Voltage step-down and electrolysis...

[MDJH]

Now that it's been summertime for weeks I'd like to do what I was waiting to do during the school year... apply what I learned in Physics and Chemistry during the summer. Trouble is, I'm not sure if I adequately remember Physics and Chemistry, so I'd like to check here to make sure I'd know what I'm doing.

 

First off, with volts, amps, and watts; Volts is joules per coulomb, meaning energy divided by charge; am I to assume this means voltage refers to "how much energy the charge has"? Amps is coulombs per second, so how much charge is flowing per second, and watts refers to joules per second, so how much energy is flowing; so in a modified version of "Faraday's Iron Ring" (IIRC it used coils of wire on one side to create a magnetic field in the iron that was pushed to the other side where it induced current in coils of another wire) where there's more coils on one side than the other, using the side with more coils as the current input and the side with less as the current output would step-down the voltage and therefore step-up the current to still have about the same amount of power... am I to assume this means that as a result of the iron ring transfer, the same amount of charge doesn't have as much energy, but there's more charge?

 

Ok, so now to chemistry, where I learned about electrolysis of water and of aqueous solutions. The quantity of products, according to Faraday's Law, depends on the charge, not the voltage, right? So is determining whether or not electrolysis occurs the only effect voltage has on electrolysis? If not, what other effects does voltage have? If so, does this mean that as long as the voltage is past a certain threshold (I think it's called the cell potential) increasing the voltage won't affect the electrolysis so long as the current is the same? If so, what happens to the extra energy being supplied? Does it heat the aqueous solution?

 

Anyway, I still have my chemistry textbook, it says the cell potential for pure water is 1.23 volts. Was that a typo? Does that mean I can electrolyse water, without an electrolyte, from a 1.5 volt cell? If so, would the extra .27 volts of electric potential overheat the cell or is it safe? Also, would how big the water container is affect it, or would that only affect the resistance and therefore the current drawn?

 

I think I had more to ask but I'll leave it there anyway, as that's a lot of questions already.

[Reaper]

This would probably help you with the electrolysis of water: http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/electrol.html

 

The negative end will form hydrogen bubbles, while at the positive end will form oxygen bubbles. The water completes the circuit, and you have a charge going through it. Size of the water container is not terribly important, but the energy that your battery supplies does.

 

 

The site also has good info on electro-chemistry, so check it out.

 

Anyway, I still have my chemistry textbook, it says the cell potential for pure water is 1.23 volts. Was that a typo? Does that mean I can electrolyse water, without an electrolyte, from a 1.5 volt cell? If so, would the extra .27 volts of electric potential overheat the cell or is it safe? Also, would how big the water container is affect it, or would that only affect the resistance and therefore the current drawn?

 

All the potential really is is the minimum amount of energy needed for electrolysis to take place. The same applies to specific heat of materials too. Also, you need an electrolyte in order for it to work because that is what will deliver the energy necessary. Sulfuric Acid and Sodium hydroxide are commonly used as electrolytes.

[MDJH]

This would probably help you with the electrolysis of water: http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/electrol.html

 

The negative end will form hydrogen bubbles, while at the positive end will form oxygen bubbles. The water completes the circuit, and you have a charge going through it. Size of the water container is not terribly important, but the energy that your battery supplies does.

 

 

The site also has good info on electro-chemistry, so check it out.

 

 

 

All the potential really is is the minimum amount of energy needed for electrolysis to take place. The same applies to specific heat of materials too. Also, you need an electrolyte in order for it to work because that is what will deliver the energy necessary. Sulfuric Acid and Sodium hydroxide are commonly used as electrolytes.

Ok, so the energy the battery supplied is what's important... but the electric energy is a product of the voltage and the charge, right? So what determines how much charge will flow per second? According to Ohm's Law it would be the voltage divided by the resistance, right? So would the electrolyte used determine the resistance and therefore the current flowing, therefore the energy transferred per second? Otherwise, does the electrolyte used alter the voltage required at all? And what about my earlier question of the remaining volts, would those return to the battery and damage it? Also, if I used a higher-voltage source, what effect would that have?

 

If I were to use sulfuric acid, where could I get it to known concentration? I think I remember reading before about auto repair shops selling sulfuric acid for refilling car batteries, is that the case? Would that be the most convenient source of sulfuric acid, or are there other ways? Also, what would happen to the sulfur in the sulfuric acid? Would it produce any harmful gases, and if so, at which electrode? And lastly, if I were to use copper wire, wouldn't copper react with sulfuric acid to form copper sulfate? If so, in which case I'd use a graphite rod from a pencil, how much resistance would the graphite rod from a pencil have if I separated the graphite from the wood by burning the pencil?

 

EDIT: Oh, and where can I get a hydrogen fuel cell?