Any Ideas?

[zking786]

I have heard about the separation of water into hydrogen and oxygen underwater by short circuiting a charge. The process uses most metals (I've heard of aluminum and carbon as common) as electrodes. One touches the two electrodes together underwater and applies a current through the two electrodes (either DC or AC). If there's enough power, the electrodes will spark and bubbling will proceed.

 

I've done the reaction with carbon rods and have produced hydrogen alone. It seems the oxygen might have combined with the carbon which was melted to form CO2 (I'm not exactly sure whether this is happening). The discoveries have led me to the following questions:

 

1. What is this process called and how is it occuring?

2. I've been told the metals (carbon in this case) are "melting" at around 2000 Degrees Celsuis. Is this true? Is there a chart that discusses the melting temperatures of various metals?

3. How can I determine which metals will "melt" most readily to absorb the oxygen and liberate the hydrogen?

4. How can I measure, quantitaitively, the amount of energy being consumed in the various components of the reactions? From a thermodynamics standpoint, I can measure the total amount of energy inputed, but how can I calculate the amount of energy utilized in melting the metal? How much is utilized by heat, by spark, etc.?

5. Is it necessary for both electrodes to be composed of the decided metal? Can I use one of carbon and one that is less consumable?

6. What equations would enable me to deduct the amount of current and charge necessary for differing metals of varying sizes.

 

 

I would greatly appreciate your assistance in my research. Also, please try to keep explanations basic when possible (I am a HS student ).

 

Thanks!

[insane_alien]

1. Electrolysis

2. Carbon isn't a metal. It's a non metal. It also doesn't melt at 2000*C Nor does it melt at all at standard pressure. It sublimes(goes straight to a gas)

3. if the metal is molten then its more likely to react with both hydrogen and oxygen and cause a reaction between the hydrogen and oxygen. try separating the electrodes and shoving a test tube over the negative electrode.

4.not a clue. too tired just now.

5.nope as long as the electrode conducts.

6.whats the question. for some reason i can't work out what you want to know.

[zking786]

Maybe the carbon is subliming? All I know is that the electrode is consumed (relatively rapidly) in the process and there is no precipitation. The end result is a H2 and no O2--that's why I'm assuming CO2 is formed.

 

But why would it be considered electrolysis if the electrodes are touching?

 

Also, what about for aluminum? Surely it can't be called electrolysis, since I'm told that the temperature is causing the aluminum to melt. Also, I'm told in the reaction with aluminum that a precipitate (an oxide of aluminum) is formed along with hydrogen gas.

 

Regarding the equations, I'm refering to thermodynamics equations that would apply here and how I'd use them.

[silkworm]

Regarding the equations, I'm refering to thermodynamics equations that would apply here and how I'd use them.

 

I'd look at Faraday's Laws. This website claims to be a website explaining Faraday's Laws that is appropriate for HS students: http://www.ausetute.com.au/faradayl.html. I'd explain them myself but I have no idea where you are in your understanding and a complete explaination is beyond my efforts right now. However, if you try to learn it and have some questions I'm sure we'll be happy to help. If you don't get anything from that site just go to any library and look up Michael Faraday, he's a superstar in the science world and made an enormous contribution to mankind with his work.

 

The process uses most metals (I've heard of aluminum and carbon as common) as electrodes.

 

Carbon is not a metal, however a few allotropes of it are known to conduct electricity, especiallly graphite (pencil lead). But this is a property of the hyperconjugation of the allotrope and not the element itself.

 

One touches the two electrodes together underwater and applies a current through the two electrodes (either DC or AC).

 

For the love of Edison, please use DC and know what you're running before you get yourself hurt. DC is used in the proper procedure of electrolysis anyway.

 

I've done the reaction with carbon rods and have produced hydrogen alone.

 

What do you mean by carbon rods? Something like this? .

 

How did you know it was hydrogen?

 

What do you mean the carbon has melted?

 

1. Appears to be electrolysis (if you're decomposing water).

2. http://www.webelements.com, stay to the left of B, Si, As, Te, At, and you'll be okay.

3. The metal doesn't need to melt to react with oxygen. The metals on the far left are very reactive with oxygen and the heavier ones react explosively with water. In electrolysis you don't have to worry about this, you're using electricity to physically seperate water into hydrogen and oxygen.

4. Faraday's Laws.

5. I don't know what you're using right now, but platinum and pallidium are the electrodes of choice do to their inertness.

6. Faraday's Laws.

 

I'd also research electroplating if I were you, as it appears it might be what you're interested in.

[zking786]

Thanks for the informative response. I thought electrolysis wouldn't apply since the electrodes are actually touching (not close together actually touching).

 

Also, I assume it's an allotrope of carbon, since it is a physical rod. It's the type of carbon rod used in welding.

 

You discuss not using AC voltage, why? I read in a booklit on arcing (with carbon rods) that one could use either AC or DC. Why do you mention Edison? I haven't conducted the experiment with AC (only DC), but don't see why AC wouldn't work.

 

I have produced hydrogen, for sure. I'm confident that I did not produce oxygen, since the gasses produced will not combust without additional air. I know the byproduct of the combustion is emmision-less water. As a result, I believe it's producing H2 and possibly CO2.

 

Also, the carbon wasn't melted, it probably sublimed.

 

Why do you mention using an inert electrode? I want something that will react with the oxygen to form an oxide or some gas that won't affect combustion.

 

 

Lastly, thanks for all your information! I will check out the site on Faraday's laws and will research electroplating.

 

Btw: I am a first-year chemistry student. I've only taken a Chem 101 course at my local Junior College.

[YT2095]

so lets see if we`ve got this straight, you`re in effect "Arc welding" underwater (without the actualy welding bit), and you`re interested in finding out what the gas products are?

 

is this correct?

[silkworm]

I haven't conducted the experiment with AC (only DC), but don't see why AC wouldn't work.

 

Don't use AC, you'll get yourself killed. Just use DC and make sure you know what you're using. I think there is a reason why AC doesn't work, but I can't remember why.

 

I thought electrolysis wouldn't apply since the electrodes are actually touching (not close together actually touching).

 

The electrodes touching should complete the circuit, but I don't know what the terminology is.

 

I have produced hydrogen, for sure. I'm confident that I did not produce oxygen, since the gasses produced will not combust without additional air.

 

If the sides weren't touching you'd produce both, one at each electrode.

 

Also, the carbon wasn't melted, it probably sublimed.

 

How do you know? Are you saying that you're massing it or is it noticeably deformed?

 

Why do you mention using an inert electrode? I want something that will react with the oxygen to form an oxide or some gas that won't affect combustion.

 

Inert electrodes are used to keep them out of the product, because they're needed only to supply the DC. I don't understand what you're trying to make, but understand that H2 is will effect combustion and providing a spark for it will make H2O with explosive violence. Combustion reactions have water vapor as a product, and its production is very exothermic, so be careful.

[zking786]

YT2095:

I think it is "arc welding without the welding", but am not sure since it's completing a circuit.

 

silkworm:

Okay, I won't use AC. Anyone know why, though?

 

Also, it is completing a circuit and in doing so it does noticably consume the carbon rod. Within about 15 minutes, the carbon rod is almost completely consumed (8-10in).

 

Here's a background on my research. I'm trying to inexpensively and efficiently produce hydrogen (for use in combustion).

 

Technique 1:

The first technique (using a carbon welding rod to arc with a carbon disc or another carbon rod underwater) was recommended by a friend who knew someone who was able to produce large amounts of hydrogen efficiently. I've seen a prototype for this machine and noted the design of the apparatus. It consists of a large tank of water (1ft^3) and a pair of electrodes which are connected to a source of emf (car battery in this case). These electrodes are touching and DO complete a circuit. When the circuit is complete, one notices large underwater sparks and bubbling. Eventually, the carbon welding rods are consumed and the reaction concludes. Throughout the process, hydrogen is definitely produced (vapor is a result when the gas is mixed with air). Carbon dioxide might be produced, as the gas derived by the experiment is NOT combustible on its own. Anyway, the closed reaction chamber (in this case a fish tank) does not heat up significantly probably due to the localized heat in a large chamber of water.

 

I'm wondering, firstly, what the process is called. I still am unsure as to whether it's electrolysis, since you don't need an electrolyte solution and you actually connect the two electrodes together. Some have told me it's the heat (from the short circuit) that separates the water, causing the carbon to bind with oxygen to form carbon dioxide, while liberating hydrogen.

 

 

Technique 2:

I only briefly discussed this technique, as I have only heard of it's viability. It discusses a simillar procedure, except using aluminum rods. I was told that the heat liberates the hydrogen and causes the melted aluminum to form an oxide with oxygen. While I have not tried this method, early experimenters seem to have had some sucess with it.

 

For this experiment, the same questions I had for Technique 1 apply. Also, I was wondering why aluminum and whether it relates to the "melting point" of aluminum. How does this work?

 

 

 

Hope this is enough clarification. Thanks for all the responses. It really does help.

[zking786]

After reading and attempting to comprehend the site on Faraday's Laws, I had one question. By Faraday's first law, the amount of a substance produced (I'd say hydrogen, in this case) is proportional to the quantity of electricity (Q in coulombs). Since the quantity of electricity is determined by current and time, it seems voltage is inconsequential. Please correct me if I'm wrong (which I probably am), but does that mean I could produce a large amount of hydrogen using a small amount of power. That just doesn't seem to make sense. Can someone explain Faraday's laws in terms of voltage?

[silkworm]

Please correct me if I'm wrong (which I probably am), but does that mean I could produce a large amount of hydrogen using a small amount of power.

 

Yes, it will just take you longer. The voltage is not inconsequential, you still need it, but you don't need a whole lot of it to get results. You seem to have it figured out.

 

It has made sense to you. Pay attention to Faraday, this is his thing. Also look up Humphry Davy, he's somewhat of a wildman who discovered/isolated 6 (I think, maybe more and all in 1808) elements by electrolysis. Mainly they were group 1 and 2 metals. I'm pretty sure you don't have to complete the circuit or use an electrolyte solution, in fact I think either would be counterproductive when decomposing water.

 

Also if your doing this also be cautious with the O2 and the H2.

 

I consider the foregoing investigation as sufficient to prove the very extraordinary and important principle with respect to WATER, that when subjected to the influence of the electric current, a quantity of it is decomposed exactly proportionate to the quantity of electricity which has passed, notwithstanding the thousand variations in the conditions and circumstances under which it may at the time be placed....[url']http://dbhs.wvusd.k12.ca.us/webdocs/Chem-History/Faraday-electrochem.html[/url]

 

 

I have no clue what is happening to your carbon rods, but if you haven't boiled your water it should have some CO2 already dissolved into it.

 

In case you don't know, this whole area is callled electrochemistry and it's good to see you're taking an interest in it, nobody seems too but it's pretty cool.

[silkworm]

Sorry, there was a weird hiccup and it posted that message twice.

[xeluc]

The carbon Electrode is eaten away by O2 producing CO2. If you notice, It should only be the Anode that is attacked. H2 produced at teh cathode does not react with the carbon

 

EDIT: I forgot that you have a different setup here... Im not even sure if there IS a cathode/anode if the carbon rods are connected.. who knows.. I still believe that CO2 is produved though. Not ALL O2 should react though...

[woelen]

What the OP does has nothing to do with electrolysis. The carbon electrode becomes VERY hot and at the water/carbon interface there will be an hefty main reaction:

 

C + H2O --> CO + H2

 

At the temperatures involved, CO2 is not stable and CO will be formed. This is a dangerous gas and may kill you without you even noticing anything (it is odorless, colorless and tasteless and does not give any irritation or discomfort while breathed, as opposed to most other toxic gases, like Cl2, HCl, etc.).

 

So, indeed, here you cannot talk of anode/cathode, there simply is a rod of carbon at red heat in the water.

 

Before the above mentioned reaction occurs, however, the temperature should be really high. I don't know the exact figures, but it is not a matter of 100 degrees centigrade or so. It must be 1000 or even 2000 degrees.

 

The entire carbon electrode disappears. It, together with the water is converted to gaseous products only.

 

You cannot give a precise relation between the amount of electric power or charge taken and the amount of hydrogen produced. With true electrolysis there is a very precise relation, because each molecule of hydrogen requires two electrons. Here, however, the driving force is heat, not charge/electrons. ANy heat source would do, you now just use the electric current for heating the carbon rod.

[zking786]

I've been told that this method is "pollution free". There can't be ANY carbon monoxide. How could I test for CO? Also, I though (from a few other posts) that carbon will sublime and then go back to its solid state (not bonding with any thing). I'm starting to become confused.

 

The mechanism seems to be what you're describing.The rods are completing a circuit and touching, generating a very high temperature where we can notice lots of underwarer sparks and bubbling. Why can't this be precisely predicted?

[zking786]

Also, thanks for the clarification silkworm. So voltage does affect electrolysis, by determining the speed at which the "amount of a substance" is produced. So, then the amount of hydrogen produced in a set time is determined by BOTH voltage and current (basically power). So, why would you prefer a low voltage high current set up? If this works simillarly to power, why wouldn't you be able to use high voltage low current to produce the same amount of hydrogen?

[silkworm]

So, then the amount of hydrogen produced in a set time is determined by BOTH voltage and current (basically power).

 

Yes, I found it useful to think of the whole thing as simply the product of volts and seconds. Just calculate the total volts you are going to use equal to volts that you're applying multiplied by seconds and adjust accordingly.

 

So, why would you prefer a low voltage high current set up?

 

You don't need much voltage to get results, you may have a power limitation, and it's safer (which is important when you're just starting).

 

If this works simillarly to power, why wouldn't you be able to use high voltage low current to produce the same amount of hydrogen?

 

Okay, I think there may be a bit of confusion here. I think I can clear it up with a section of a paper I wrote. Hopefully it makes sense, but it is a little bit out of context (that the rest of the paper builds) and the superscript and subscript are made normal by posting.

 

The most useful principles that govern electrodeposition are the Faraday laws' date=' which are simplified here due to space limitations. They establish the theoretical quantitative relationship between the electric current and material converted by electrodeposition. Before these laws make sense, a few basic concepts must be introduced. The SI unit for electric charge is the coulomb ©. The charge of one electron = 1.6 x 10-19 C. A mole is equal to Avogadro’s number, 6 x 1023, in atoms, molecules, or ions. Multiplying these two numbers we find the charge of one mole of electrons = 96, 500 C, which is known as the Faraday constant (F). Equivalent weight (Echcm) can be calculated by dividing the molar mass by valency. For example, the Echcm of Ag+ is simply the molar mass of silver. The Echcm of Cu2+ is the molar mass of copper divided by 2, or 63.6 grams / 2 electrons = 31.8 grams.

Faraday’s second law says that electrochemical equivalents (Eel) are proportional to Echcm with the constant being Faraday’s constant, mathematically expressed as EelF = Echcm. One amp (A) = 1 C/second (s). Faraday’s first law says that the quantity of a substance (Q) converted by 1 A in 1 s is Eel, mathematically expressed as Q = EelAs. It should be noted that 1 As = 1 C. For practical purposes 1 Amp hour is used for electroplating purposes. 1 Amp x (60 minutes x 60 seconds) = 3600 C (Raub, 25). These relationships apply to both electrodeposition at the cathode and dissolution at the anode (Raub, 28) and assumes that the efficiency of the current is 100% (Raub, 26).

Now let’s walk through the hypothetical situation of replacing the light copper jacket of a US penny made after 1983 with a winter coat. The average US penny has a mass of approximately 2.5 grams with its jacket. To give it a true winter coat, let’s give this penny a 0.5 gram jacket which is the same as a 50 pound child wearing a 10 pound coat. To get rid of the penny’s old jacket simply immerse the coin in a solution HNO3 or H2SO4. Depending on the concentration of the acidic solution, this will occur very quickly and you’ll be left with a zinc penny and copper ion in the acidic solution. Take this zinc penny use it as the cathode. Use a penny made before 1982 as the anode. The pennies before 1982 are pure copper, so dissolution at the anode will provide more copper. For the electrolyte bath use copper sulfate, CuSO4, to provide the Cu2+, in a dilute solution of a strong acid, like HNO3 or H2SO4, to provide electrolytes, and a buffer like acetic acid, CH3COOH, to buffer the solution and keep the pH below neutral (pH = 7) to avoid the formation of copper hydroxide, Cu(OH)2. From above we know that the Echcm of Cu2+ = 31.8 grams. So Eel of Cu2+ = 3.3 x 10-4 grams by taking Echcm / F because EelF = Echcm is the same as Eel = Echcm / F. Using Faraday’s first law, assuming that the current efficiency is 100%, if we use 1 Amp to plate this zinc penny with a copper jacket of 0.5 grams will take a little bit longer than 1500 s or around 25.5 minutes because Q = EelAs here is 0.5 grams = 3.3 x 10-4 grams x 1 Amp x s is the same as 0.5 grams / 3.3 x 10-4 x 1 Amp = 1515.2 s because Q / EelA = s.

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[YT2095]

btw, you DO realise you`re just re-inventing the exposed element electric kettle don`t you

and although it Does work, it was found to be somewhat less than practical.

 

maybe I should have saved this news until After xmas?

 

oh well, nevermind

[zking786]

What's the exposed element electric kettle?

 

Also, why would you say that it's impractical?