VCE Chemistry Question Thread

[sweetcheeks]

Hi guys,
The question is about an electrolytic cell with electrodes of iron and magnesium, in a solution of magnesium sulfate. The electricity is flowing towards iron electrode.
My question is:
Because there is no spontaneous reaction occurring in the system anyway, would the electrolytic cell still work should the electrons be pumped towards the magnesium electrode?

However, one thing I realise is that Fe2+ ions will be free floating in the solution and possibly (would it?) be reduced to Fe(s) on magnesium electrode instead of magnesium. This, if true, I believe would mean that there would be deposits of both Fe and Mg on Mg electrode. Can someone clarify these two questions?

Thanks in advance

Reversing the polarity (given sufficient voltage is supplied), you will still get a reaction. In this case the iron electrode will be the anode and the magnesium electrode is the cathode. The reaction at the anode would be Fe --> Fe2+ + 2e-. The reaction at the cathode will be a little more difficult. Initially you will have a solution of Mg 2+, H2O and SO42-. The strongest oxidant of the three will be the water, where the reaction occurring is 2H2O + 2e- --> 2OH- + H2. After a while, when the concentration of Fe2+ builds up, it will also be reduced back to iron metal. The magnesium electrode will be coated in iron metal (electroplating)

Magnesium metal cannot be formed via electrolysis of aqueous solutions, as magnesium (ii) ions are weaker oxidants than water.

[yplee0926]

Reversing the polarity (given sufficient voltage is supplied), you will still get a reaction. In this case the iron electrode will be the anode and the magnesium electrode is the cathode. The reaction at the anode would be Fe --> Fe2+ + 2e-. The reaction at the cathode will be a little more difficult. Initially you will have a solution of Mg 2+, H2O and SO42-. The strongest oxidant of the three will be the water, where the reaction occurring is 2H2O + 2e- --> 2OH- + H2. After a while, when the concentration of Fe2+ builds up, it will also be reduced back to iron metal. The magnesium electrode will be coated in iron metal (electroplating)

Magnesium metal cannot be formed via electrolysis of aqueous solutions, as magnesium (ii) ions are weaker oxidants than water.

thank you so much!

[Cassidyhogi]

Hi
Do we need to memorise ions like ammonium ions are NH 4+ and sulfate ions are SO4 2- for UNIT 3/4 SACs and exams? Is the exam papers I have seen, they always give us the chemical formula of a compound they want us to use in the question.

[dream chaser]

Hi Guys,

Need help with this question. I don't understand why you have to double the volume. Could someone please explain me why. The question is in the attachment of this post.

Also, could someone please explain to me what molar volume is. I don't understand by what they say in the book that it is 'the amount of space, or volume, occupied by 1 mole of any gas at a particular pressure and temperature'.

All help will be much appreciated. Thanks.

[peachxmh]

Hey guys! Just wanted to clarify my understanding of an electrolyte - so it's basically a chemical compound that can conduct electricity because of its dissociation into ions, when dissolved or melted into a solution. I don't really understand how it can conduct electricity as a result of the dissociation - is it because the ions are charged particles and the movement of charged particles creates an electrical current? But I thought the electrons lost at the anode that flow through the external circuit of a galvanic cell were responsible for the generation of electricity in a galvanic cell, and the ions in the electrolyte simply neutralised the build up of charges. Are both the ions and the electrons generating electricity and how? I'm really confused, please help!

[sweetcheeks]

Hi Guys,

Need help with this question. I don't understand why you have to double the volume. Could someone please explain me why. The question is in the attachment of this post.

Also, could someone please explain to me what molar volume is. I don't understand by what they say in the book that it is 'the amount of space, or volume, occupied by 1 mole of any gas at a particular pressure and temperature'.

All help will be much appreciated. Thanks.

Avogadro's Law (not Avogadro's number) states that "equal volumes of all gases, at the same temperature and pressure, have the same number of molecules." What this means is that if you have 50 mL of hydrogen at standard conditions, it will have the same number of molecules as 50 mL of nitrogen under the same conditions. What this allows us to do, is rather than having to work with moles, we can simply do stoichiometry using volumes.

In the case of your question, the ratio between oxygen to methane is 2:1. This means that for every mole of methane you want to fully combust, you need 2 moles of oxygen. Using Avogadro's law, 100 mL of oxygen has twice the number of molecules as 50 mL of methane.

Molar volume is based upon this principle.  The volume occupied by a gas does not depend on the nature of the gas. Under standard conditions (273 K, 1 atm), the molar volume of a gas is 22.4 L. What this means is that any gas you have 1 mole of, under these standard conditions, will occupy a volume of 22.4 L.

*The assumption made here is that the gases being looked at are 'ideal'.

[TheBigC]

Avogadro's Law (not Avogadro's number) states that "equal volumes of all gases, at the same temperature and pressure, have the same number of molecules." What this means is that if you have 50 mL of hydrogen at standard conditions, it will have the same number of molecules as 50 mL of nitrogen under the same conditions. What this allows us to do, is rather than having to work with moles, we can simply do stoichiometry using volumes.

In the case of your question, the ratio between oxygen to methane is 2:1. This means that for every mole of methane you want to fully combust, you need 2 moles of oxygen. Using Avogadro's law, 100 mL of oxygen has twice the number of molecules as 50 mL of methane.

Molar volume is based upon this principle.  The volume occupied by a gas does not depend on the nature of the gas. Under standard conditions (273 K, 1 atm), the molar volume of a gas is 22.4 L. What this means is that any gas you have 1 mole of, under these standard conditions, will occupy a volume of 22.4 L.

*The assumption made here is that the gases being looked at are 'ideal'.

Well explained, however, I should just clarify that VCE uses 24.8 mol per L at SLC (298.15K and 100kPa).

Also, for a proof of:

"In the case of your question, the ratio between oxygen to methane is 2:1. This means that for every mole of methane you want to fully combust, you need 2 moles of oxygen. Using Avogadro's law, 100 mL of oxygen has twice the number of molecules as 50 mL of methane."

Consider the following:


Since V_{m} is a constant at SLC, then:

Thus, the mole ratio is equal to the volume ratio.

EDIT: Accidentally wrote that SLC was at 1 atm (which is incorrect), it is 100kPa (sorry).

[dream chaser]

Well explained, however, I should just clarify that VCE uses 24.8 mol per L at SLC (298.15K and 101.3kPa).

Also, for a proof of:

"In the case of your question, the ratio between oxygen to methane is 2:1. This means that for every mole of methane you want to fully combust, you need 2 moles of oxygen. Using Avogadro's law, 100 mL of oxygen has twice the number of molecules as 50 mL of methane."

Consider the following:


Since V_{m} is a constant at SLC, then:

Thus, the mole ratio is equal to the volume ratio.

Does the n1/n2 = v1/v2 formula only work at SLC or at any temperature and pressure where they are the same for all the gases calculated? And is it at only SLC and STP where the temperature and pressure is constant?

[huity]

Does the n1/n2 = v1/v2 formula only work at SLC or at any temperature and pressure where they are the same for all the gases calculated? And is it at only SLC and STP where the temperature and pressure is constant?

For any situation where temperature and pressure are constant (doesn't have to be at SLC/ STP) 

As stated by sweetcheeks, Avagadro's law states that the amount (n) of a substance is directly proportional to the volume (V).

Note: You do not need to memorise the name of the law i.e. Avagadro's law. Instead, you should understand the concept.   

[dream chaser]

For any situation where temperature and pressure are constant (doesn't have to be at SLC/ STP) 

As stated by sweetcheeks, Avagadro's law states that the amount (n) of a substance is directly proportional to the volume (V).

Note: You do not need to memorise the name of the law i.e. Avagadro's law. Instead, you should understand the concept.   

Okay. Thank you for the help liz.h. Also, thank you TheBigC and sweetcheeks for helping as well. Much appreciated.

I have another question. When a volume is given at SLC for instance, to find the number of moles, should I use n=V/Vm or n=PV/RT because I get slightly different answers for both?

[sweetcheeks]

Okay. Thank you for the help liz.h. Also, thank you TheBigC and sweetcheeks for helping as well. Much appreciated.

I have another question. When a volume is given at SLC for instance, to find the number of moles, should I use n=V/Vm or n=PV/RT because I get slightly different answers for both?

I would use n=V/Vm, where the molar volume is 24.8 L/mol, which is given in the data booklet. It used to be 24.5 L/mol, but that was when the pressure at SLC was said to be 101.325 kPa. Now they use 100 kPa instead, which is probably why you get a different volume (plug 100 kPa into the ideal gas equation, rather than 101.325 and you get 24.78 L which rounds to 24.8 L).

[peachxmh]

Another question to add to my previous one lol -
Why is it that when primary cells reach equilibrium that they become flat? I read something online which explained that at equilibrium the voltage drops to 0 which explains why they can no longer run an electrical current, but I don't get the concept of equilibrium with regards to galvanic cells and how exactly it (i.e. what processes occur with regards to the electrons and ions) causes the voltage to drop?

Thank you!!

[dream chaser]

Hi Guys,

Could someone please identify what the answer to this question is. I calculated that in the reaction, the volume of C3H8 is 100mL and the volume of O2 is 400mL. Since 500ml>400mL, this means that the excess reagent is O2 and the limiting reagent is C3H8. However, I am confused by how they state the question. Is it for all excess and limiting reagent questions that the limiting reagent is the one that does not completely react? The answer says '100 mL of oxygen in excess'.

Also, I found out which is the excess reagent and which is the limiting reagent by using Avogadro's Law. When I compare them, should I always do it in terms of moles or can I use volume to identify as well?

All help will be much appreciated. Thanks.

[sweetcheeks]

Hi Guys,

Could someone please identify what the answer to this question is. I calculated that in the reaction, the volume of C3H8 is 100mL and the volume of O2 is 400mL. Since 500ml>400mL, this means that the excess reagent is O2 and the limiting reagent is C3H8. However, I am confused by how they state the question. Is it for all excess and limiting reagent questions that the limiting reagent is the one that does not completely react? The answer says '100 mL of oxygen in excess'.

Also, I found out which is the excess reagent and which is the limiting reagent by using Avogadro's Law. When I compare them, should I always do it in terms of moles or can I use volume to identify as well?

All help will be much appreciated. Thanks.

The limiting reagent is the one that is completely used up/reacted. That is why it is limiting, as it limits the extent of the reaction. In the case of the question you attached, the propane is the limiting reagent, as it gets used up before all the oxygen does and no more reaction can occur.

As for giving the answer, it depends on what the question wants you to find. It may ask for it in moles, it may ask for it in volume. If it simply asks you how much is leftover/ in excess, give the answer in moles (when in doubt always give moles).

@peachxmh I'll try to get to your questions, it's been a while since I've studied electrochemistry and I need to look over my notes. Your question about equilibrium is a bit difficult to answer at a VCE level (it involves something known as Gibb's free energy).

[dream chaser]

The limiting reagent is the one that is completely used up/reacted. That is why it is limiting, as it limits the extent of the reaction. In the case of the question you attached, the propane is the limiting reagent, as it gets used up before all the oxygen does and no more reaction can occur.

As for giving the answer, it depends on what the question wants you to find. It may ask for it in moles, it may ask for it in volume. If it simply asks you how much is leftover/ in excess, give the answer in moles (when in doubt always give moles).

@peachxmh I'll try to get to your questions, it's been a while since I've studied electrochemistry and I need to look over my notes. Your question about equilibrium is a bit difficult to answer at a VCE level (it involves something known as Gibb's free energy).

Thanks sweetcheeks. Got a couple more questions here. Sorry for the series of questions. For Q26, how would I identify what the reactant is?
For Q29, are we even able to attempt the question. I don't think enough imformation has been provided.