Chemistry 3/4 2013 Thread

[barydos]

During discharge, electrons go towards the positive half-cell, which is the LiCoO2. The half equations given are when the cell is being recharged.

But I thought the electrode's "polarity" switches when switching from recharge to discharge (and vice versa)..?
As in, I thought during discharge, the positive electrode is the electrode at which reduction is occurring.

[lzxnl]

To quote my electrochemistry thread, which is in the stickied chemistry resources page in the chemistry forum:

I've seen many people ask questions about these things online, so I figured I may as well write something to answer the same question once and for all.

FAQs:

1. What does the polarity of a cell mean?
"Positive" and "negative" are misleading ways of thinking about these cells if you don't know what they actually represent. They're not actually charged; you have the salt bridge to make sure that they are not charged. So what are these things?
The "positive" electrode means the electrode at higher electric potential. NOTE: electric potential is not electric potential energy. Electric potential is potential energy per charge. This means that a positive charge will lose potential energy and thus speed up when moving from higher potential to lower potential, whereas a negative charge will gain potential energy and thus slow down when moving from higher potential to lower potential. This agrees with the common observation that negative charges move from "negative" to "positive", i.e. lower to higher electric potential.

2. Why the heck do the polarity designations change in galvanic and electrolytic cells? AKA why are cathodes positive in a galvanic cell but negative in an electrolytic cell?

There are a few things to bear in mind here. Firstly, if we have two electrodes, electrode A and electrode B, and we assume electrode A is at higher potential, it will be at higher potential REGARDLESS if the cell is electrolytic or galvanic. What changes is the direction of the current. So you don't say "the cathode has become positive". Rather, "the positive electrode is now the cathode". As an example, suppose you had a Cu²⁺/Cu half cell and a Fe²⁺/Fe half cell. If you leave it alone, the copper half cell will attract electrons and is thus the electrode with higher electric potential. However, if you turn this into an electrolytic cell, the copper half cell will STILL have higher electric potential; it is still "positive", but the external voltage overrides these considerations and force the electrons into the iron half cell which is still at lower potential.

Now how does this external voltage work? It's actually pretty neat. Let's take the previous example of the copper and iron half cells. By design, the positive terminal of the battery is connected to the cell with higher electric potential and the negative terminal of the battery is connected to the cell with lower electric potential. You can probably see that if we reversed these terminals, the battery would support the spontaneous reaction and the extra voltage would be lost as heat.

Let us assume that initially the positive terminal of the battery and the copper half cell are at the same potential, i.e. they are connected first. Then, as the voltage drop from the positive terminal to the negative terminal of the battery must be larger than the voltage drop from the positive to negative electrode, and as the positive terminal and electrode are at the same potential, the negative terminal of the battery must be at a lower potential than the negative electrode. Think about this until you get it.

If the negative terminal of the battery is at lower potential, electrons will attempt to move from lower potential to higher potential by moving from the battery to the negative electrode. Remember; electrons move from lower to higher potential because they have a negative charge. When electrons leave the battery, an electron deficit is created at the negative terminal. Electrons then flow from the positive terminal to meet this deficit, and electrons from the positive electrode then flow to the positive terminal to meet the electron gap created. Hence, we have a circuit running!

If the negative terminal of the battery is connected to the negative electrode first so that they are at the same potential, then as the voltage rise from the negative terminal to the positive terminal is higher than that from negative electrode to positive electrode, the positive terminal is at higher potential than the positive electrode. Electrons flow from the positive electrode to the positive terminal and by similar reasoning the electrons still flow from the negative terminal to the negative electrode.

Now, as the electrons leave the positive electrode, the positive electrode is the anode, and the negative electrode is the cathode. In a conventional galvanic cell, the electrons leave the negative anode because electrons normally go from lower potential to higher. Here, the external voltage has not changed the order of potentials, but it has changed the electron flow to cause the electrons to move from the positive electrode, the new anode, and the negative electrode, the new cathode.

So...remember that "positive" and "negative" designations do not change; the "anode" and "cathode" designations do, as the external voltage reverses the current flow, meaning the electrons come from different places.

I hope this made sense.

You're correct, the positive electrode is the electrode at which reduction occurs, but the electrode is the same electrode as the positive electrode which is the anode during recharge. The polarities do NOT switch; only the anode/cathode designations switch. Think about it. If you have copper metal at an electrode, and it becomes the anode to be oxidised to Cu(II) during recharge, it makes sense that it becomes the cathode during discharge, otherwise what does the Cu(II) do during discharge?

[barydos]

To quote my electrochemistry thread, which is in the stickied chemistry resources page in the chemistry forum:

You're correct, the positive electrode is the electrode at which reduction occurs, but the electrode is the same electrode as the positive electrode which is the anode during recharge. The polarities do NOT switch; only the anode/cathode designations switch. Think about it. If you have copper metal at an electrode, and it becomes the anode to be oxidised to Cu(II) during recharge, it makes sense that it becomes the cathode during discharge, otherwise what does the Cu(II) do during discharge?

Ahhh rookie mistake, my bad haha
Thanks a lot, makes sense!

[Scooby]

For hydrolysis of uncharged acids, should the acid be in a liquid or aqueous state?

What colour is an indicator in its pH range? Is a reaction mixture containing bromophenol blue green (mixture of yellow and blue) at pH 3.0-4.6?

If we've got two Ag⁺/Ag half cells, with the Ag⁺ solution in one half cell being 0.5 M and the solution in the other being 5.0 M, why is there a spontaneous reaction? How're we supposed to figure this out?

There was this question that asked why potassium chloride solution would be unsuitable for a salt bridge in a Cu²⁺/Cu II Ag⁺/Ag cell, and then the answers said that it was because chloride ions would precipitate with silver ions. But wouldn't these chloride ions migrate to the anode (in the copper half cell) and not precipitate with the silver ions?

In a button cell, why does hydroxide need to be included in the equation for oxidation of zinc?

Thanks

[lzxnl]

For hydrolysis of uncharged acids, should the acid be in a liquid or aqueous state?

What colour is an indicator in its pH range? Is a reaction mixture containing bromophenol blue green (mixture of yellow and blue) at pH 3.0-4.6?

If we've got two Ag⁺/Ag half cells, with the Ag⁺ solution in one half cell being 0.5 M and the solution in the other being 5.0 M, why is there a spontaneous reaction? How're we supposed to figure this out?

There was this question that asked why potassium chloride solution would be unsuitable for a salt bridge in a Cu²⁺/Cu II Ag⁺/Ag cell, and then the answers said that it was because chloride ions would precipitate with silver ions. But wouldn't these chloride ions migrate to the anode (in the copper half cell) and not precipitate with the silver ions?

In a button cell, why does hydroxide need to be included in the equation for oxidation of zinc?

Thanks

Depends on the acid. Hydrogen chloride would be gaseous, benzoic acid would be solid, ethanoic acid would be liquid.

In its pH range, the colour is a mixture of the acid and base forms; the exact colour will depend on the pH and the composition, but you don't need to worry about that.

If you have two Ag+ cells, but one is at a higher concentration than the other, you can note that when both cells have the same concentration, there will be no tendency to react with one another. That's sort of it. This is where the VCE course doesn't really address things well.
When we say "a reaction won't happen", we really mean "the equilibrium constant for this reaction at standard conditions is less than 1", meaning that the forward reaction is more favoured at standard conditions. That's it. If you have 1 M sodium chloride and you electrolyse it, sure, water will be preferentially oxidised, but that does not mean chloride ions will not be.
Also, oxidant and reductant strengths depend on their concentrations. If you looked at sulfuric acid, you may have been taught that concentrated sulfuric acid is a strong oxidant. Yes, CONCENTRATED sulfuric acid. Here, silver(I) in 5.0 M form is more concentrated than the 0.5 M one. The former is a stronger oxidant.

As for your potassium chloride solution question:
Yes, the silver cell forms the cathode and potassium ions are supposed to be attracted there. However, initially, before the cell operates, your salt bridge is uniformly spread out. There, any chloride ions that happen to be attracted by the silver solution...you get the point.

Hydroxide needs to be included in the oxidation of zinc because the electrode potential is affected by the hydroxide ions. It's just like how oxygen becomes a stronger oxidant in concentrated acid than in water.

[Scooby]

Thanks nliu1995!

[Scooby]

How do we deduce half equations in fuel cells? How are we expected to know that methanol is oxidised to carbon dioxide in a methanol-oxygen fuel cell?

I got a question that asked to write the equation for the reaction occurring at an impure copper electrode in an electrolytic cell. The electrode contained copper, nickel and gold. I wrote that nickel (the strongest reductant) would be oxidised, but the answers had copper being oxidised. Am I wrong here?

Thanks

[brightsky]

1. well given that most reaction that occur in fuel cells are combustion reactions...you probably would be expected to deduce that methanol is oxidised to carbon dioxide. (you only really need to write out the equation for the combustion reaction of methanol and then work out which species is oxidised and which reduced, etc.)

2. i'm assuming this question is about electroplating? if so, yes nickel would be oxidised, but it will never nickel solid would never be deposited at the cathode. not sure if that answers your question though...

[Scooby]

1. well given that most reaction that occur in fuel cells are combustion reactions...you probably would be expected to deduce that methanol is oxidised to carbon dioxide. (you only really need to write out the equation for the combustion reaction of methanol and then work out which species is oxidised and which reduced, etc.)

2. i'm assuming this question is about electroplating? if so, yes nickel would be oxidised, but it will never nickel solid would never be deposited at the cathode. not sure if that answers your question though...

Yeah, it just asked for the reaction occurring at the anode, and I just assumed that since nickel was the strongest reductant in that impure sample it would be oxidised preferentially

Thanks

[Scooby]

Alright, so we've got an NaOH solution of pH 10 (prior to dilution) which is diluted from 1 mL to 100 mL with deionised water, and we need to find the pH after the dilution

Sooooo I found the concentration of hydronium in the original solution using [H₃O⁺] = 10^-pH, and then I used the dilution factor (10^-pH x1/100) to find the concentration of hydronium in the diluted solution... and then the pH I ended up getting was 12, which of course isn't right at all... since we've got a basic solution the pH should decrease after dilution

Then instead of doing it that way I found the concentration of hydronium in the original solution, used the Kw of water to find the concentration of hydroxide in that solution, found the concentration of hydroxide in the diluted solution using the dilution factor (1/100), found the concentration of hydronium using the Kw of water... then used the pH formula to find the pH, and ended up with the right answer

Anyway, just wanted to know where exactly I stuffed up using the first method... I don't really see why it didn't work 

Thanks

[Alwin]

Haven't been on in a while, so just a few things from me:

This was in the VCAA 2010 Exam 1 where you had to write an equation for the ionisation of ethanol in mass spec:

Spoiler

I'm just wondering what the importance of the square brackets is, would we lose marks if we did not include the square brackets?

Personally I believe this is because the 'missing' electron could be any electron in the original molecule. Thus, the [ ] brackets indicate the molecular has an overall charge of + irrespective of which atom the electron was 'knocked' from.
@nliu It's like how people put [ ] around complex ions, but not always I guess. eg [Cu(H₂O)₆]²⁺

In comparison, CH3COO^- the minus (-) sign represents the O has a negative charge having lost a proton (hydrogen). More commonly, on vcaa exams they have the sodium salt of salicylic acid:
C₆H₄(OH)COO^- Na⁺ indicating the (-) is on the Oxygen and (+) on the Sodium

But that's just being pedantic. I don't think it actually matters

How do we deduce half equations in fuel cells? How are we expected to know that methanol is oxidised to carbon dioxide in a methanol-oxygen fuel cell?

I got a question that asked to write the equation for the reaction occurring at an impure copper electrode in an electrolytic cell. The electrode contained copper, nickel and gold. I wrote that nickel (the strongest reductant) would be oxidised, but the answers had copper being oxidised. Am I wrong here?

Thanks

Yes. Remember balance Carbon with CO₂, balance Oxygen with H₂O etc etc like normal half equation rules

I think if this is for electro-refining perhaps? if so you should use impure copper electrode (anode) , pure copper electrode (cathode) and electrolytic usually acidified metal to be collected.
btw, from wikipedia:

Some metals, such as nickel do not electrolyze out but remain in the electrolyte solution. These are then reduced by chemical reactions to refine the metal. Other metals, which during the processing of the target metal have been reduced but not deposited at the cathode, sink to the bottom of the electrolytic cell, where they form a substance referred to as anode sludge or anode slime. The metals in this sludge can be removed by standard pyrorefining methods.
http://en.wikipedia.org/wiki/Electrowinning

[thushan]

Alright, so we've got an NaOH solution of pH 10 (prior to dilution) which is diluted from 1 mL to 100 mL with deionised water, and we need to find the pH after the dilution

Sooooo I found the concentration of hydronium in the original solution using [H₃O⁺] = 10^-pH, and then I used the dilution factor (10^-pH x1/100) to find the concentration of hydronium in the diluted solution... and then the pH I ended up getting was 12, which of course isn't right at all... since we've got a basic solution the pH should decrease after dilution

Then instead of doing it that way I found the concentration of hydronium in the original solution, used the Kw of water to find the concentration of hydroxide in that solution, found the concentration of hydroxide in the diluted solution using the dilution factor (1/100), found the concentration of hydronium using the Kw of water... then used the pH formula to find the pH, and ended up with the right answer

Anyway, just wanted to know where exactly I stuffed up using the first method... I don't really see why it didn't work 

Thanks

You forget that the system re-equilibriates. There is a reversible reaction going on:

2H2O <--> H3O+ + OH-

When the solution is pH = 10, [OH-] = 10^-4 M and [H+] = 10^-10 M, so there is a lot more OH- than H+ originally.

When you dilute the solution, [OH-] initially decreases 100fold to 10^-6 M, and so does [H+] to 10^-12 M.

However, the system has to reach equlibrium, with K = 10^-14 and the reaction quotient at that time Q = 10^-18 M.

Hence, the system will shift to the right (also consistent with Le Chatelier's principle, given we have a dilution and the system favours the reaction with more aqueous particles).

However, since there is 1000000 times more OH- than there is H+, the amount of OH- formed will be negligible compared to its initial diluted concentration 10^-6 M.

Hence, we can say that [H+] would increase from 10^-12 to about 10^-8 M. This is the same as the amount of OH- formed, which would be much smaller than 10^-6 M. So,

Final:
[OH-] = 10^-6 M
[H+] = 10^-8 M

[lzxnl]

Or, in a simpler way:
When you dilute the NaOH solution from 1 mL to 100 mL, you are essentially adding a solution of 10^-7 M H+ and OH-. You initially had [OH-]=10^-2. The addition of the hydroxide to this amount is tiny. But you only had 10^-10 M H+ initially; 10^-7 M is a HUGE concentration. Therefore you can't think of it as a dilution of the H+.

[Jaswinder]

http://www.vcaa.vic.edu.au/Documents/exams/chemistry/2010chem1-w.pdf

From Section A:

Question 12 - Why can't it be A? and could you also explain why D is the correct answer. I tried reading the assessment report and I kinda get it but not too sure.

All from Section B:

Question 2e - if I drew the structure as shown in the attachment, would that still be right?

Question 3a - I wrote NO2 -> NO2+ + e- Why is the answer 2e-?

Question 3b- I said, Spectrum A. Nitrogen dioxide cannot form a peak at 31 m/z. Would this yeild full marks?

Question 7bii - Why is the answer C22H32O2? I get C22H34O2, which I know is incorrect but I don't know why.

Also what exactly is fermentation? Which chemicals undergo fermentation? What are the products of this process?
Thanks 

[ECheong]

Hi! I'll try my best to help out.

A12:
A is not possible because the compound needs to react with both NaOH and HCl, the OH- ions from NaOH will react in a substitution reaction however, the HCl will not. (The mechanism as to why I'm not 100% sure of so I won't post, hopefully someone can help out) As for why D is the correct answer. D contains both a COOH(Proton Donator -Acidic) and NH2(Proton Acceptor - Basic) functional group. When placed in HCl, the NH2 will accept a proton, when placed in NaOH the COOH will donate a proton. i.e. the compound is amphiprotic.

B
2e:
I'm probably derping but I don't see an attachment

3a:
This is because you need to take into account the electron fired at the NO2. So, you end up with the electron you knocked off the NO2 and the one that was initially fired at it.

3b:
NO2 can in fact form peaks at 31m/z due to (although very rare) isotopes of Nitrogen, which admittedly probably isn't what they're looking for. Having said that though, in an exam, my mantra is better safe than sorry. I would explain explicitly that it's not possible for NO2 to break up into so many types of fragments; there are only a small number of combinations 3 atoms can make.

7bii:

From the previous question we know there are 6 double bonds.

My process of thinking when it comes to these types of questions is this:
1. I take out the COOH group first, so now I know I have a 21 Carbon long chain with a COOH attached at the end
2. In a Saturated fatty acid, each C would have 2 hydrogens attached to it except the one on the far end, which has 3 and the Carbon which is part of COOH.
3. Therefore, in this specific instance, disregarding the COOH, I'd have 21*2 hydrogens + 1 on the end which gives me 43 hydrogens. Each double bond removes 2 hydrogens so now we have 43 - 6*2 =31 hydrogens on the fatty chain.
4. Finally I combine this and end up with C₂₁H₃₁COOH or C₂₂H₃₂O₂

Finally, fermentation, at its core, is the conversion of sugars into acids, gases, or alcohol. (By enzymes or organisms)
With respect to VCE Chem, this is usually to do with the conversion of Glucose (C6H12O6), into CO2 and CH3CH2OH.

Hope this was helpful