Stankovic123's chem q's

[lzxnl]

I just cant see how theyre strictly connected though. So youre saying that electrons flow through the solutions of each cell as well??

Also, according to TSSM, fuel cells can be rechargeable, like the vanadium redox cell apparently?

Cheers nliu

There is probably some electrical connection between the cells that allows charge to flow.

Some fuel cells can be recharged; you could make a hydrogen fuel cell in which the recharging is the electrolysis of water. It's not impossible, but generally they're not designed to be recharged.

[zvezda]

There is probably some electrical connection between the cells that allows charge to flow.

Some fuel cells can be recharged; you could make a hydrogen fuel cell in which the recharging is the electrolysis of water. It's not impossible, but generally they're not designed to be recharged.

Ah ok, fair enough.
Also, the same TSSM exam has stated that given an unknown monoprotic acid reacting with sodium carbonate, a sharp endpoint produced when adding methyl red indicator suggests that the monoprotic acid is strong?
Now i thought sodium carbonate was a pretty strong base, so couldnt the acid be weak? Strong bases reacting with weak acids produce pretty sharp endpoints dont they?
Thanks

[lzxnl]

Ah ok, fair enough.
Also, the same TSSM exam has stated that given an unknown monoprotic acid reacting with sodium carbonate, a sharp endpoint produced when adding methyl red indicator suggests that the monoprotic acid is strong?
Now i thought sodium carbonate was a pretty strong base, so couldnt the acid be weak? Strong bases reacting with weak acids produce pretty sharp endpoints dont they?
Thanks

Sodium carbonate is a weak base, but it's not too weak of a base.
Methyl red indicator changes colour at around 4.2-6.3 pH. Changing colour at that pH implies we have a strong acid against a relatively strong base, which is what we have here. The conjugate acid of the carbonate ion, HCO3 -, has an acidity constant on the order of 10^-10, so CO3 - is a moderately strong base.


If you had a strong base against a weak acid titration, at the equivalence point, you'd only have a weak base left, so the solution would be basic. It's not just the sharpness of the endpoint; it's the pH of the endpoint that matters too.

[zvezda]

Sodium carbonate is a weak base, but it's not too weak of a base.
Methyl red indicator changes colour at around 4.2-6.3 pH. Changing colour at that pH implies we have a strong acid against a relatively strong base, which is what we have here. The conjugate acid of the carbonate ion, HCO3 -, has an acidity constant on the order of 10^-10, so CO3 - is a moderately strong base.


If you had a strong base against a weak acid titration, at the equivalence point, you'd only have a weak base left, so the solution would be basic. It's not just the sharpness of the endpoint; it's the pH of the endpoint that matters too.

Ok fair enough.
3 things now hahah:
1 - when writing the equation out for the electroylsis of molten potassium hydroxide for example, would the potassium atom be a solid or a liquid?
2 - a situation in a tssm paper outlined that a potassium hydroxide electrolysis cell (molten) no longer operates. Basically the potassium now reacts with the water that was produced from the electrolysis. Now would the potassium ion that is being reproduced be in an aqueous or liquid form?
3 - is it true that for any strong acid/base, its conjugate is weak? Similarly, for any weak acid/base, is its conjugate strong?

Cheers nliu

[lzxnl]

Ok fair enough.
3 things now hahah:
1 - when writing the equation out for the electroylsis of molten potassium hydroxide for example, would the potassium atom be a solid or a liquid?
2 - a situation in a tssm paper outlined that a potassium hydroxide electrolysis cell (molten) no longer operates. Basically the potassium now reacts with the water that was produced from the electrolysis. Now would the potassium ion that is being reproduced be in an aqueous or liquid form?
3 - is it true that for any strong acid/base, its conjugate is weak? Similarly, for any weak acid/base, is its conjugate strong?

Cheers nliu

Metallic bonding is weaker than ionic generally, so you can expect the resultant potassium atoms to be liquid as well.

I think your TSSM question is really dodgy. Liquid potassium hydroxide requires very high temperatures to form, and I highly doubt water can be present in liquid form. I'd have to see the question, but it seems very dodgy.

As for your first acid base question, that is correct. Think about it; a strong acid is strong because the forward dissociation process goes to completion; it implies the backwards process, the protonation of the base, does not occur to any noticeable extent. However, the conjugate of a weak base could be a weak acid. For example, methanoic acid is a weak acid, and methanoate ion is also a weak base. Only the conjugate of a VERY weak acid or base is strong, like chloride ion or hydrogen sulfate ion.

[zvezda]

Metallic bonding is weaker than ionic generally, so you can expect the resultant potassium atoms to be liquid as well.

I think your TSSM question is really dodgy. Liquid potassium hydroxide requires very high temperatures to form, and I highly doubt water can be present in liquid form. I'd have to see the question, but it seems very dodgy.

As for your first acid base question, that is correct. Think about it; a strong acid is strong because the forward dissociation process goes to completion; it implies the backwards process, the protonation of the base, does not occur to any noticeable extent. However, the conjugate of a weak base could be a weak acid. For example, methanoic acid is a weak acid, and methanoate ion is also a weak base. Only the conjugate of a VERY weak acid or base is strong, like chloride ion or hydrogen sulfate ion.

Yeah some of these company papers are really irritating.
I see now with the acids/bases.
Cheers nliu. Always giving great help

[zvezda]

With GLC,
Why wouldnt the mobile phase be detected at the end of the column, thereby contributing to the peak area at a certain retention time? Or does it but is taken into account?
Thanks

[zvezda]

Also,
People have been saying that the potential differences of two half cells provide a sort of equilibrium constant. Does this really mean that the higher the potential difference, the more products that are formed??
Cheers

[lzxnl]

With GLC,
Why wouldnt the mobile phase be detected at the end of the column, thereby contributing to the peak area at a certain retention time? Or does it but is taken into account?
Thanks

I think they take the mobile phase into account. They would have to.

Also,
People have been saying that the potential differences of two half cells provide a sort of equilibrium constant. Does this really mean that the higher the potential difference, the more products that are formed??
Cheers

Not quite. The standard electrode potential difference, the one given by the electrochemical series, is directly proportional to the natural log of the equilibrium constant. However the cell potential is different. The actual cell potential depends on the reaction quotient as well. Look up Nernst equation for more details.
larger standard electrode potential means higher equilibrium constant, but you'll find that most reactions have high K values

[zvezda]

I think they take the mobile phase into account. They would have to.
Not quite. The standard electrode potential difference, the one given by the electrochemical series, is directly proportional to the natural log of the equilibrium constant. However the cell potential is different. The actual cell potential depends on the reaction quotient as well. Look up Nernst equation for more details.
larger standard electrode potential means higher equilibrium constant, but you'll find that most reactions have high K values

Ok fair enough. So does that mean that redox reactions are effectively equilibrium reactions?

Also, why must electrodes in a fuel cell be catalysts? Arent the reactions already spontaneous?

Thanks nliu

[brightsky]

Ok fair enough. So does that mean that redox reactions are effectively equilibrium reactions?

Also, why must electrodes in a fuel cell be catalysts? Arent the reactions already spontaneous?

Thanks nliu

1. yeah redox reactions are equilibrium reactions. it is when reaction reaches equilibrium that the cells needs to be either recharged or disposed of, not necessarily when the reaction reaches completion. if you're using a reactive anode, not the whole electrode needs to be used up for the cell to stop working.

2. the electrodes are catalytically impregnated because (you guessed it) the reaction is way too slow. normally for a fuel cell, you have a combustion reaction occurring. now we all know that combustion reactions usually needs some sort of spark to get started. but in a fuel cell, you don't have that. so you need huge electrodes and catalytic impregnation to ensure decent reaction rate.

[Mao]

With GLC,
Why wouldnt the mobile phase be detected at the end of the column, thereby contributing to the peak area at a certain retention time? Or does it but is taken into account?
Thanks

I think they take the mobile phase into account. They would have to.

It depends on the detection method.

We can tack on a number of detection instruments to the end of a chromatograph. Mass spec, UV-Vis, or other detection methods (a common one is the Flame Ioniser, which we don't touch on in VCE) can be attached to the output. Whether or not there is a peak for the mobile phase depends on whether or not the chemical is detectable by that particular detection method.

e.g. In HPLC, if we were to use UV-Vis to detect large organic compounds (e.g. caffeine or food dye), our mobile phase (e.g. ethanol) will be transparent to the UV-Vis detector, and so won't show up.

In GC, if we were to use N2 as our mobile phase, but we use a flame ioniser to detect the product (i.e. tries to combust the eluent), N2 will be transparent to the detector.

In GC-MS, by setting the ionisation energy lower (how harshly we ionise the analyte molecules), we can make N2 transparent to the detector too.

So, the mobile phase does pass through, but we generally choose detection methods that are transparent to the mobile phase, and so avoid inteference.

[Mao]

What's the chemical basis for the connection of two electrolytic cells? Doubt and explanation of one will be needed for the exam, but i was just curious.
Thanks

What i meant was that i doubt that we will be asked to explain the chemistry behind the 2 connected electroyltic cells, but i was just curious as to how they work anyway.

For example, let's consider a cell like this:

Code: [Select]

  ______(-->)______
  |    _______    |
  |   |       |   |
| A   | |   | B   | |
|       |   |       |
|  X2+  |   |   Y+  |
|_______|   |_______|
   (1)         (2)
With a small section of the electrochemical series arranged as such:

Code: [Select]

A2+(aq) + 2e-  <--> A(s)
X2+(aq) + 2e-  <--> X(s)
B2+(aq) + 2e-  <--> B(s)
Y+(aq)  + 2e-  <--> Y(s)

What happens?

1. at the cathode of (2), Y+(aq) + e- --> Y(s).
2. B and X2+ will react spontaneously anyways, and so that reaction happens without much effort. (B spontaneously gives electrons to X2+, via the wire that connects them) This reaction is limited or accelerated by the rate of Y+(aq) --> Y(s), as the cell must maintain neutral.
3. Due to the voltage difference applied betweent the anode of (1) and cathode of (2), we cause the non-spontaneous reaction A(s) --> A2+(aq) + 2e-

Here, the key is that B(s) and X2+(aq) will react spontaneously. What if they didn't?

Code: [Select]

A2+(aq) + 2e-  <--> A(s)
B2+(aq) + 2e-  <--> B(s)
X2+(aq) + 2e-  <--> X(s)
Y+(aq)  + 2e-  <--> Y(s)
Here, the middle reaction is no longer spontaneous. What happens?

1. highly energetic electrons arrive at the cathode of (2), causing the non-spontaneous reaction Y+(aq) + e- --> Y(s).
2. Cell (2) becomes more and more negatively charged
3. Due to the voltage difference applied betweent the anode of (1) and cathode of (2), we cause the non-spontaneous reaction A(s) --> A2+(aq) + 2e-
4. Cell (1) becomes more and more positively charged
5. As the cell becomes charged, we leave standard conditions, and at some point, the cells are charged enough such that the B and X reactions swap in their order in the electrochemical series (think: electrolysis of brine, we already know the relative ordering can swap based on concentration and other factos).
6. B(s) --> B2+(aq) and X2+(aq) --> X(s)

In both cases though (and especially the second case), the resistance of the cells are huge. Both are complicated multi-step processes that do not have very good charge-conduction properties, especially because of the reaction between B and X, so we cannot pass too high currents or achieve very high reaction rates. To give you an idea, the resistance is in the range of or greater. That is, in the voltage range of these electrolytic cells (several volts), we can only achieve several nanoamps of current. We can of course pump much higher voltages, but then we'll start seeing an entirely different class of reactions.

[Mao]

Also,
People have been saying that the potential differences of two half cells provide a sort of equilibrium constant. Does this really mean that the higher the potential difference, the more products that are formed??
Cheers

Not quite. The standard electrode potential difference, the one given by the electrochemical series, is directly proportional to the natural log of the equilibrium constant. However the cell potential is different. The actual cell potential depends on the reaction quotient as well. Look up Nernst equation for more details.
larger standard electrode potential means higher equilibrium constant, but you'll find that most reactions have high K values

One quick addition, the equilibrium constant, and indeed the cell potential, depends on the concentration of the species. This is why we can make batteries of the same voltage from different materials. E.g. Alkaline and Lithium AA batteries have the same marked voltage, 1.5V, but have different reactions, and so different standard electrode potentials.

[lzxnl]

One quick addition, the equilibrium constant, and indeed the cell potential, depends on the concentration of the species. This is why we can make batteries of the same voltage from different materials. E.g. Alkaline and Lithium AA batteries have the same marked voltage, 1.5V, but have different reactions, and so different standard electrode potentials.

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