Ratios of produced gases in water electrolysis

Question

I just did a little water electrolysis experiment at home by immerging two aluminum electrodes in water with a small amount of table salt. The voltage of the battery was 6 V.

What surprised me is that I was able to collect a much bigger volume of gas at the electrode connected to the - than to the + (about a factor 10 by eye). I would have expected a ratio of 1/2 (2 times more hydrogen than oxygen).

• Why did I get less gas than expected at the electrode connected to the +?

• Was the oxygen consumed by oxydising the electrode? (It appeared quite damaged at the end.)

• How can I get a ratio of 1/2?

Answer 1

We know from stoichiometry that you did produce twice as many moles of $\ce{H}$ at the negative electrode as $\ce{O}$ at the positive electrode. However, we don't know a few important things:

1. What were the actual volumes of the gases that were finally collected? Volume is notoriously hard to estimate by eye, unless you are comparing the levels in two identical containers. So, we'll take your word for it that it was "more", but we won't place any bets on the factor of 10.
2. What happened to the oxygen and hydrogen after the water split? Did it react?
3. What else was in the water that could have reacted? We know there is $\ce{NaCl}$, but maybe there are other dissolved compounds as well?

Question 1 can't really be answered here - the best we can do for now is assume that the volume of hydrogen was 2-10 times larger than that of oxygen. To find out for sure, you would need to conduct another experiment using graduated collection vessels.

Question 2 we can partially answer. If the electrode was corroded, then something reacted - either chlorine, oxygen, or some other more complicated set of reactions. We can get a rough estimate of the volume of oxygen this might consume by assuming it was oxidized by oxygen. If that 0.1 g of the electrode was oxidized, and that it was oxidized into $\ce{Al(OH)3}$, then at 1 atm and 298 K it would consume roughly 137 mL of $\ce{O2}$:

$\ce{Al^3+ + 3OH^- -> Al(OH)3}$

$0.1 \space \rm{g \space Al} * \frac{1\rm{\space mol \space Al}}{27\rm{\space g \space Al}} * \frac{3 \space mol \space O}{1 \space mole \space Al} * \frac{1 \space mol \space \ce{O2}}{2 \space mol \space O} = 0.0056 \space mol \space \ce{O2}$

$PV = nRT$ (oxygen is close enough to ideal under normal lab conditions)

$V = \frac{nRT}{P} = \frac{(0.0056 \space \rm{mol} \space \ce{O2})(0.08206 \space L \space atm \space mol^{-1}K^{-1})(298 \space K)}{1 \space \rm{atm}} = 137 \space \rm{mL}$

$\ce{Al(OH)3}$ is a likely candidate for a reaction product because of the white flakes you observed - aluminum hydroxide is insoluble in water. Aluminum oxide ($\ce{Al2O3}$) is also a potential candidate, and would consume 68 mL of $\ce{O2}$ under the same conditions.

So, assuming 0.1 g is a reasonable estimate for the mass of corroded aluminum, it seems very possible that what happened is the oxygen reacted with the aluminum electrode, reducing your yield of oxygen gas.

How can you get around this? Well, my five minutes of google research did not turn up any good candidates for inert electrodes. Usually graphite electrodes are recommended for other electrolysis setups, but they will react with oxygen as well. I have seen stainless steel recommended, but I would think that gold/platinum/silver are probably your best bets.

Question 3 is also hard to answer. We know that the $\ce{NaCl}$ will react and produce hydroxide ion (further supporting our hypothesis for aluminum hydroxide formation) at low concentrations, and chlorine gas at higher concentrations. If it were producing chlorine gas, you would know - it would be a pale yellow/green and it would make you sick as soon as you smelled it. So it is unlikely that your concentration of table salt was high enough for that to be the competing reaction. However, it would contribute to the aluminum oxidation. The easiest way to rule this out is to not use table salt. I have seen baking soda (sodium bicarbonate) recommended instead, and that would probably work. However, it might still produce hydroxide in solution. What about other unknown substances? If you used tap water, it is likely that there is additional chloride, possibly fluoride, and probably various magnesium and calcium salts. All of these would react during the hydrolysis, and would affect yields. The way to eliminate them is to buy or make distilled or deionized water.

Summary:

Why did I get less gas than expected at the electrode connected to the +? Was the oxygen consumed by oxidising the electrode? (It appeared quite damaged at the end.)

The answer is most likely yes, the oxygen reacted with the aluminum. I would bet that it formed aluminum hydroxide, which then precipitated into the white flakes you observed.

How can I get a ratio of 1/2?

You might not be able to get it to exactly 1:2, but if you use sodium bicarbonate or a weak acid (vinegar would probably work) instead of salt as an electrolyte, and an inert electrode - like gold, or platinum, you can probably get close. Keep in mind - hydrogen gas is usually the desired product in hydrolysis. Oxygen is cheap if you don't need it to be pure! And even pure oxygen is relatively cheap compared to other gases.

Answer 2

Unfortunately, the previous answers never gives real answers. I will try to explain it now, from the very beginning.

Let's first start with Tony's experiment. He electrolyzed a solution of table salt with two aluminum electrodes at 9 V, and obtained about $10$ times more hydrogen than oxygen. Well ! This is not too difficult to understand, as the usual aluminum metal is always covered by a thin, waterproof, colorless and insulating layer of alumina (aluminum oxide $\ce{Al2O3})$. When dipped into water, aluminum metal is separated from water by this protecting layer. If this layer is rubbed, it will be immediately restored by contact with air. So aluminum pieces cannot be used for electrolyzing water under the usual tensions (about 2 Volts). But Tony used as high a tension as 9 V. This is enough for breaking the protecting alumina layer, and allowing the aluminum atoms to touch water. So now the following reaction occurred : $$\ce{2 Al + 6 H2O -> 2 Al(OH)3 + 3 H2}$$ which has nothing to do with any electrolytic reaction. As a consequence, most of the hydrogen gas obtained at the cathode was mainly due to the previous reaction and not to the electrolytic process. And the precipitate observed must have been $\ce{Al(OH)3}$, as this hydroxide is insoluble, white and flaky. At the anode, the alumina layer was also broken, but water was not oxidized into $\ce{O2}$ as expected. On the contrary, both the chloride ion and the aluminum metal will be discharged in priority according to : $$\ce{2 Cl^- -> Cl2 + 2 e-}$$ $$\ce{Al -> Al^{3+} + 3 e-}$$ so that practically no Oxygen gas will appear on this electrode. Also chlorine will react with aluminum of the anode to produce $\ce{Al^{3+}}$ ions $$\ce{2 Al + 3 Cl2 -> 2 Al^{3+} + 6 Cl^-}$$ This is why the electrode was damaged at the end of the experiment.

At the end, Tony asks how to obtain the ratio $\ce{H2/O2 = 2:1}$. The answer is that with aluminum electrodes, it is simply impossible. He has to use graphite or platinum electrodes, and replace $\ce{NaCl}$ by $\ce{Na2SO4}$ which will not produce chlorine gas at the anode.

Now let's comment Peter Peterson experiment. He does not give as many details as Tony, but he used electrodes of magnesium, aluminum and stainless steel, and pure water. Well ! Pure water is an insulating material. No electrolysis is possible with pure water at low tensions. But he used a rather high tension : $12 V$. As he obtained the same gas at both electrodes, this shows that the main reactions were not due to electrolysis, but to a pure chemical reaction of metal + water. For example $$\ce{Mg + 2 H2O -> Mg(OH)2 + H2}$$ Here too, alumina protective layer on aluminum has probably been destroyed by the high tension, as described in the previous lines. And he should remember that stainless steel resists to oxidation by air, but not to anodic oxidation.

Now let's comment Thomij answer. He has no real explanations to give. He proposed to do more experiments, to work with another electrolyte, and to use platinum electrodes. This is correct, but it does not explain Tony's observations.

Answer 3

It looks like chlorine from table salt, rather than oxygen from water, is created at the positive electrode (I believe the reason is given in http://en.wikipedia.org/wiki/Electrolysis_of_water#Electrolyte_selection ), and the chlorine reacts with aluminum (http://www.chemicalforums.com/index.php?topic=3673.0 ).

Answer 4

I've done exactly the same electrolytic demonstrations not just with aluminium but also with stainless steel and magnesium materials for both anode and cathode. I had not used any electrolyte because one doesn't need to. On all occasions gases were produced from the process. From doing these demonstrations, both anode and cathode electrodes decomposed over time and what's more interesting is that the same gases were produced at the anode and the cathode - hydrogen. Using magnesium ribbon as the electrodes was more interesting because after doing the electrolytic process on three separate occasions, once using a 12v battery, there was always more gas produced at the anode than the cathode and this again was hydrogen. Another interesting point was that after collecting the gas from the anode and cathode using magnesium ribbon and aluminium, both gases ignited and a clear yellow flame was seen. In my understanding using the same material for both anode and cathode electrodes one will observe them decompose during the electrolytic process. This would happen with any material used for electrodes but given the way some materials are manufactured the decomposition rate could take much longer.