Colloidal Gold
The purpose of this page is to demonstrate how to make approximately 40 ppm Colloidal Gold from .999 fine (24 Karat) gold wire, using an electrolytic process to make AuCl3, and then to reduce the resultant gold chloride to colloidal gold by means of a reducing agent and stabilizer.
WARNING: Ingest Colloidal Gold at your own risk. This information is presented for educational purposes only. Ignore this at your own peril!
Gold nanoparticles were used to make the beautiful red and purple colors in stained glass found in the Medieval Churches. According to Professor Zhu Huai Yong of Queensland University, gold stained glass windows are able to purify the air as well as look pleasing to the eye.
The Electrolytic Process:
The electrolytic process (electrolysis) consists of placing two electrodes in a solution containing an electrolyte, and passing electrical current through the electrodes and solution. The electrode connected to the positive voltage is called the anode, and the one connected to the negative electrode is called the cathode. Electrolytic processes are normally used for electroplating one metal on top of another, and are the processes which make batteries work. In this process, gold will be removed from the anode and enter the solution as gold chloride, and then be converted to gold particles by a reducing agent.
Materials:
You will need the following materials.
1 ) .999 fine gold wire anode, 22 guage, of about 6″ to 1 foot in length.
3 ) Sodium Chloride (NaCl) (Ordinary table salt) for the electrolyte.
4 ) Sodium Ascorbate (buffered vitamin C) for the reducing agent.
5 ) DC Power supply capable of supplying 500ma @ 30volts
6 ) 500 ml pyrex beaker
7 ) Hot plate, bunsen burner, or alcohol burner capable of boiling 250 ml of water.
8 ) Distilled water.
9 ) Lab supplies, scale, graduated cylinders, ring stand, etc.
Warning:
If you are not experienced working with electricity, get assistance, or do not perform this procedure. While 30 volts dc is not lethal, short circuits can ruin the power supply.
Stock Solutions:
We will make stock solutions of our chemicals in .05M strength1
To make 100ml of stock:
NaCl: 0.3 grams of NaCl dissolved into 100 ml of distilled H2O
Sodium Ascorbate: 1.0 grams dissolved into 100ml of distilled H2O (must be made fresh, as Na-Ascorbate rapidly decomposes).
The Electrodes:
For the positive electrode, a piece of .999 pure gold wire is needed, preferable 22 guage or larger, and long enough to submerge at least an inch into the water.
A plain 14 gauge copper wire works well as the cathode (negative electrode), and may be bent so that it forms a hook allowing it to be hung on the edge of a 500ml beaker.
Power Supply:
The power supply should be able to supply at least 500ma 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 purchased from Radio Shack. The power supply voltage itself is not critical, and any supply capable of supply 500 ma at 20 to 40 volts should work. 30 volts was used here because that is what was available.
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 or copper electrode to the negative. Make sure that there are no short circuits, and that there is no exposed wiring which could cause a short or a 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.
Add 10ml of stock .05M NaCl solution to the beaker.
Add 8ml of the stock .05M Sodium Ascorbate solution to the beaker.
Energize the power supply. Bubbles should start forming on the electrodes, with approximately twice as much gas evolving from the negative electrode.2
Within minutes, a red tint should start to appear.
Continue to add distilled water as needed to maintain 250ml.
When the desired color depth is produced, or no more color change is detected, turn off the power first, then remove the electrodes. You should now have red colloidal gold. Filter, and bottle.
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 submerged surface area of the electrode also affects the voltage somewhat; more electrode area requires less power supply voltage. Both of these effects are to be expected. These variables are canceled out when using a constant current power supply.
Actual voltage is inconsequential to the electrochemistry. What matters is current (amperes). (See Faraday’s Laws)
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 (Na-Ascorbate) 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 sodium atoms are essentially trapped at the cathode. As soon as they are reduced to sodium metal, they immediately react with water to again produce sodium hydroxide which immediately ionizes, allowing the sodium 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.
Notes:
1) .05M means .05 Molar. A 1 molar concentration means that there is the equivalent molecular weight of the molecule in one liter of solution. A Mole of any substance contains the same number of molecules of that substance as any other substance. So, a .05M solution has 1/20th of a molecular weight of that substance per 1 liter of water.
2) The negative electrode produces twice as much gas because the electrolysis breaks down 2 hydrogen atoms to 1 oxygen atom.