|In my school science project, I have a submerged steel mesh structure and titanium mesh strip. There is a charge running through this seawater, the titanium's the anode, and the steel mesh is the cathode. From what I already know, the current through the sea water causes the dissolve carbon dioxide to become carbonate ions and attach onto the cathodic steel mesh structure. I was hoping someone could give me the balanced chemical and redox reaction for this process. The steel mesh structure is rusting so maybe this has something to do with it? I didn't really understand the answer to why the cathodic steel mesh structure rusted despite being the cathode nor why the calcium carbonate formed on the steel structure instead of just anywhere.
|Question Date: 2006-06-08|
I appreciate your patience and perseverance in trying to understand what is happening in your experiment! Indeed, that is the very essence of science. I have made an attempt below to explain what is happening as best I can. If you still do not understand, please ask again and I will be happy to continue this discussion.
The first thing I should mention is that seawater has a lot of dissolved salts and ions, including calcium and magnesium salts, chlorides, carbonates, bicarbonates, and so on. Therefore the reactions involved in seawater electrolysis are more complex than say, the electrolysis of water or of NaCl solution.
The basic electrolysis reactions are most likely as below. The essential process is the electrolysis of the water to form hydroxyl ions and gaseous hydrogen and oxygen.
I have denoted the charge on a species within brackets.
2 H20 + 2 e(-) -> H2 + 2 OH(-)
2 H20 -> 4 H(+) + O2 + 4 e(-)2 Cl(-) -> Cl2 + 2 e(-)
The OH(-) ions are generated at the cathode. Hence the solution near the cathode is very alkaline, and this results in the precipitation of calcium and magnesium salts from solution.
First, bicarbonate ions, HCO3(-) are present in water due to CO2 dissolution, which typically forms carbonic acid as below: CO2 + H20 -> H2CO3
which dissociates as below:
H2CO3 -> HCO3(-) + H(+)
then precipitation of salts takes place as below:
HCO3(-) + Ca(2+) + OH(-) -> CaCO3 + H20
Mg(2+) + 2 OH(-) -> Mg(OH)2
Again, these alkaline conditions are only present near the cathode.Hence these salts are generated at the cathode and hence deposit on the cathode, at the site of their production. It is highly unlikely that they deposit on the walls of the container or other locations because they are not being produced there.
Note:These reactions are all individually balanced; however, they are not balanced with respect to overall charge (the reaction should be electrically neutral overall).
I think you are right; the cathode should be protected from corrosion. However the steel mesh may have a layer of rust on it, BEFORE the electrolysis. Some amount of rust will form immediately as the metal is submerged in seawater. When current is passed thru the cathode, the first reaction (before the other electrolysis reactions) will be that rust (iron oxide) is reduced. This, along with the production of H2 gas at the cathode, may result in the formation of the red foam you described.
Once the electrolysis proceeds, the steel cathode should no longer rust. Indeed, the rust should be reduced to iron as mentioned above. You may still have the foam on the surface, as H2 bubbling will continue and there may still be some oxide in solution. However, as the reaction proceeds, take a look at the cathode color. Is it red, or is it black/grey? Red would indicate rust formation, whereas black/grey would indicate deposition of iron. Further on, as the calcium/magnesium salts deposit on the cathode, it should turn whitish.
I hope you understood my previous reply about the effect of different currents. To repeat, a different current will result in a different rate of material deposition at the cathode due to a different rate of electrolysis. However, I should add, that in the complex case of seawater electrolysis, it may also result in a slightly different set of materials being deposited on the cathode due to different growth rates of calcium or magnesium salts. For example, you may have more calcium salts and less magnesium salts. This may result in different physical properties of the deposited material, e.g. different hardness, texture or color. Check for these after running your experiment at different currents.
I'd appreciate it if you could get back to us with your feedback. Is it clear to you what is happening? is there indeed deposition of Ca and Mg salts?
Carbon dioxide, when it dissolves in water, forms a compound called carbonic acid, H2CO3. True to its name, this compound is an acid, meaning that it will dissociate to form bicarbonate ion, HCO3-, and hydrogen ion, H+. The H+ will accept an electron from the cathode; I would expect that this creates hydrogen gas, H2. This leaves a bicarbonate ion lying around in solution, which precipitates out. I suspect that it is precipitating on the cathode because everywhere else in the solution, there are enough hydrogen ions left to turn it back into carbonic acid, so it can't accumulate and remains in low enough concentration that it is still soluble in water.
The negative charge on the cathode will also have a tendency to pull the hydrogen off of carbonic acid, so that it doesn't just dissociate passively. Having an acid there corrodes the iron by shifting electrons onto the iron. The iron can also take the oxygen needed to rust right out of the bicarbonate as well, releasing carbon dioxide and more hydrogen gas (in other words, the carbon dioxide may be acting as a catalyst).
I am mystified by the titanium anode not corroding. Possibly it is because the solution is already acid enough from the carbonic acid (and what is happening on the cathode) that oxygen in the water can't readily give up its hydrogen and bind to the titanium to make titanium oxide. I suggest you take a closer look at the titanium, too, though, and see if it's corroding.
H20 + CO2 -> H2CO3
2H2CO3 + 2e- -> H2 + 2HCO3-
Fe + H2CO3 -> FeO + H2 + CO2
Ti + H2O -> H2 + TiO
And, of course...
2H2 + O2 -> 2H2O
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