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Edexcel A2 Chemistry Exams -6CH04 (14th June) and 6CH05 (22nd June) Discussion Thread watch

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    (Original post by Funky_Giraffe)
    Hey, please can someone just explain how to get the mole ratio in this calculation? I'm really struggling!! Thanks

    Cr2+ --> Cr6+ + 4e-
    Mn7+ + 5e- --> Mn2+

    therefore

    5Cr2+ : 4Mn7+

    Be careful in the calculation with the moles of chromium ion in solution from the chromium ethanoate
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    (Original post by TeachChemistry)
    Cr2+ --> Cr6+ + 4e-
    Mn7+ + 5e- --> Mn2+

    therefore

    5Cr2+ : 4Mn7+

    Be careful in the calculation with the moles of chromium ion in solution from the chromium ethanoate
    Ah yes - thanks very much!
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    (Original post by ayvaak)
    Attachment 542917
    i think i understand why it would be A. Is it due to the auto ionisation of water being slightly endothermic meaning that an increase in temperature would favour ionisation? Causing an increase in Kw. The pH would still be neutral as number of H+ would be the same because H20 is in surplus?

    Thanks in advance
    because Kw=[H+][OH-] as water dissociates the concentration of H+ increases and so does the concentration of OH- there are no extra H+ or OH- present from anywhere to make it acidic or alkaline as they increase in concentration to the same extent hence its neutral.
    H2O---> H+ + OH-

    Hope this makes sense
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    (Original post by ayvaak)
    Attachment 542917
    i think i understand why it would be A. Is it due to the auto ionisation of water being slightly endothermic meaning that an increase in temperature would favour ionisation? Causing an increase in Kw. The pH would still be neutral as number of H+ would be the same because H20 is in surplus?

    Thanks in advance
    Think about it this way - although your pH is decreasing due to an increase in temperature, the ratio of [OH-]:[H+] will still remain the same. This is from chemguide: "Although the pH of pure water changes with temperature, it is important to realise that it is still neutral. In the case of pure water, there are always going to be the same number of hydrogen ions and hydroxide ions present. That means that the pure water remains neutral - even if its pH changes."

    (Original post by samb1234)
    Yeah there's still a fair amount of time before unit 5 so you should be good. Are you going for the A or the A*?
    I'll be trying to do my best - I have U4 under control, it's just U5 that needs taming. I'll be happy with an A in chemistry, though - I won't need to do it university anyway. Which is a bad thing since I've really come to enjoy the subject.
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    (Original post by Ayman!)
    Think about it this way - although your pH is decreasing due to an increase in temperature, the ratio of [OH-]:[H+] will still remain the same. This is from chemguide: "Although the pH of pure water changes with temperature, it is important to realise that it is still neutral. In the case of pure water, there are always going to be the same number of hydrogen ions and hydroxide ions present. That means that the pure water remains neutral - even if its pH changes."



    I'll be trying to do my best - I have U4 under control, it's just U5 that needs taming. I'll be happy with an A in chemistry, though - I won't need to do it university anyway. Which is a bad thing since I've really come to enjoy the subject.
    Yeah unit 5 has way way too much content in it unfortunately, i've literally spent like 75% of the last week or so just on unit 5 so hopefully will be fine but gonna need to shift my priorities a bit I think lol. What are you doing at uni? Some form of engineering/physics type thing i'm guessing
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    Hey guys, I was wondering, what would happen to the electrode potential if current was allowed to pass through the cell? Because I know that normally you use a high-resistance voltmeter to prevent this from happening but I think I remember seeing a question asking what would happen if current was to pass through?

    Thanks
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    (Original post by Don Pedro K.)
    Hey guys, I was wondering, what would happen to the electrode potential if current was allowed to pass through the cell? Because I know that normally you use a high-resistance voltmeter to prevent this from happening but I think I remember seeing a question asking what would happen if current was to pass through?

    Thanks
    You get electron flow from where there are lots of them to where there are less of them, so can you work out what will happen from there?
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    (Original post by samb1234)
    You get electron flow from where there are lots of them to where there are less of them, so can you work out what will happen from there?
    Ah okay soooo electrode potential would decrease!
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    (Original post by Ayman!)
    Think about it this way - although your pH is decreasing due to an increase in temperature, the ratio of [OH-]:[H+] will still remain the same. This is from chemguide: "Although the pH of pure water changes with temperature, it is important to realise that it is still neutral. In the case of pure water, there are always going to be the same number of hydrogen ions and hydroxide ions present. That means that the pure water remains neutral - even if its pH changes."



    I'll be trying to do my best - I have U4 under control, it's just U5 that needs taming. I'll be happy with an A in chemistry, though - I won't need to do it university anyway. Which is a bad thing since I've really come to enjoy the subject.
    So we kinda assume that they're talking about pure water right?
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    (Original post by Don Pedro K.)
    Ah okay soooo electrode potential would decrease!
    Yes but there are more extreme things that happen as well. At each electrode you have the equilibrium X --->> X+ +e- (insert number of charges and electrons accordingly). When electrons move from the negative electrode (both are technically negative but voltages is a relative measurement) at that electrode there are now less electrons so by le chatelier's eq moves to the right, and the converse will happen at the positive electrode
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    (Original post by samb1234)
    Yes but there are more extreme things that happen as well. At each electrode you have the equilibrium X --->> X+ +e- (insert number of charges and electrons accordingly). When electrons move from the negative electrode (both are technically negative but voltages is a relative measurement) at that electrode there are now less electrons so by le chatelier's eq moves to the right, and the converse will happen at the positive electrode
    Wait, I'm not quite sure what you mean by that :s

    So, let's say you have Mg2+ + 2e- ---> Mg, for which the SEP = -2.37V.
    and Al3+ + 3e- ---> Al, for which the SEP = +0.80V.

    Could you try to explain what you're trying to explain (lol xD) in this context?
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    (Original post by Don Pedro K.)
    Wait, I'm not quite sure what you mean by that :s

    So, let's say you have Mg2+ + 2e- ---> Mg, for which the SEP = -2.37V.
    and Al3+ + 3e- ---> Al, for which the SEP = +0.80V.

    Could you try to explain what you're trying to explain (lol xD) in this context?
    Of course. Essentially the standard electrode potential is a measurement of the position of equilibrium for the reaction for each electrode. this reaction here :

    Mg2+ + 2e- ---> Mg, for which the SEP = -2.37V, is in reality an equilibrium, so at that electrode we have the equilibrium
    Mg2+ + 2e- <---> Mg going on.
    The negative electrode potential tells us that, relative to a SHE, this equilibrium lies more towards the side with the electrons on. In comparison at the aluminum electrode we have a similar equilibrium going on, except this time the position of equilibrium is much more on the side of the elemental Al, hence the positive SEP (it's important to realise that in reality all of these are technically negative, but a voltmeter measures the potential difference so being less negative than an SHE or other electrode means it is the 'positive' electrode). When we allow current to flow, the electrons will flow from the Mg side as there are more of them there, to the Al side where there are less. As each of these reactions are in equilibrium, by removing electrons from the Mg side the position of equilibrium is going to move even further to the left to produce more electrons, meaning it pretty much turns into the one way reaction
    Mg---->Mg2+ +2e-

    At the Al side, we are adding electrons so to counteract this the position of equilibrium is going to shift even further to the right so that the reaction essentially becomes one way and is Al3+ +3e- ---->Al
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    (Original post by samb1234)
    Of course. Essentially the standard electrode potential is a measurement of the position of equilibrium for the reaction for each electrode. this reaction here :

    Mg2+ + 2e- ---> Mg, for which the SEP = -2.37V, is in reality an equilibrium, so at that electrode we have the equilibrium
    Mg2+ + 2e- <---> Mg going on.
    The negative electrode potential tells us that, relative to a SHE, this equilibrium lies more towards the side with the electrons on. In comparison at the aluminum electrode we have a similar equilibrium going on, except this time the position of equilibrium is much more on the side of the elemental Al, hence the positive SEP (it's important to realise that in reality all of these are technically negative, but a voltmeter measures the potential difference so being less negative than an SHE or other electrode means it is the 'positive' electrode). When we allow current to flow, the electrons will flow from the Mg side as there are more of them there, to the Al side where there are less. As each of these reactions are in equilibrium, by removing electrons from the Mg side the position of equilibrium is going to move even further to the left to produce more electrons, meaning it pretty much turns into the one way reaction
    Mg---->Mg2+ +2e-

    At the Al side, we are adding electrons so to counteract this the position of equilibrium is going to shift even further to the right so that the reaction essentially becomes one way and is Al3+ +3e- ---->Al
    Holy crap you just explained that in the most masterful way ever. That makes perfect sense; thank you so much! You're sorted for chemistry aren't you hahaha xD!
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    (Original post by Don Pedro K.)
    Holy crap you just explained that in the most masterful way ever. That makes perfect sense; thank you so much! You're sorted for chemistry aren't you hahaha xD!
    We'll see lol, haven't quite finished notes for unit 5 yet because they take too long but still a while until the exam. And thanks, glad it helped.
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    (Original post by samb1234)
    We'll see lol, haven't quite finished notes for unit 5 yet because they take too long but still a while until the exam. And thanks, glad it helped.
    One last thing, in exam questions, they say that a reaction won't be energetically feasible if E(cell) < 0V, such as here:

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    But in the textbook, we're told that if -0.6< E(cell) < 0, the reactants predominate and only when E(cell) < -0.6V is there no reaction.

    Who do we trust, Edexcel's textbook or Edexcel's papers xD?!
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    (Original post by Don Pedro K.)
    One last thing, in exam questions, they say that a reaction won't be energetically feasible if E(cell) < 0V, such as here:

    Name:  l.JPG
Views: 92
Size:  18.0 KB

    But in the textbook, we're told that if -0.6< E(cell) < 0, the reactants predominate and only when E(cell) < -0.6V is there no reaction.

    Who do we trust, Edexcel's textbook or Edexcel's papers xD?!
    Depends what you mean by no reaction. For exam purposes I would stick with negative means reaction isn't feasible, in reality there will be some negligible amounts of reaction taking place. This can relatively easily be shown assuming you do A2 maths, we know that Ecell is proportional to total entropy change which is proportional to ln(K), where K is equilibrium constant so if you say that Ecell=Alnk you can solve that for K and see that it will never actually be 0, even if the ecell was massively massively negative
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    (Original post by samb1234)
    Depends what you mean by no reaction. For exam purposes I would stick with negative means reaction isn't feasible, in reality there will be some negligible amounts of reaction taking place. This can relatively easily be shown assuming you do A2 maths, we know that Ecell is proportional to total entropy change which is proportional to ln(K), where K is equilibrium constant so if you say that Ecell=Alnk you can solve that for K and see that it will never actually be 0, even if the ecell was massively massively negative
    Ah yeah makes sense haha but as you said I will just stick to what the mark scheme says, because I guess that's what matters at the end of the day (annoyingly, to be honest haha).
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    (Original post by Don Pedro K.)
    Ah yeah makes sense haha but as you said I will just stick to what the mark scheme says, because I guess that's what matters at the end of the day (annoyingly, to be honest haha).
    Yeah it's a bit annoying which is why i personally don't like the text book, I just make notes for every spec point off chemguide which takes ages but tends to give you a better idea of what is going on
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    https://a5c076379da85d2c0b501a60f1a4...0Chemistry.pdf

    Could someone please explain how to do 24di,ei and f please :/ So confused
    -thanks
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    (Original post by ST_123)
    https://a5c076379da85d2c0b501a60f1a4...0Chemistry.pdf

    Could someone please explain how to do 24di,ei and f please :/ So confused
    -thanks
    Alright well for part di):

    Try to work out the number of moles of hydrogen gas reacted and the number of moles of carvone reacted. Look at the ratio between the two.

    Once you've done this, do you remember what happens when you react something with double bonds with hydrogen gas? What kind of reaction occurs? Knowing this, you should be able to draw the product's skeletal structure!
 
 
 

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