Stankovic123's chem q's

[Mao]

Wait, I thought equilibrium constants were dependent on temperature only. Do you mean reaction quotients?

Up until a point where activity is linear with concentration. In heavily concentrated systems such as batteries, it stops being so clear-cut.

(Fine you got me, I did mean reaction quotient. But I can cheat my way out of it so nerrr)

[lzxnl]

Up until a point where activity is linear with concentration. In heavily concentrated systems such as batteries, it stops being so clear-cut.

(Fine you got me, I did mean reaction quotient. But I can cheat my way out of it so nerrr)

But even then the equilibrium constants are constant; we just can't approximate with concentrations


Which leads me to ask a question of my own.
In the VCE course, we are told that brine, a 5 M solution of sodium chloride, will electrolyse to form sodium hydroxide and chlorine gas. Now, the Nernst equation in its standard form doesn't predict this, so I'm presuming the discrepancy arises from the 5 M concentration and the non-ideality of the solution?

[Mao]

But even then the equilibrium constants are constant; we just can't approximate with concentrations

You are very correct. I attempted two or three times to write something about counter-cases, but they are very nit-picky and not very useful, so I deleted them.

The point is though, equilibrium constants are taught and used because it is a simple tool that is surprisingly accurate in its predictions. When its use become so complex that it is no longer a simple tool, it becomes a rather irrelevant concept. E.g. we wouldn't calculate cell voltage in batteries via the equilibrium constant, because calculating/measuring activity coefficients in a MnO2 sludge is difficult. Instead, we'll simply build the battery and test its voltage.

Which leads me to ask a question of my own.
In the VCE course, we are told that brine, a 5 M solution of sodium chloride, will electrolyse to form sodium hydroxide and chlorine gas. Now, the Nernst equation in its standard form doesn't predict this, so I'm presuming the discrepancy arises from the 5 M concentration and the non-ideality of the solution?

Yes, but also importantly, the Nernst equation describes the equilibrium cell potential when the solution is static (i.e. reaction rate is zero). While the cell is not in equilibrium static (which, in this case it is not), the Nernst equation does not directly apply.

Note for the general readers: if you don't quite understand what I mean, don't worry. This isn't even 1st year university material. Only people who specialise in surface chemistry and diffusion processes in research eventually learn about methods to fix this problem. That makes this... at least Master's or PhD level stuff.

[lzxnl]

Yes, but also importantly, the Nernst equation describes the equilibrium cell potential. While the cell is not in equilibrium (which, in this case it is not), the Nernst equation does not directly apply.

I thought that by the Gibbs energy relation we'd have zero cell potential at equilibrium. Isn't the point of the Nernst equation to calculate non standard cell potentials?
I mean we use standard electrode potentials to calculate the cell potential at standard conditions and that generally isn't at equilibrium either.

[Mao]

I thought that by the Gibbs energy relation we'd have zero cell potential at equilibrium. Isn't the point of the Nernst equation to calculate non standard cell potentials?
I mean we use standard electrode potentials to calculate the cell potential at standard conditions and that generally isn't at equilibrium either.

Sorry. I really haven't been paying enough attention to the reading. You are very correct.

I did not mean equilibrium. I meant static. Nernst equation applies in a static, homogenous liquid. The Wikipedia page on Nernst equation mentions the limitation that when there is a net current through the electrode (a reaction is happening), the Nernst equation no longer applies. The problem is not so much the net current, but rather the diffusion-limited motion of ions, and so creating a concentration gradient of ions near the electrode. This can often cause problems when the current is high (e.g. industrial-scale systems).

In molecular modelling, I use the more complex equations, http://en.wikipedia.org/wiki/Nernst-Planck_equation , coupled with http://en.wikipedia.org/wiki/Poisson%E2%80%93Boltzmann_equation to describe the effects of ionic diffusion and changes to chemical/cell potential due to the concentration gradient. As you can see, vastly different.

I cannot confidently say whether or not a more sophisticated method like this can account for the large discrepancy between the Nernst relation and the chloralkali process, but I would imagine that large scale industrial processes would also have other important considerations which I don't know about. The answer is never so clear cut.

Let's hope I didn't state the completely wrong thing again this time. Sorry about that

[lzxnl]

Those equations. All I can do with them is probably substitute things into them let alone manipulate them.

That sort of makes sense...I remember my dad, who has a PhD in chemical engineering, telling me that solids don't reach equilibrium due to similar surface issues.

[zvezda]

On another note.... :p
So when we say that when secondary cells come to equilibrium, the cell is no longer able to be. recharged and used much like the primary cells, this strictly means equilbrium as i have come to know it yeah?
So how do redox reactions come to equilbrium anyway? Because i always thought that due to a strong reductant/oxidant, a reaction onlt occurs one way.
Cheers

[lzxnl]

Reducing and oxidising strength depends on concentration; the relative ordering of reduction and oxidation ability in the electrochemical series only holds for standard conditions.

[zvezda]

Ahh yeah fair enough.
Also, came across this question on a neap paper where there is an aluminium air cell. It says that an early problem with this cell was that the aluminium hydroxide that formed was a gel-like substance that formed on the anode. Apparently this would decrease the productivity of the reaction and the production of energy by the fuel cell as the reactants are prevented from coming into contact with the electrodes (namely hydroxide as the anode is made from aluminium). However, even though the oxidation reaction involves hydroxide ions, whats stopping the aluminium oxidising to aluminium ions instead??
Cheers

[zvezda]

Another thing,
When writing the equilibrium reaction of a particularly weak acid, is the initial molecule in a liquid or aqueous state?

[lzxnl]

Ahh yeah fair enough.
Also, came across this question on a neap paper where there is an aluminium air cell. It says that an early problem with this cell was that the aluminium hydroxide that formed was a gel-like substance that formed on the anode. Apparently this would decrease the productivity of the reaction and the production of energy by the fuel cell as the reactants are prevented from coming into contact with the electrodes (namely hydroxide as the anode is made from aluminium). However, even though the oxidation reaction involves hydroxide ions, whats stopping the aluminium oxidising to aluminium ions instead??
Cheers

If the oxygen can't make contact with the anode, the reaction can't take place as readily.

Another thing,
When writing the equilibrium reaction of a particularly weak acid, is the initial molecule in a liquid or aqueous state?

Depends on what the acid is. Generally, acids will be aqueous.

[zvezda]

If the oxygen can't make contact with the anode, the reaction can't take place as readily.

Depends on what the acid is. Generally, acids will be aqueous.

Why would the oxygen make contact with the anode? Its reducing?

So in terms of the vce, the acids will be aqueous?

[lzxnl]

Obviously I'm not thinking today.





The aluminium's effective surface area decreases if it is covered by a layer of aluminium hydroxide.

[zvezda]

Obviously I'm not thinking today.





The aluminium's effective surface area decreases if it is covered by a layer of aluminium hydroxide.

But then that brings me back to the question:
Whats stopping the aluminium from oxidising to aluminium ions? Because the anode is made out of aluminium, who cares if its covered in something else?

[lzxnl]

Oxidation only occurs if it can get rid of the electrons. If it's covered in aluminium hydroxide, the aluminium can't really oxidise.