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Author Topic: VCE Chemistry Question Thread  (Read 2848955 times)  Share 

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yplee0926

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Re: VCE Chemistry Question Thread
« Reply #7785 on: January 20, 2019, 03:34:02 am »
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Hey guys,

I’ve two questions regarding electrolysis.
1. The textbook provided a scenario for an electrolytic cell but its details confuse me. In a cell electrolysing aqueous solution of zinc sulfate using copper electrodes, I understand that the overall reaction is Cu + Zn(2+) —> Cu(2+) + Zn. But would this not imply that Zn is in direct contact with Cu(2+) which is dissolved in the solution? This means that the reverse reaction will occur, thus rendering the electrolytic cell very inefficient. It doesn’t make sense to me why the textbook would use this example. Is what I’ve said correct or am I missing something?

2. If electrolysis is occurring and I cease the power supply (and assuming that there is no direct contact between products, but the products are in contact with their respective electrodes), would the reverse reaction occur due to its spontaneity? I’m not sure why the textbook has emphasised a lack of direct contact as a factor that prevents reverse reaction from occurring, because from what I imagine, even being in contact with their electrodes would induce reverse reaction to occur (as no power is supplied).

Thank you :))
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Yertle the Turtle

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Re: VCE Chemistry Question Thread
« Reply #7786 on: January 20, 2019, 03:53:26 am »
+3
Hey guys,

I’ve two questions regarding electrolysis.
1. The textbook provided a scenario for an electrolytic cell but its details confuse me. In a cell electrolysing aqueous solution of zinc sulfate using copper electrodes, I understand that the overall reaction is Cu + Zn(2+) —> Cu(2+) + Zn. But would this not imply that Zn is in direct contact with Cu(2+) which is dissolved in the solution? This means that the reverse reaction will occur, thus rendering the electrolytic cell very inefficient. It doesn’t make sense to me why the textbook would use this example. Is what I’ve said correct or am I missing something?

2. If electrolysis is occurring and I cease the power supply (and assuming that there is no direct contact between products, but the products are in contact with their respective electrodes), would the reverse reaction occur due to its spontaneity? I’m not sure why the textbook has emphasised a lack of direct contact as a factor that prevents reverse reaction from occurring, because from what I imagine, even being in contact with their electrodes would induce reverse reaction to occur (as no power is supplied).

Thank you :))
Welcome to AN!

1. You are correct that the Zn is in contact with the Cu2+ solution, but the flow of the charge prevents the reverse reaction occurring. Thus the electrolytic cell will produce Zn(s) and Cu2+ ions.

2. You seem to be correct, as removing the power supply will always cause the spontaneous reaction to occur, and I believe that your statement would be correct, so long as you are using two half-cells each with only one electro-chemical series equation occurring, which it sounds like you do.

Hope this helps! :D
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yplee0926

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Re: VCE Chemistry Question Thread
« Reply #7787 on: January 20, 2019, 01:23:55 pm »
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Welcome to AN!

1. You are correct that the Zn is in contact with the Cu2+ solution, but the flow of the charge prevents the reverse reaction occurring. Thus the electrolytic cell will produce Zn(s) and Cu2+ ions.

2. You seem to be correct, as removing the power supply will always cause the spontaneous reaction to occur, and I believe that your statement would be correct, so long as you are using two half-cells each with only one electro-chemical series equation occurring, which it sounds like you do.

Hope this helps! :D

Thanks for that! it cleared up a lot of my thoughts :)
But regarding question 1, what did you mean by flow of charge? It just got me thinking that in the situation where Cu2+ is freefloating in the solution, it would actuallly start getting reduced at the cathode instead of zinc (or react directly with zinc itself).
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DBA-144

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Re: VCE Chemistry Question Thread
« Reply #7788 on: January 21, 2019, 06:34:46 pm »
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Why is it that the hydronium ion and hydroxide ion concentration decrease as the other increases? My understanding is that although at 25 C, the pH would be 7, I would expect that the hydronium ion concentration remains constant with the hydroxide ions. However, this is not the case. Why is this so? I mean why does the concentration of each not stay the same?

Intuitively, it makes sense that the protons would transfer from the acid to the water, eg HCl, increasing the overall concentration of H30+ ions and thus decreasing pH. However, what I am really asking is what would occur if a base was present and if there was just water?

Any help is appreciated.  :)
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Quinapalus

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Re: VCE Chemistry Question Thread
« Reply #7789 on: January 21, 2019, 09:16:45 pm »
+2
Why is it that the hydronium ion and hydroxide ion concentration decrease as the other increases? My understanding is that although at 25 C, the pH would be 7, I would expect that the hydronium ion concentration remains constant with the hydroxide ions. However, this is not the case. Why is this so? I mean why does the concentration of each not stay the same?

Intuitively, it makes sense that the protons would transfer from the acid to the water, eg HCl, increasing the overall concentration of H30+ ions and thus decreasing pH. However, what I am really asking is what would occur if a base was present and if there was just water?

Any help is appreciated.  :)

[H3O+] x [OH-] = 10-14
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dream chaser

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Re: VCE Chemistry Question Thread
« Reply #7790 on: January 21, 2019, 09:34:07 pm »
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Hi Guys,

Please check the question in the attachment of this post. I'm interested to part (a) of this question. The answer to part (a) states that no reaction would occur. It says because both Br2 and Mg2+ are oxidising agents.

I still don't get why though.

This is what I think:
Let Br2(aq) + 2e- --> or <---  2Br-(aq) be reaction 1
Let Mg2+(aq) + 2e- --> or <--- Mg(s) be reaction 2

As reaction 1 is higher up the electrochemical series, reaction 1 would be: Br2(aq) + 2e- --> 2Br-(aq)
Therefore, reaction 2 would be: Mg(s) -> Mg2+(aq) + 2e-

Looking at both reactions, reaction 1 undergoes reduction whilst reaction 2 undergoes oxidation. Hence the Mg2+(aq)/Mg(s) half cell would be the anode and the 2Br-(aq)/Br2(s) half cell would be the cathode. Hence electrons would move from the anode to the cathode and thus, since there is a transfer of electrons between half cells, a reaction takes place.

I don't understand how I am wrong with this. Please someone explain.

All replies will be much appreciated. Thanks.  :)



sweetcheeks

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Re: VCE Chemistry Question Thread
« Reply #7791 on: January 21, 2019, 09:39:25 pm »
+2
Hi Guys,

Please check the question in the attachment of this post. I'm interested to part (a) of this question. The answer to part (a) states that no reaction would occur. It says because both Br2 and Mg2+ are oxidising agents.

I still don't get why though.

This is what I think:
Let Br2(aq) + 2e- --> or <---  2Br-(aq) be reaction 1
Let Mg2+(aq) + 2e- --> or <--- Mg(s) be reaction 2

As reaction 1 is higher up the electrochemical series, reaction 1 would be: Br2(aq) + 2e- --> 2Br-(aq)
Therefore, reaction 2 would be: Mg(s) -> Mg2+(aq) + 2e-

Looking at both reactions, reaction 1 undergoes reduction whilst reaction 2 undergoes oxidation. Hence the Mg2+(aq)/Mg(s) half cell would be the anode and the 2Br-(aq)/Br2(s) half cell would be the cathode. Hence electrons would move from the anode to the cathode and thus, since there is a transfer of electrons between half cells, a reaction takes place.

I don't understand how I am wrong with this. Please someone explain.

All replies will be much appreciated. Thanks.  :)




There is no magnesium metal present. This question isn't about electrochemical/galvanic cells, there isn't and anode or cathode or half cells, it is about simply adding bromine to a solution of magnesium ions (e.g. having a beaker containing a solution of magnesium nitrate and adding liquid bromine to it will yield no reaction.).

lzxnl

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Re: VCE Chemistry Question Thread
« Reply #7792 on: January 21, 2019, 09:41:58 pm »
+2
Thanks for that! it cleared up a lot of my thoughts :)
But regarding question 1, what did you mean by flow of charge? It just got me thinking that in the situation where Cu2+ is freefloating in the solution, it would actuallly start getting reduced at the cathode instead of zinc (or react directly with zinc itself).

I would redirect you to my 'Some thoughts on electrochemical cells' post here http://www.atarnotes.com/forum/index.php?topic=148802.0
Without shamelessly advertising myself, it's a summary of a few misconceptions of electrolytic cells.

So, what part of that is useful here? We have an electrolytic cell with copper and zinc. Evidently, the zinc electrode is lower on the electrochemical series and thus is the negative electrode, while the copper electrode is the positive electrode. For electrolysis to work, we have to connect the negative electrode of an external battery (with bigger voltage than produced by the cell) to the copper electrode and the positive electrode to the zinc electrode; positive to negative.

Now, the tendency for zinc metal to react with copper ions can be thought of as a voltage, as it releases energy per electron transferred in the reaction. However, the external battery provides a bigger voltage that overrides the intrinsic tendency for copper ions to react with zinc metal, and FORCES the charge to go the other way (i.e. forces the reaction between copper metal and zinc ions to force the electrons to move from copper to zinc, or positive to negative).

Why is it that the hydronium ion and hydroxide ion concentration decrease as the other increases? My understanding is that although at 25 C, the pH would be 7, I would expect that the hydronium ion concentration remains constant with the hydroxide ions. However, this is not the case. Why is this so? I mean why does the concentration of each not stay the same?

Intuitively, it makes sense that the protons would transfer from the acid to the water, eg HCl, increasing the overall concentration of H30+ ions and thus decreasing pH. However, what I am really asking is what would occur if a base was present and if there was just water?

Any help is appreciated.  :)
Protons and hydroxide ions are in equilibrium with liquid water. If you add base, the base reacts with the protons and decreases the proton concentration, so by Le Chatelier, the system will attempt to compensate for the reduction in proton concentration by decomposing more water to replenish the proton concentration. This ends up increasing the hydroxide ion concentration.

Alternatively, look at the expression for the equilibrium constant for water.
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dream chaser

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Re: VCE Chemistry Question Thread
« Reply #7793 on: January 21, 2019, 09:51:31 pm »
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There is no magnesium metal present. This question isn't about electrochemical/galvanic cells, there isn't and anode or cathode or half cells, it is about simply adding bromine to a solution of magnesium ions (e.g. having a beaker containing a solution of magnesium nitrate and adding liquid bromine to it will yield no reaction.).

What about part(c) then? The solution for part (c) is based of what I said for part (a)? Why would part (c) form a reaction and part(a) doesn't?

MindCabbage

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Re: VCE Chemistry Question Thread
« Reply #7794 on: January 21, 2019, 09:56:14 pm »
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Hello! can anyone provide examples and/or definitions for the following?
1. An indication of the size of particles compared to everyday objects (absolute size)
2. An indication of the size of particles compared to each other (relative size)
3. What a nanoparticle and nanostructure is
4. What small and giant molecules are (include examples)
I just need the basics because I can't find anything online for most of these.
Thank you in advance! ;D

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Re: VCE Chemistry Question Thread
« Reply #7795 on: January 21, 2019, 10:02:09 pm »
+2
What about part(c) then? The solution for part (c) is based of what I said for part (a)? Why would part (c) form a reaction and part(a) doesn't?

Always look at the species that you have currently. In part c, you have a reducing and oxidising agent that can react (according to the electrochemical series, note that this gives you no information about the rate of the reaction).

In part a, you only have two oxidising agents. Don't look at their equations, look at what species you have been given. However, be aware that once a reaction has proceeded, new species can form.
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sweetcheeks

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Re: VCE Chemistry Question Thread
« Reply #7796 on: January 21, 2019, 10:06:52 pm »
+1
What about part(c) then? The solution for part (c) is based of what I said for part (a)? Why would part (c) form a reaction and part(a) doesn't?

You have magnesium metal in part C, whereas in A you have magnesium ions Mg(2+).

yplee0926

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Re: VCE Chemistry Question Thread
« Reply #7797 on: January 22, 2019, 04:35:34 pm »
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I would redirect you to my 'Some thoughts on electrochemical cells' post here http://www.atarnotes.com/forum/index.php?topic=148802.0
Without shamelessly advertising myself, it's a summary of a few misconceptions of electrolytic cells.

So, what part of that is useful here? We have an electrolytic cell with copper and zinc. Evidently, the zinc electrode is lower on the electrochemical series and thus is the negative electrode, while the copper electrode is the positive electrode. For electrolysis to work, we have to connect the negative electrode of an external battery (with bigger voltage than produced by the cell) to the copper electrode and the positive electrode to the zinc electrode; positive to negative.

Now, the tendency for zinc metal to react with copper ions can be thought of as a voltage, as it releases energy per electron transferred in the reaction. However, the external battery provides a bigger voltage that overrides the intrinsic tendency for copper ions to react with zinc metal, and FORCES the charge to go the other way (i.e. forces the reaction between copper metal and zinc ions to force the electrons to move from copper to zinc, or positive to negative).
Protons and hydroxide ions are in equilibrium with liquid water. If you add base, the base reacts with the protons and decreases the proton concentration, so by Le Chatelier, the system will attempt to compensate for the reduction in proton concentration by decomposing more water to replenish the proton concentration. This ends up increasing the hydroxide ion concentration.

Alternatively, look at the expression for the equilibrium constant for water.

Ahhhhh thanks sooo much. Sorry for all these chains of questions but I got one more 😬

How do I decide which electrode is cathode or anode in electrolytic cell? Is it dependent solely on what element the electrode is made of, or on the position on electrochemical series of the substance which is reacting on the electrode (for e.g Pt is often used as inert electrode for other substances to react on it)
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Re: VCE Chemistry Question Thread
« Reply #7798 on: January 22, 2019, 05:58:14 pm »
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Hey, I don’t think I’ve posted on the chemistry thread before so HI  ;D

I’ve just started some old day revision and I’m a lil confused by this question, so I would be so grateful for any help :)

Which of the following lists show the four liquids in order of decreasing heat capacity?
A. Ethanol, water, octane, pentane
B. Ethanol, water, pentane, octane
C. Water, ethanol, pentane and octane
D. Water, ethanol, octane and pentane

I was only given the data booklet to complete this question and there wasn’t a chart, so I’m not sure how to figure this out “closed-book” if that makes sense... is there some kinda trick I don’t get??

Thank you! :)
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Re: VCE Chemistry Question Thread
« Reply #7799 on: January 22, 2019, 06:11:54 pm »
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Can someone please help me with these questions?

1. Calculate the final concentration of 500mL of 0.500M hydrochloric acid after the addition of 200mL of water.

2. Calculate the final concentration of 1000mL of 0.500M potassium iodide after the evaporation of 10.0mL of water.

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