Stankovic123's chem q's

[lzxnl]

Q19 on multi choice 2009 exam 2. Theres a solution of KI CaBr2 and both are at 1.0M. Now theres a electrolysis being carried out with inert electrodes. Apparently, iodide ions oxidise to iodine?? Doing this questions, i was thinking about how Cl- ions oxidise at high concentrations in preference to water, but this wouldnt occur with iodide would it because it is a much stronger reductant than water??
Help appreciated

Hmm. We have 1.0 M K+, Ca 2+ and I- but 2.0M Br-
The electrochemical series gives iodide as a vastly stronger reductant than any of the other ions. Looks at the series. The reduction of oxygen gas in acid to form water is at 1.23 V. The reduction of solid iodine to iodide is 0.54V. That's a huge difference.

And well, the fact that we have 2.0 M bromide isn't enough to make up for the difference.

So I don't see what your problem is...iodide would perfectly naturally oxidise to form solid iodine in this situation.

[zvezda]

Hmm. We have 1.0 M K+, Ca 2+ and I- but 2.0M Br-
The electrochemical series gives iodide as a vastly stronger reductant than any of the other ions. Looks at the series. The reduction of oxygen gas in acid to form water is at 1.23 V. The reduction of solid iodine to iodide is 0.54V. That's a huge difference.

And well, the fact that we have 2.0 M bromide isn't enough to make up for the difference.

So I don't see what your problem is...iodide would perfectly naturally oxidise to form solid iodine in this situation.

Neither do I tbh. I understand this now. I think there was just a series of seriously unfortunate events that lead me to some other option.
Apologies for wasting your time

[zvezda]

Hey,
A question on determining changes in pH.
Say there was a solution of KOH (0.100M and 10.0mL). Now the solution is diluted to 1.00L. So obviously, [OH-] decreases significantly so the pH will decrease. But technically, isnt the [H3O+] also decreasing. If we consider the equilibrium of waters self ionisation, the concentration of both ions will initially decrease and then partially increase yeah?
So why is the normal calculation for this type of question valid, wherby we use
[OH-]x[H3O+]=10^-14??

[SocialRhubarb]

Ahhh yeah thats what i was missing; the fact that the water used to dilute the solution has [H3O+] = 10^-7.
Cheers guys

[zvezda]

LOL!
My word. Cheers

[zvezda]

Say for example that there is an electrolytic cell and we have 4M NaCl. When voltage from an external power source is applied to the cell, and when the oxidation/reduction reactions proceed, does water oxidise at the same time as chloride ions or is it only the chloride ions that oxidise??
Thanks in advance

[lzxnl]

Say for example that there is an electrolytic cell and we have 4M NaCl. When voltage from an external power source is applied to the cell, and when the oxidation/reduction reactions proceed, does water oxidise at the same time as chloride ions or is it only the chloride ions that oxidise??
Thanks in advance

If you JUST have 4 M NaCl with no chlorine gas, initially you should just get chloride ions oxidising as the standard reduction potential of chlorine gas to chloride ion only applies when the chlorine gas is at 1 atm and the ions at 1 M. Here, with no chlorine gas, the chloride ions are much more readily oxidised. The standard reduction potentials are really a measurement of the equilibrium constant of the reactions.

However, soon water will begin to be oxidised as well; both processes will occur together to some extent as more chlorine gas is formed. If you have an open system where the chlorine and oxygen gases are allowed to escape...I don't know what will happen then...

[zvezda]

If you JUST have 4 M NaCl with no chlorine gas, initially you should just get chloride ions oxidising as the standard reduction potential of chlorine gas to chloride ion only applies when the chlorine gas is at 1 atm and the ions at 1 M. Here, with no chlorine gas, the chloride ions are much more readily oxidised. The standard reduction potentials are really a measurement of the equilibrium constant of the reactions.

However, soon water will begin to be oxidised as well; both processes will occur together to some extent as more chlorine gas is formed. If you have an open system where the chlorine and oxygen gases are allowed to escape...I don't know what will happen then...

What do you mean by you dont know what will happen? Wouldnt both reactions continue to proceed?

[lzxnl]

Both reactions will continue to proceed, but I'm not sure to what extent as equilibrium isn't reached.

[zvezda]

It would depend on whether the reactants escape the cell or not wouldnt it?

Also, how do we know whether 3-ethylhexane or hexane has a higher boiling point? Hexane has a smaller molar mass, but 3-ethylhexane is pretty branched as well.
Without data, how do we determine the strengths of their dispersion forces?
Cheers

[zvezda]

Are the values given on a calibration curve included in the determination of sig figs in the final answer?

[lzxnl]

It would depend on whether the reactants escape the cell or not wouldnt it?

Also, how do we know whether 3-ethylhexane or hexane has a higher boiling point? Hexane has a smaller molar mass, but 3-ethylhexane is pretty branched as well.
Without data, how do we determine the strengths of their dispersion forces?
Cheers

Just assume hexane has weaker dispersion forces; I'm sure the larger numbers of electrons compensate for the branching.

Are the values given on a calibration curve included in the determination of sig figs in the final answer?

Generally yes.

[zvezda]

Just assume hexane has weaker dispersion forces; I'm sure the larger numbers of electrons compensate for the branching.

Generally yes.

Ok fair enough. Cheers nliu.
Also two things:
1 - whats the chemical basis behind the connection of two blocks of metal (say copper and nickel) and putting them in an acidic solution where only hydrogen forms on one metal (copper in this example) whilst if the 2 blocks were separate, the other metal block would oxidise (nickel)? Ive come across this sort of situation in a company exam and i dont feel as if the solutions properly explained this.

2 -  there was a question in a company paper where 2 compounds are combusted (Hydogen gas and i think methanol). The q asked to compare the efficiency of them in terms of energy released per gram. Shouldnt they have asked for per mol? Doesnt energy released per mol make more sense as a measure of efficiency?? 

[lzxnl]

Ok fair enough. Cheers nliu.
Also two things:
1 - whats the chemical basis behind the connection of two blocks of metal (say copper and nickel) and putting them in an acidic solution where only hydrogen forms on one metal (copper in this example) whilst if the 2 blocks were separate, the other metal block would oxidise (nickel)? Ive come across this sort of situation in a company exam and i dont feel as if the solutions properly explained this.

2 -  there was a question in a company paper where 2 compounds are combusted (Hydogen gas and i think methanol). The q asked to compare the efficiency of them in terms of energy released per gram. Shouldnt they have asked for per mol? Doesnt energy released per mol make more sense as a measure of efficiency??

This question was answered before somewhere but I'll answer it again.
If you connect the two blocks of metal, they become electrically connected too. Put them in acidic solution and the nickel will react with the acid hydrogens. However, when the nickel reacts, it forms nickel cations, which repel the hydrogen ions. As the two blocks of metal are electrically connected, the hydrogen ions can grab electrons from any part of the metal surface. In this case, the positive hydrogen ions go towards the neutral copper metal and grab electrons from the copper, which are really provided by the nickel. The net result? Nickel is still oxidised, but the oxidation takes place at the copper surface.

If the blocks were separated, nickel is still reactive enough to react with protons, but copper isn't, so gas only evolves at the nickel surface; there is no electrical connection anywhere this time.



As for your second question, think about a rocket; every gram of fuel carried costs money. Also, energy released per mole makes LESS sense because some molecules are larger. Are you telling me that glucose is a more efficient fuel source simply because its molar heat of combustion is of higher magnitude than that of hydrogen? To adequately compare different fuel sources, energy per gram is used as now we're really measuring the energy output per atom (the atomic masses don't change noticeably). Isn't that what fuel efficiency is?

[zvezda]

So what specifically prevent the nickel from directly reacting with the hydrogen ions? What specifically about copper makes it attract electrons from the nickel (the nuclei of the copper atoms amongst the delocalised electrons??)?

Apologies, but i wasnt able to follow that very last bit about energy output per atom.