine volt batteries do not have enough capacity to supply the needed current. Inexpensive constant current power supplies are often sold on eBay, as tattoo machine power supplies, as well as lab supplies. Used electrophoresis power supplies are ideal, as they usually have constant current circuitry, and higher voltages allowing the electrodes to be placed far apart. If possible, and your power supply does not have a built in meter, use a milliameter to measure the current flow and adjust as necessary to stay within the ratings of your power supply.
2) It is not necessary to use the glass rod to hold the gold anode. You can just ball up a couple of inches of gold wire with a 4 inch straight section to connect to the power supply, and hang it from a string. The danger though is that it may inadvertently touch the cathode and short out the power supply.
3) If you don't have the ability to measure small amounts of liquids, 1cc is approximately 15 drops from an eyedropper. Alternatively, make your stock solutions weaker, or make a liter instead of 100ml to make the quantity of chemicals higher and thus easier to measure.
4) If the wire is too springy, anneal it by passing it through the flame of an alcohol lamp to soften it, and let it cool slowly (don't cool it in water). It will then be dead soft.
5) It is very important to use the least amount of NaCl possible. The reason is that the Sodium will reduce at the cathode to pure sodium. Sodium reacts with water to produce sodium hydroxide. Sodium hydroxide will react with some of the gold chloride produced at the anode to produce gold hydroxide which is insoluble, and contaminates the CG. Over time, the gold hydroxide will settle out of the solution, and if the sodium level is too high, it will make the solution cloudy or even oily looking. The amount suggested in the article is the absolute maximum which should be used.