Ibpointless2 Crystal Cells

[triffid]

That is a completely missleading thing to say. The ammount of energy you can make to affect the two surounding plates totaly depends on how close they are. You could fill a large room full of devises which had one centimeter in distance between the plates and get zero energy extracted from it.[/size]
You could also fill the room with devises that had only a few nanometers in distance between the plates and get a usefull amount of energy out of it. This is the reason to why it is missleading to say that so and so much region of space ( the earth for example) only can create a fixed output of energy. It totally depends on the technology we use to extract it with. As nanotechnology in the future will be many times more advanced, and able to be mass distributed, it will surly make this form of energy extraction become many times more efficient. --Nabo0o (talk) 10:54, 29 April 2008 (UTC)[/size]

[jbignes5]

http://www.av8n.com/physics/battery.htm


We have to understand that it isn't the plates that cuase the potential diffrerence it it the area in between the plates. It has to have the ability to polarize and allow a flow in one direction. I am thinking making the capacitor with active plates with a potential difference allows a certain amount of flow through the polarized medium in between those plates.


When we use electrolytes in water as the medium, it polarizes the medium based on the potential difference of the plates. During this process it also involves flow of the medium. This is where current comes from. Water is what is causing the effect of current. It flows and rubs against the plate in the direction of the polarization.


When we use the solid electrolytic as they setup we loose current capability because it is locked into the crystal structure and most of the free flowing water can not impart it's current ability to the electrode. This is evident by the action of Bedini's battery chargers as proof of this flow. When you apply the pulses back to the battery it causes super flows into the battery both cleaning the plates of sulfated crystals and imparting a flow to the chemicals and water inside of the battery. I have always believed that batteries worked as such and this finally proves it. The evidence is overwhelming by the simple fact that in the case of the bedini chargers continue to charge the battery as the flows are winding down in the battery. These flows I suspect are driven by the plates themselves. When the pulses hit the plates it changes the charge level of that plate to a lower value. This is like squeezing the plate like a sponge. This maintains the flow and slowly ebbs as the plate expands out as in the sponge analogy.


In the case of our cells we could use this method to increase the separation of the plates we choose. This would increase the voltage and may even increase the current to a small degree. It could be that we could use these cell exclusively with the pulse system to maintain the separation even while using the spring back to maintain the plates separation in potential. As Bedini has shown the benefits of using such a system from experimental data of his special unit he uses with his test cells.

[Peanutbutter29]

First @ Phi, I wasn't necessarily meaning sacrificial in terms of it being destroyed necessarily;  more of a reference to being the oxidative electrode in comparison to a reductive electrode;  sorry to confuse. 

Also, I'd gotten confused and incorrectly stated which electrode that is, from the Electro-potential POV.  I'd forgotten that Current in one direction is galvanic and current the other direction is electrolysis.  E.g.  Oxidative electrode is Opposite in the two cases.  So, ya;  without terminal connection but WITH 2 electrodes (Al C) the electo-potential makes Carbon the oxidative pole.  When a circuit is connected (wire) then current is flowing opposite and the Al becomes the Oxidative pole.  An electrolysis condition would not occur, specifically from electrodes, if both plates are alike. 

Now for some good stuff!  I went ahead and analyzed / tested this reaction further to fully understand what's going on.  I was able to learn a couple things too.

-Okay so first, the reaction I elaborated on with Durhams and MgSO4 should be right.
-Second, adding an Al electrode to the (still) chemical only equation.  Changes nothing.  Mg is more reactive than Al and Ca is more reactive than Mg NO REACTION.
-Third, adding a second electrode of C DOES (proven) produce a flow of electrons and corresponding electrolysis.  This is easily figured and quantified in the equation below.  Note, since current (in electrolysis) is proportional to the reaction rate.  You could expect to see a reduction (electrolysis) on the Al plate in approximately 3 months (104 days, first pic) So, again with BOTH Al and C left in H20 +MGSO4 +(CaSO4 not reactive in this case) = reduction on Al and Oxidation of C.
-Fourth, I must say good job to you Phi!  you found a very unique case where a "closed loop" reaction can occur with a electro-potential difference!!  Meaning, MgSO4 + H20 +( Al + C)2e- = the ability to charge!!!  So, this is good knowing durhams is not required or involved in this reaction AND another method of a closed- loop cycle to permit charging.  Nice one!

-Fifth a KEY note, this combination;  when assembled represents a discharged cell.  The electo-potential setup by the Al C electrodes (.6v @ 20uA w 2tsp in 300ml) is actually "charging" the cell by electrolysis (this would be extremely slow).  However, this IS then the reason for an apparent "re-charging" or bounce back of certain cells!!!  Connection to a load will attempt oxidation of the negative pole (galvanic), whereas an open connection (and 2 dissimilar metals) attempts an oxidation of the Positive pole (electrolysis).  If a cell has any ability for any level of closed-loop reaction, you will have a apparent "bounce back" but is in fact an electrolysis of the 2 metals and a reversal of current flow.

Noting that MgSO4, Al, C and H20 will accept a charge and knowing there must be a closed-loop reaction;  I took the time to figure it out.  You can see what I mean by these reactions getting complex quickly, lol.  Anyway here you go

Charge cycle
-Cathode: Al + MgSO4 + H20 + (2 H2O(l) + 2e-) = H2(g) + 2 OH-(aq) = 2 Al(OH)3 + MgSO4 + H20 >>>>
+Anode: C + MgSO4 + H20 + (4 OH- (aq) ) = O2(g) + 2 H2O(l) + 4 e-) = MgO + H2O + H2 + MgSO4= Mg(OH)2 + MgSO4 + H2S04 >>>>
>>>>(2 Al(OH)3 + 3 H2SO4) = Al2(SO4)3 (solid O,P,Hydrate) + Mg(OH)2 (s) + MgSO4 (diluted) + H20
-----
Discharge cycle
Al2(SO4)3 +Mg(OH)2 + MgSO4 +H2o = MgSO4 + Al2O3 + H20 + 2e- (overall water is lost)

Molar quantities and Hydrate forms are not shown (thus not balanced);  but this is the reaction shown in total.  The inclusion of electrolysis effect and products were accounted.  If you can follow the equation it does indeed show a closed-loop chargeable cell!!!!  Again, nice find Phi!  The only down side is you do loose some water and Al is oxidized during the discharge (galvanic) cycle.  The good thing is Al will not form thick O2 layers and may conduct for a long time.

This reaction HAS BEEN TESTED and all output products accounted for, yay! No if's and's or but's.  Attached at the bottom are 2 pics of the aluminum electrode following a short period of electrolysis or long duration of sitting (proportional to current).  You can see the formation of Al2(SO4)3 on the plates from the charge cycle (electrolysis). 

Onto some testing results: (note I will determine proper current for least Gas lost with a variable DC supply and then use a digital current controlled charger for proper charging)  Dc supply V & A adjustable 0-18v.  DC charger Integy 16x7 Microprocessor charger (few things I kept from R/c days , Integy Rocks!
*Typically charge rates are around C/10, so it will be pretty low

Okay, Al C electrodes.  MgSO4 @ 2tsp in 300ml H20.
-Start .6V and 20uA
-Charge @ 300ma with Dc supply V @ 8.6-7.4v <<(this proved it was taking a charge and led to deducing equation ;P)
-Charge duration 15 min (first photo below, beginning of Al2(SO4)3 reduction on plate
-Charge duration 1 hr (second photo below, good thick coating of Al2(SO4)3
-End Charge 2.56v and 30ma dropping fast
-Electrodes then were shorted through meter on ma and left shorted (galvanic)
-Time 0 min = 30ma
-Time 5 min = 2.5ma
-Time 15 min = 2.0ma
-Time 22 min = 2.9ma*
-Time 37 min = 3.1ma*
-Time 62 min = 3.51ma*
-Time 87 min = 3.54ma*
-Time 102 min = 3.45ma*
-Time 117 min = 3.46ma*
-Time 137 min = 3.3ma
-Time 157 min = 3.25ma
-Time 457 min = 0.65ma
-Time 517 min = 0.58ma
-Time 727 min = 0.49ma
-Time 787 min = 0.48ma Current time now

Power input - 2.25whr, recovered after 787 min - 17.6maH approximately (plotting output on graph looks similar to discharge cycle for SCE Ni-Cd)
There is no way here to quantify power here as shorted voltage is 0 (electrically no work being done, lol), but you can record maH.  Once current is determined the Integy charger records maH input and a resistive load will be used @ C or less to determine maH out.

Most of the Al2(SO4)3 has been oxidized off the Al plate and (from formula above) AlO2 is being formed.  (again Al is nice not being able to have a thick coating ).  I'll continue to record until back to discharged state (as assembled) of 20uA to determine recovered maH.

This reaction, of course, would be limited to cycles based on Al02 produced and H20 lost from overall reaction;  but decidedly this can be charged multiple times.  overall efficiency can't be determined without proper charge rate though I don't expect it to be high.

Oh also, in my findings to help;  Sulfates are more reactive than Hydroxides (great to know more reactivities of anions)

I hope this info can help some for cell design.  I believe that reaction above (or similar) may prove to be a valuable one.  From the finding I can think of quite a few more ways to accomplish this with different electrolytes an metals (ordering goodies for one soon).  I'll try to see if I can find a good x/C charge rate for this solution and report back if y'all like.

Hope this helps,
Thanks
PB

EDIT:  I also went to see what total power we are trying to "beat".  Looking at a AA Extended duracell;  we have 2800maH and (assume)avg voltage of 1.3v.
this equals appx. 3.64 watt/hours.  So that would seem the first target to beat.  Assuming then, an average crystal cell discharge voltage of 1.1v and 3 ma;  the cell would need to run for 46 days to equal 3.64 watt/hours.  Assuming 1.1V and 1ma then, would be 138 days for 3.64 watt/hours. So, these don't have to last forever at all to beat an existing cell, but at 200uA it would need to last 1.89 years!  Just FYI

Magnification is 400x

[PhiChaser]

@PB: Wow! Okay, I will have to read that again... By your estimates I should have a dead cell in four months using Durham's, epsom salt, and H20? Well, I hope they last longer heh heh... If not, we will have to see what to do to make that happen.
Now, as far as voltage/current to beat. If I can make small (like mini ice cubes) cells and make them work together in some sort of case or whatever, I don't care if they produce only .5v each, as long as they last and last and last... Long term low numbers are better than short term high numbers (my line of thinking anyways). Fantastic work BTW!! I wish I could appreciate it fully with my meager chemistry knowledge... Now what would happen if you used silica sand as an electrolyte in that equation? Would it somehow increase recharge time or up the vA numbers?
I'm trying to find the 'tipping point' where the amount of epsom salt will make the cell harder as it dries instead of softer... It is funny how a 1/4tsp of water in either direction makes for a never-setting semi-soup or a hard to work with stiff putty... Getting closer though... This last cell looks nice anyways LOL!! Now I am truly scraping the bottom of the barrel on supplies...
Right now I am looking at a 4:1 ratio of epsom salt to Durham's in this last cell, and .75 H2O in regards to the Durhams (I hope that makes sense). Does this affect your equations (is that the right word?) up there? What should the balance between gypsum and epsom be for best results?
Tomorrow I'm hoping this cell is insanely hard AND giving me over a volt AND doesn't drop off fast.
Oh, I hooked up another cell to the 6v batteries in parallel and kept it's twin (for lack of a better term) unhooked for a baseline on that formula.
Why haven't chemists found this already?
Happy experimenting,
PC

[PhiChaser]

This morning I unhooked the cell I had on the lantern batteries all night. It read just over 2.7 volts (about half of one of the batteries?). I hooked it back up and have decided to leave it for the rest of the day. I will look at it later this evening and see if there is an improvement. It looks like there is an upper limit to how much voltage these cells can hold, but perhaps long term hookup will increase that number? Maybe I should hook it up to my car battery LOL!!! (I'm pretty sure I'm kidding...)
@PB I re-read your post again. You said I should see a drop off in about 100 days on the aluminum electrode? I'm not sure I understood why but I appreciate your work!! Too bad we all don't live in the same city, we could start a new r&d company with federal grants... (Yeah, kidding there...mostly...)
I've got the day off today so I will be taking some notes/measurements so I can see if more epsom is the way to go, maybe more isn't necessarily better? Perhaps dissolving the salts in hot water will give me better results? I can imagine that the Durham's will set off WAY fast with steamy water... Lots of testing still to do...
Would the basic ingredients in Durham's be easy to combine into a hard electrolyte? What I mean to say is; If I used powdered gypsum and amylose (corn starch?) and powdered limestone, how hard do you think it would be to make my own version of Durham's? Cost would be an issue maybe... I don't have THAT much free time, just thinking out loud here...
And I still need to get some wax (parrafin?) eventually, so I can 'dip' my finished dried cells to enclose them. Seems to be helping out with triffid's cells at any rate...
I think I need an adjustable DC power supply... Ah, isn't it fun being a backyard scientist? 

Happy experimenting,
PC