
#1
Dec1213, 01:08 PM

P: 339

The partial pressures of O_{2} and H_{2} in the atmosphere above water are typically very low indeed, so why is it we don't see the equilibria 2 H_{2}O = O_{2} (g) + 4H^{+} + 4e^{} and 2 H^{+} + 2e^{} = H_{2} (g) occurring whenever we have a beaker of water (if this was observed then the H^{+} concentration would be changing even in pure water)? Since K=1 for the 2 H^{+} + 2e^{} = H_{2} (g) and the partial pressure of H2 in air is incredibly small, shouldn't this reaction be moving forward?
And what about the redox reactions water can undergo involving hydroxide ions rather than protons? Don't these equilibria exist too? (e.g. 4H_{2}O + 4e^{} = 2 H_{2} (g) + 4 OH^{}) 



#2
Dec1213, 02:21 PM

Sci Advisor
P: 3,375

The concentration of H2 and O2:
http://en.wikipedia.org/wiki/Atmosphere_of_Earth You probably know the redox potential of oxygen. So you can compare. 



#3
Dec1213, 02:50 PM

P: 339

Ok so I used this website to provide standard electrode potentials: http://web.missouri.edu/~puckettj/Ch...ntials298K.pdf and your link for the partial pressures. Then put them in the Nernst equation.
I found that O_{2} (g) + 4H^{+} + 4e^{} > 2 H_{2}O should be occurring spontaneously and H2 (g) > 2 H^{+} + 2e^{} should be going spontaneously. This doesn't answer my questions though: 1) Why don't we observe this change happening in the direction for which we have predicted spontaneity? Does it just take ridiculously long for some of these reactions to go to equilibrium? 2) The value of E for these reactions is not that small  O_{2}/H_{2}O has E = +0.805 V for example  so why is it that, in aqueous solutions with electrodes measuring redox potential, we seem to be able to treat the effect of these water equilibria as negligible (so that the experimental value of E is taken as a decent estimate of the potential of the electrode equilibrium, even though water is also present to set up an equilibrium and add to the potential observed). 3) If we try to understand the beaker of water as an equilibrium system, we have two more redox reactions possible for the pure water (those involving hydroxide ions). How do we factor this in? i.e. does the value of E for these reactions from the Nernst equation accurately predict the spontaneous direction of these equilibria, and how do we get away with always treating these reactions as negligible when measuring electrode potentials of other redox equilibria? 



#4
Dec1213, 05:14 PM

Sci Advisor
P: 3,375

Redox reactions in water
Come on, you have to look at a table of standard electrode potentials to tell me that burning of hydrogen gas should be a spontaneous process?
Well, as you have looked up these values you could calculate, e.g. the equilibrium concentration of H2 in the atmosphere given the concentration of O2. Would be interesting to compare with the actual values from wikipedia. 



#5
Dec1213, 06:39 PM

P: 339

Can you please answer my questions first?
1) Why don't we observe this change happening in the direction for which we have predicted spontaneity? Is kinetics the main barrier to watching these reactions happen? In which case nature itself is unlikely to have reached equilibrium. (and the calculations also agree with this) 2) The value of E for these reactions is not that small  O_{2}/H_{2}O has E = +0.805 V for example  so why is it that, in aqueous solutions with electrodes measuring redox potential, we seem to be able to treat the effect of these water equilibria as negligible (so that the experimental value of E is taken as a decent estimate of the potential of the electrode equilibrium, even though water is also present to set up an equilibrium and add to the potential observed). 3) If we try to understand the beaker of water as an equilibrium system, we have two more redox reactions possible for the pure water (e.g. 4H_{2}O + 4e^{} = 2 H_{2} (g) + 4 OH^{}). Then how do we come up with a good way of modelling the equilibrium system of pure water across all these reactions? e.g. if the system were to reach equilibrium, however long that takes, how do we calculate equilibrium concentrations/partial pressures, given the values of K for each of these reactions? 



#6
Dec1313, 01:26 AM

Sci Advisor
P: 3,375

I fear I don't understand your questions very well. All we have predicted is that even small amounts of hydrogen will react with oxygen to form water.
We observe this whenever we provide hydrogen and oxygen and some decent electrodes to catalyze the process. Maybe the last statement is what you are looking for. The reaction of oxygen to water is kinetically inhibited at many metal surfaces. In German this is called "Ueberspannung" http://de.wikipedia.org/wiki/%C3%9Cb...ektrochemie%29 I am not sure about the proper translation. This is the reason why you can often neglect the effect of oxygen in electrochemical measurements. To your question 3 I think that taking e.g. OH into account is rather trivial as it is linked to H+ via the autoprotolysis constant Kw. 



#7
Dec1413, 05:20 AM

P: 339

Ok perhaps if I asked a couple broader questions the whole thing would make more sense to me. If you take a beaker of pure water, or any system where there are a few possible redox reactions and equilibria (such as water which has 2 redox equilibria and Kw), and say it goes to equilibrium, and you know these 3 equilibrium constants, how do you write the equations to calculate equilibrium concentrations (of H+ and OH) and partial pressures (of O2 and H2)?




#8
Dec1413, 11:03 AM

Sci Advisor
P: 3,375

Basically, we use Nernst equation
http://en.wikipedia.org/wiki/Nernst_equation to determine the equilibrium constant i.e.## E=E^0_O+E^0_H (RT/4F)\ln (p(O2)[H+]^4)(RT/2F)\ln(p(H2)/[H+]^2)=E^0_O+E^0_H(RT/4F)\ln(p(O2)p(H2)^2)=0## you see that (I set [H2O]=1) you don't need Kw and only the partial pressures of oxygen and hydrogen are left. 



#9
Dec1413, 07:46 PM

P: 339

I see. And performing the same manipulation with the other two redox reactions of water (which involve OH^{}) gives the same result (to be expected since they are not independent equilibria but rather can be found by combining K_{w} with the proton redox reactions of water).
1) Given that we know all equilibrium constants (let's say so), shouldn't we be able to find the equilibrium partial pressures individually? Not just the p(O_{2})p(H_{2})^{2} value, as you have shown ... 2) What if, say, there was also some Fe metal in the solution, so now the two equilibria Fe^{3+} + e^{} = Fe^{2+} and Fe^{2+} + 2e^{} = Fe also are established along with the previous ones. Given that the aqueous solution already has some redox equilibria (as discussed above), how do we calculate the equilibrium concentrations now? You can assume any equilibrium constant values needed are known. I added this complication to see how we try to deal with multiple possible redox equilibria. In fact, now I think about it, let's say I know the concentrations of each species in the above mixture at a certain point in time; how can I work out E accurately? (This is a separate question to the one above on calculating equilibrium concentrations.) 



#10
Dec1513, 02:06 PM

Sci Advisor
P: 3,375

To 1: Of course, if you start out e.g. from pure water you have 2p(O2)=p(H2) and the system if fully determined.
In your original question I would rather take p(O2) as given and see whether the hydrogen concentration in the atmosphere corresponds remotely to the calculated equilibrium value. To 2: Basically, you can treat all redox equilibria like any other equilibrium. The only difference is that instead of the standard Gibbs enthalpies you are given standard potentials. 


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