How-to make electro-colloidal gold
Making Colloidal Gold
Purpose:
see video here https://youtu.be/HkljxNU2aZY
The purpose of this page is to show how to make Colloidal Gold. using an electrolytic process to make gold chloride (AuCl3), and then to reduce the resultant gold chloride to colloidal gold by means of a sugar based reducing agent and sodium citrate stabilizer.
Materials:
You will need the following materials.
1) .999 fine gold RIBBON, about 6" to 1 foot in length.
http://www.riogrande.com/Product/22K-Yellow-Gold-1-8-Plain-Bezel-Wire-Dead-Soft/609218
1) .999 fine gold RIBBON, about 6" to 1 foot in length.
http://www.riogrande.com/Product/22K-Yellow-Gold-1-8-Plain-Bezel-Wire-Dead-Soft/609218
3) Sodium Chloride (Table salt)
3) Sodium Citrate (Also called tri-sodium citrate) 1lb $5.95 http://thechemistryconnection.com
3) Sodium Citrate (Also called tri-sodium citrate) 1lb $5.95 http://thechemistryconnection.com
4) Ordinary Corn Syrup.
5) Power supply capable of supplying ordinary house current 110vac
6) common cheap coffee maker
7) Distilled water.
5) Power supply capable of supplying ordinary house current 110vac
6) common cheap coffee maker
7) Distilled water.
If you are not experienced working with electricity, get assistance, or do not perform this procedure. 110vac is lethal.
Stock Solutions: To make the Sodium Chloride solution, weigh 0.3 grams of table salt, and dissolve it in distilled H20 to make 100 ml of stock solution.
To make our Citrate stock solution, weigh out 1.47 grams of Sodium Citrate and dissolve it in distilled H2O to make 100 ml of stock solution.
The Electrodes:
For the electrodes use .999 pure gold ribbon. IONS flow best off sharp edges and points. ANY gold will work but ribbon works best.
Power Supply:
Arrange the apparatus so that the carife is on the hotplate.
Add 250ml of distilled water to the coffee maker.
Suspend the electrodes in the carife and bring the water to a boil. Leave the power supply off. i simply bend the elements over the edge of the carife.
With the power supply turned off, connect the gold electrodes to the supply terminals, 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.
Bring water to a boil.
add 15 ml ( 3tsp ) of stock table salt solution to the beaker. stir . i use a swizzle stick or a chop stick. non-conducting material just to be safe.
Add 2 drops syrup to the beaker Or about the amount picked up by immersing the end of a chopstick 1/4 inch into the syrup and letting the excess drip off.
Add 5 ml ( 1tsp ) of the stock Sodium Citrate solution to the beaker.
Half Reactions:
Cathode:
6NaCl + 6e --> 6Na +6Cl- Chloride ion travels to Anode where it combines with gold
6Na + 6H20 --> 6NaOH +3H2 (which bubbles off as hydrogen gas)
6NaOH --> 6Na+ +6OH- New hydroxyls journey to anode continuing electrolysis of water.
Au+++ +3e --> Au We don't want this to happen as it merely produces gold sludge on the cathode
4H20 +4e --> 2H2 + 4OH- This reaction may not happen because of the preference to electrolyze the potassium
Anode:
6Cl- --> 3Cl2 + 6e
2Au +3Cl2 --> 2AuCl3 Gold is liberated as gold chloride
4OH- --> O2 + 2H2O + 4e Oxygen gas is liberated as hydroxyl ions are oxidized
Secondary Reactions:
AuCl3 is known to react with the Cl- of the NaCl to produce Chloroauric acid
- So -
AuCl3 + NaCl +H2O --> HAuCl4 +NaOH
Sodium Citrate Chloroauric Acid Reduction Reaction:
4HAuCl4 + 12C6H7NaO7 +6H2O --> 4Au + 12NaCl +4 HCl + 12C6H8O7 +3O2
Further Observations:
The power supply should be able to supply ordinary house current of 110vac. literally all that's needed is two wires with a house plug on one end and alligator clips on the other.
There is a minimum voltage for the process to work, based on the electrochemical series. It is only about 30 volts.
MAKE CERTAIN that you plug in the electricity LAST and NEVER TOUCH THE DEVICE WHILE IT IS ENERGIZED.
i prefer to put a SWITCH in the circuit just for added safety.
The Process:
There is a minimum voltage for the process to work, based on the electrochemical series. It is only about 30 volts.
MAKE CERTAIN that you plug in the electricity LAST and NEVER TOUCH THE DEVICE WHILE IT IS ENERGIZED.
i prefer to put a SWITCH in the circuit just for added safety.
The Process:
Arrange the apparatus so that the carife is on the hotplate.
Add 250ml of distilled water to the coffee maker.
Suspend the electrodes in the carife and bring the water to a boil. Leave the power supply off. i simply bend the elements over the edge of the carife.
With the power supply turned off, connect the gold electrodes to the supply terminals, 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.
Bring water to a boil.
add 15 ml ( 3tsp ) of stock table salt solution to the beaker. stir . i use a swizzle stick or a chop stick. non-conducting material just to be safe.
Add 2 drops syrup to the beaker Or about the amount picked up by immersing the end of a chopstick 1/4 inch into the syrup and letting the excess drip off.
Add 5 ml ( 1tsp ) of the stock Sodium Citrate solution to the beaker.
turn on the power source
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.
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.
Half Reactions:
Cathode:
6NaCl + 6e --> 6Na +6Cl- Chloride ion travels to Anode where it combines with gold
6Na + 6H20 --> 6NaOH +3H2 (which bubbles off as hydrogen gas)
6NaOH --> 6Na+ +6OH- New hydroxyls journey to anode continuing electrolysis of water.
Au+++ +3e --> Au We don't want this to happen as it merely produces gold sludge on the cathode
4H20 +4e --> 2H2 + 4OH- This reaction may not happen because of the preference to electrolyze the potassium
Anode:
6Cl- --> 3Cl2 + 6e
2Au +3Cl2 --> 2AuCl3 Gold is liberated as gold chloride
4OH- --> O2 + 2H2O + 4e Oxygen gas is liberated as hydroxyl ions are oxidized
Secondary Reactions:
AuCl3 is known to react with the Cl- of the NaCl to produce Chloroauric acid
- So -
AuCl3 + NaCl +H2O --> HAuCl4 +NaOH
Sodium Citrate Chloroauric Acid Reduction Reaction:
4HAuCl4 + 12C6H7NaO7 +6H2O --> 4Au + 12NaCl +4 HCl + 12C6H8O7 +3O2
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. However, closer electrodes are detrimental in that the gold ions have less distance to travel before contacting the negative electrode. Any gold ion that contacts the negative electrode is wasted. We wish the gold ions to encounter and react with the citrate or peroxide reducing agents before touching the cathode. Therefore higher voltage is beneficial in that it allows the electrodes to be spaced farther apart while maintaining a reasonable electrical current. that is why i use 110vac. not only does it afford more rapid making but also emmits from BOTH elements.
Actual voltage is inconsequential to the electrochemistry. 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 agents (Corn Syrup, and Na-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 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 ion 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 sodium cycle to the gold cycle.
A similar occurence seems to happen with the Cl- ion at the anode. As soon as the Cl- reacts with gold to produce AuCl3 it would immediately ionize, again producing Cl- ions attracted to the anode.
The stoichiometry above assumes all of the chlorine will combine with the gold to produce gold chloride. Experimentation shows that is not so. In fact, it is necessary to at least double the amount of NaCl to achieve CG with the color density of a known 50ppm sample produced by reacting commercial gold chloride with citrate per the Turkevich method. Therefore, it must be assumed that the chlorine is sequestered somewhere or it escapes with the oxygen produced at the anode before it can react with the Au.
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 agents (Corn Syrup, and Na-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 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 ion 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 sodium cycle to the gold cycle.
A similar occurence seems to happen with the Cl- ion at the anode. As soon as the Cl- reacts with gold to produce AuCl3 it would immediately ionize, again producing Cl- ions attracted to the anode.
The stoichiometry above assumes all of the chlorine will combine with the gold to produce gold chloride. Experimentation shows that is not so. In fact, it is necessary to at least double the amount of NaCl to achieve CG with the color density of a known 50ppm sample produced by reacting commercial gold chloride with citrate per the Turkevich method. Therefore, it must be assumed that the chlorine is sequestered somewhere or it escapes with the oxygen produced at the anode before it can react with the Au.
STUDIES
