Battery chemistry...

[Externet]

Hello.

 

I do not get it; can anyone explain on layman terms ?

 

This reaction on a Mg - C cell uses the dissolved oxygen :

 

http://www-odp.tamu.edu/publications/191_IR/chap_03/c3_5.htm

 

From the galvanic series table Mg and C are at the extremes (I assume to yield maximum potential difference):

 

http://www.corrosion-doctors.org/Definitions/galvanic-series.htm

 

---> Why are the most abundant Cl and Na ions not considered better to be used for the reaction ?

Edited September 19, 2009 by Externet

[Melvin]

Na and Cl2 aren't used because such a cell would require a pure sodium anode, which would instantly react with water anyway and would be consumed too fast. It doesn't work with ions; it only would work with the pure elements. Since sodium has to be produced with electricity anyway, there's no sense in doing this. (To be honest, using the Mg doesn't seem great to me either, because Mg needs to be produced electrolytically as well)

[Externet]

Thanks, Melvin.

 

The point is not the economics of the cell construction at all. The Mg-graphite on link above seems to be the most convenient and effective way to produce electricity (~1.5V) from seawater and am happy with it.

 

These cells are used in open sea bouys that provide telemetry for years between servicing, at the expense of Mg metal as a 'fuel', and am happy with that.

 

Cost of Mg is peanuts compared to moving a ship and crew 2000 miles for servicing shorter duration cells.

 

My lack of understanding is the (chosen?) use of dissolved oxygen which is present in limited amounts on the electrolyte that has (apparently to me) much higher availability of Cl, Na ions to engineer a reaction.

 

Let me put in other words. Using Mg; what other energic electrode instead of graphite coud interact with the ions instead of oxygen to produce long life electricity ?

For sure developers of such cell had it very thought. I just do not get it.

Another cell could be made with Al and Cu electrodes (none being pure sodium), but the generated potential is only ~0.3 V and does use Cl and Na ions for the reaction, not oxygen.

 

So, could the voltage be raised to nearly 1 Volt and use the ions ?

[Melvin]

The graphite doesn't actually react; it just serves as a "return point" for the electrons. The Mg loses electrons, which flow through your circuit to the graphite, where the electrons react with the oxygen to reduce it to oxide (and probably the MgO reacts with water to make Mg(OH)2)

 

I think the main problem with the Al/Cu cell is that the amperage is very low (because the reaction occurs slowly). I think if you used a more alkaline solution (which I believe seawater ranges from 7.5 to 8.5 in pH) you would get more current, so this could be feasible. The only real reaction here is with the aluminum, which reacts with the water to produce voltage. The more alkaline the solution, the faster this reaction occurs. The other electrode only serves to return the electrons to the water and produce hydrogen. Aluminum's value on the electromotive series is 1.67v (found at http://www.diracdelta.co.uk/science/source/e/l/electromotive%20series/source.html), so a cell like this could reach up to that voltage (although it probably won't due to some resistance).

 

Magnesium dissolves in acid solutions, so if you used a very dilute acid solution with magnesium and carbon electrodes, you could be able to make a crude cell that could just be replenished with acid (and Mg when necessary).

 

You could always make some kind of pump (non-electrical of course ) to disperse air into the water to replenish the oxygen supply.

Edited September 19, 2009 by Melvin
more info

[Externet]

Thanks again, Melvin.

 

"the Al/Cu cell is that the amperage is very low"

-->Isn't that more a function of the electrodes area? What am unhappy with Cu/Al is that would yield too low voltage.

 

"if you used a more alkaline solution"

-->No, the whole thing is about seawater.

 

"you could be able to make a crude cell that could just be replenished with acid"

-->Am not trying to make a cell on my workbench; it is more like trying to learn what voltage would yield dipping a 4'x8' Mg and a 4'x8' stainless steel plates in the real sea. Yes, that big.

 

I cannot forecast results from the electromotive series and galvanic series tables, I even do not know why are they different nor which one to use and how.

[Melvin]

It doesn't make sense to me why your Al/Cu cell gives such a low voltage (maybe oxidation of copper producing reverse voltage?).

 

I was saying that seawater was slightly alkaline, which would make it more effective than a plain salt solution.

 

I wasn't aware that you wanted a scale that big . The size of the plates only determines the current (bigger plates = faster reaction = more current). The voltage is determined only by what is reacting; in the case of the "4'x8' Mg and a 4'x8' stainless steel plates in the real sea" part of it would be the reaction of Mg (2.4v) and dissolved oxygen (-.397v). Combined gives you a theoretical maximum of 2.797v (note that since oxygen is reduced you negate the voltage). With the reaction using just the dissolved salts in the seawater, the reaction is between Mg (2.4v) and water (-1.23v), giving 3.63v (although the current is much less because the reaction is pretty slow). All of these values came from the electromotive series in the link I posted earlier.

 

In reality, there will be resistance in the system so your actual voltage and current would be much less. The only real way to get exact values is to test it .

 

Hope this helps a bit