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    I don't understand, in my CGP book it states: " Iron(III) iodide solution is unstable, because iodide ions are strong enough reducing agents to reduce the Fe3+ ions to Fe2+. Iron (III) chloride however, is stable because Cl- ions aren't powerful enough reducing agents to reduce Fe3+ ions. "

    How does the reducing agent relate to stability, and why are Iodide ions used for iron iodide, and Chloride ions used for iron chloride?
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    (Original post by I <3 WORK)
    I don't understand, in my CGP book it states: " Iron(III) iodide solution is unstable, because iodide ions are strong enough reducing agents to reduce the Fe3+ ions to Fe2+. Iron (III) chloride however, is stable because Cl- ions aren't powerful enough reducing agents to reduce Fe3+ ions. "

    How does the reducing agent relate to stability, and why are Iodide ions used for iron iodide, and Chloride ions used for iron chloride?
    You can consider the standard electrode potentials.

    The \mathrm{E^{\circ}(V)} of \mathrm{Cl_2} reduction is 1.36
    For \mathrm{Fe^{3+}} to be reduced, the \mathrm{E^{\circ}(V)} value would have to be greater than 1.36. It isn't.
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    (Original post by Kvothe the arcane)
    You can consider the standard electrode potentials.

    The \mathrm{E^{\circ}(V)} of \mathrm{Cl_2} reduction is 1.36
    For \mathrm{Fe^{3+}} to be reduced, the \mathrm{E^{\circ}(V)} value would have to be greater than 1.36. It isn't.
    What does EV mean and I don't understand, how does that relate to the stability of a compound?
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    (Original post by I <3 WORK)
    What does EV mean and I don't understand, how does that relate to the stability of a compound?
    Woah, never been quoted before.
    The E part means 'Electrode Potential'
    The V part means 'Voltage'


    Essentially, they're both the same thing.

    Electrode Potential is the potential of an element to accept electrons. The more positive this number is, the more it's likely to accept electrons.

    So going back to what Kvone the Arcane said
    The of reduction is 1.36

    This is pretty high. It's likely to accept electrons and be reduced to Chloride Ions.
    i.e.  \mathrm{Cl_2} + \mathrm{2e^-} \longrightarrow \mathrm{2Cl^-}

    Now the of  \mathrm{Fe^{3+}} is +0.77.

    Because it's lower than our E of Cl2, it's going to remain oxidised. It's not going to get converted into Fe2+.

    If this isn't making sense, let's look at that statement you said:
    "Iron (III) chloride however, is stable because Cl- ions aren't powerful enough reducing agents to reduce Fe3+ ions"

    We have two reactions.

     \mathrm{Cl_2} + \mathrm{2e^-} \longrightarrow \mathrm{2Cl^-}
     \mathrm{Fe^{3+}} + \mathrm{e^-} \longrightarrow \mathrm{Fe^{2+}}

    If we want to reduce the Fe3+ ions to Fe2+ ions, the bottom reaction has to go forward, agreed?
    In order to do that, we need to get electrons.
    To get electrons, we can oxidise Cl-, i.e. the top reaction has to go backwards.

    But, the higher the E, the more it's likely to accept electrons.
    In other words, the higher the E, the more forward the reaction is.
    Our top reaction, the Chlorine one has the higher E, so it's going to be forward.
    Thus, the bottom reaction is backward.
    As a result, we end up with Cl- and Fe3+. That's why it's stable, those two ions remain.

    That's the reason for that other sentence
    "Iron(III) iodide solution is unstable, because iodide ions are strong enough reducing agents to reduce the Fe3+ ions to Fe2+"

    Again our two reactions are:
     \mathrm{I_2} + \mathrm{2e^-} \longrightarrow \mathrm{2I^-}
     \mathrm{Fe^{3+}} + \mathrm{e^-} \longrightarrow \mathrm{Fe^{2+}}

    However, the E of I2 is +0.54.
    The E of Fe3+/Fe2+ is +0.77.

    So this time, the bottom reaction is going to go forward.
    This means Fe3+ is going to be reduced to Fe2+


    That's why if you try and create a solution of Fe(III) and I-, you can't get the two ions to live happily. I- is a strong enough reducing agent which causes the solution to be really unstable, because these ions are going to react. Stability depends upon the difference in E (electrode Potential) and which reactions need to go forward and backward.

    Have you not done Electrochemistry?

    Also, please ask away, I feel I've scared you with this long post
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    (Original post by RMNDK)
    Woah, never been quoted before.
    The E part means 'Electrode Potential'
    The V part means 'Voltage'


    Essentially, they're both the same thing.

    Electrode Potential is the potential of an element to accept electrons. The more positive this number is, the more it's likely to accept electrons.

    So going back to what Kvone the Arcane said
    The of reduction is 1.36

    This is pretty high. It's likely to accept electrons and be reduced to Chloride Ions.
    i.e.  \mathrm{Cl_2} + \mathrm{2e^-} \longrightarrow \mathrm{2Cl^-}

    Now the of  \mathrm{Fe^{3+}} is +0.77.

    Because it's lower than our E of Cl2, it's going to remain oxidised. It's not going to get converted into Fe2+.

    If this isn't making sense, let's look at that statement you said:
    "Iron (III) chloride however, is stable because Cl- ions aren't powerful enough reducing agents to reduce Fe3+ ions"

    We have two reactions.

     \mathrm{Cl_2} + \mathrm{2e^-} \longrightarrow \mathrm{2Cl^-}
     \mathrm{Fe^{3+}} + \mathrm{e^-} \longrightarrow \mathrm{Fe^{2+}}

    If we want to reduce the Fe3+ ions to Fe2+ ions, the bottom reaction has to go forward, agreed?
    In order to do that, we need to get electrons.
    To get electrons, we can oxidise Cl-, i.e. the top reaction has to go backwards.

    But, the higher the E, the more it's likely to accept electrons.
    In other words, the higher the E, the more forward the reaction is.
    Our top reaction, the Chlorine one has the higher E, so it's going to be forward.
    Thus, the bottom reaction is backward.
    As a result, we end up with Cl- and Fe3+. That's why it's stable, those two ions remain.

    That's the reason for that other sentence
    "Iron(III) iodide solution is unstable, because iodide ions are strong enough reducing agents to reduce the Fe3+ ions to Fe2+"

    Again our two reactions are:
     \mathrm{I_2} + \mathrm{2e^-} \longrightarrow \mathrm{2I^-}
     \mathrm{Fe^{3+}} + \mathrm{e^-} \longrightarrow \mathrm{Fe^{2+}}

    However, the E of I2 is +0.54.
    The E of Fe3+/Fe2+ is +0.77.

    So this time, the bottom reaction is going to go forward.
    This means Fe3+ is going to be reduced to Fe2+


    That's why if you try and create a solution of Fe(III) and I-, you can't get the two ions to live happily. I- is a strong enough reducing agent which causes the solution to be really unstable, because these ions are going to react. Stability depends upon the difference in E (electrode Potential) and which reactions need to go forward and backward.

    Have you not done Electrochemistry?

    Also, please ask away, I feel I've scared you with this long post
    Omg I quoted you because you were seriously amazing at your explanation the last time - so ignore my pm lol I didn't see your reply. Thank you so much for this, I haven't finished reading so I'm going to read now then reply again if that's okay.
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    (Original post by RMNDK)
    Woah, never been quoted before.
    The E part means 'Electrode Potential'
    The V part means 'Voltage'


    Essentially, they're both the same thing.

    Electrode Potential is the potential of an element to accept electrons. The more positive this number is, the more it's likely to accept electrons.

    So going back to what Kvone the Arcane said
    The of reduction is 1.36

    This is pretty high. It's likely to accept electrons and be reduced to Chloride Ions.
    i.e.  \mathrm{Cl_2} + \mathrm{2e^-} \longrightarrow \mathrm{2Cl^-}

    Now the of  \mathrm{Fe^{3+}} is +0.77.

    Because it's lower than our E of Cl2, it's going to remain oxidised. It's not going to get converted into Fe2+.

    If this isn't making sense, let's look at that statement you said:
    "Iron (III) chloride however, is stable because Cl- ions aren't powerful enough reducing agents to reduce Fe3+ ions"

    We have two reactions.

     \mathrm{Cl_2} + \mathrm{2e^-} \longrightarrow \mathrm{2Cl^-}
     \mathrm{Fe^{3+}} + \mathrm{e^-} \longrightarrow \mathrm{Fe^{2+}}

    If we want to reduce the Fe3+ ions to Fe2+ ions, the bottom reaction has to go forward, agreed?
    In order to do that, we need to get electrons.
    To get electrons, we can oxidise Cl-, i.e. the top reaction has to go backwards.

    But, the higher the E, the more it's likely to accept electrons.
    In other words, the higher the E, the more forward the reaction is.
    Our top reaction, the Chlorine one has the higher E, so it's going to be forward.
    Thus, the bottom reaction is backward.
    As a result, we end up with Cl- and Fe3+. That's why it's stable, those two ions remain.

    That's the reason for that other sentence
    "Iron(III) iodide solution is unstable, because iodide ions are strong enough reducing agents to reduce the Fe3+ ions to Fe2+"

    Again our two reactions are:
     \mathrm{I_2} + \mathrm{2e^-} \longrightarrow \mathrm{2I^-}
     \mathrm{Fe^{3+}} + \mathrm{e^-} \longrightarrow \mathrm{Fe^{2+}}

    However, the E of I2 is +0.54.
    The E of Fe3+/Fe2+ is +0.77.

    So this time, the bottom reaction is going to go forward.
    This means Fe3+ is going to be reduced to Fe2+


    That's why if you try and create a solution of Fe(III) and I-, you can't get the two ions to live happily. I- is a strong enough reducing agent which causes the solution to be really unstable, because these ions are going to react. Stability depends upon the difference in E (electrode Potential) and which reactions need to go forward and backward.

    Have you not done Electrochemistry?

    Also, please ask away, I feel I've scared you with this long post
    Ahah, to be honest when I first saw the post it really did scare me lol, but if I'm going to be honest when I read into it and understood it carefully it truly makes sense! So thank you so much you're amazing. I haven't actually learnt electrochemistry no as it doesn't come in AS, but yes I was just confused about the concept. Just one thing - so can you say that an unstable compound is one that is more likely to react - and does it always involve reduction reactions or would a compound being oxidised also be considered unstable?

    Also when they are comparing the stability of both the iron chloride and iodide why are they using the same halide as reducing agents - for e.g. why don't they give the iron chloride reacting with iodide ions to compare the same stability?
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    (Original post by I <3 WORK)
    Ahah, to be honest when I first saw the post it really did scare me lol, but if I'm going to be honest when I read into it and understood it carefully it truly makes sense! So thank you so much you're amazing. I haven't actually learnt electrochemistry no as it doesn't come in AS, but yes I was just confused about the concept. Just one thing - so can you say that an unstable compound is one that is more likely to react - and does it always involve reduction reactions or would a compound being oxidised also be considered unstable?
    No problem!
    Ah, I should have realised that you're in AS, yeah this is not an easy topic to learn at AS, let alone at A2, and condensing it into that post may still be insufficient.

    Generally speaking yes, unstable compounds are more reactive because they're in a high energy state which means the particles have more potential chemical energy to transferred into other forms.

    What I mean by that is, say somehow I add some heat energy to a box. If I add a tiny bit, there's not that much heat loss from the box, but if I add a shed load of heat energy, there's a good chance heat will escape from the box. The box is really unstable with all this energy, it can't contain it. Likewise, this unstable compound has all this potential energy that it needs to transfer.


    It doesn't always involve reduction reactions, an unstable compound can be oxidised to something more stable.

    For example, going back to your example on the Iron Iodide solution, we said Fe3+ is unstable relative to I-, so it tends to be reduced. And we can argue it the other way, I- is in it's unstable form when present with Fe3+, so it tends to be oxidised to I2.
    Any reaction with reduction must involved oxidation, so you can think of stability from both sides.

    When you get to A2, you're actually going to cover this idea of stability more and see what it means for a compound to be stable, and why unstable species tend to react.
    Spoiler:
    Show
    Read on if you're curious! Don't read if you'd rather not boggle down with A2 stuff!

    To give you a brief intro, there's two forms of stability: being kinetically stable, and being thermodynamically stable.

    Thermodynamic stability is when a compound undergoes a reaction that puts it an lower energy state.
    Why? Well, that's because exothermic reactions are favourable. All that potential energy in the molecules is more likely to be transferred as kinetic energy, the lower the potential energy, the less it's likely to be transferred as kinetic energy so the more stable it is. It's like as if the particles are relaxed.

    Kinetic stability describes the rate at which a reaction with a compound is going. It might be fast or slow.

    The two work together and overall you get your stability (of course, there's more to it, like entropy)

    For example,
    The reaction of Diamond ---> Graphite is possible, because in fact diamond is thermodynamically unstable. It would rather be Graphite.

    But we don't see diamond just decomposing to graphite on our bling do we? That's because it's kinetically stable. The rate of reaction is incredibly slow due to the high activation energy required; the effect is that the reaction just doesn't proceed.
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    (Original post by RMNDK)
    No problem!
    Ah, I should have realised that you're in AS, yeah this is not an easy topic to learn at AS, let alone at A2, and condensing it into that post may still be insufficient.

    Generally speaking yes, unstable compounds are more reactive because they're in a high energy state which means the particles have more potential chemical energy to transferred into other forms.

    What I mean by that is, say somehow I add some heat energy to a box. If I add a tiny bit, there's not that much heat loss from the box, but if I add a shed load of heat energy, there's a good chance heat will escape from the box. The box is really unstable with all this energy, it can't contain it. Likewise, this unstable compound has all this potential energy that it needs to transfer.


    It doesn't always involve reduction reactions, an unstable compound can be oxidised to something more stable.

    For example, going back to your example on the Iron Iodide solution, we said Fe3+ is unstable relative to I-, so it tends to be reduced. And we can argue it the other way, I- is in it's unstable form when present with Fe3+, so it tends to be oxidised to I2.
    Any reaction with reduction must involved oxidation, so you can think of stability from both sides.

    When you get to A2, you're actually going to cover this idea of stability more and see what it means for a compound to be stable, and why unstable species tend to react.
    Spoiler:
    Show
    Read on if you're curious! Don't read if you'd rather not boggle down with A2 stuff!

    To give you a brief intro, there's two forms of stability: being kinetically stable, and being thermodynamically stable.

    Thermodynamic stability is when a compound undergoes a reaction that puts it an lower energy state.
    Why? Well, that's because exothermic reactions are favourable. All that potential energy in the molecules is more likely to be transferred as kinetic energy, the lower the potential energy, the less it's likely to be transferred as kinetic energy so the more stable it is. It's like as if the particles are relaxed.

    Kinetic stability describes the rate at which a reaction with a compound is going. It might be fast or slow.

    The two work together and overall you get your stability (of course, there's more to it, like entropy)

    For example,
    The reaction of Diamond ---> Graphite is possible, because in fact diamond is thermodynamically unstable. It would rather be Graphite.

    But we don't see diamond just decomposing to graphite on our bling do we? That's because it's kinetically stable. The rate of reaction is incredibly slow due to the high activation energy required; the effect is that the reaction just doesn't proceed.
    Thank you so much for your explanations, even when you go in detail I prefer it to be honest because ironically a harder concept becomes rather easier to understand. I think I've finally come to grips with this topic so thank you once again - your analogies make things so simple! I know now the first person to come to next.
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    Sorry for the late reply, I've had a busy one today. I see RMNDK's done a fantastic job of explaining it all to you. It's a good thing too, he's explained it in a far better way than I could've done.

    Electrode potentials are pretty handy. Instead of wasting brainpower trying to work out how unstable a molecule/ion/whatever is by studying the structure you can just grab your data book. Better still is that it gives you a numerical value instead of the qualitative analysis that you get from staring at the structure, it saves a lot of time too.

    The limitation of electrode potentials is that it gives you a measure of stability instead of the actual causes of a species' stability/instability. Normally this is just unnecessary information, but it is sometimes helpful to have an idea of what's causing instability. In these cases you basically just have to study someone else's work or go and start running experiments yourself. Obviously the latter isn't really doable for 6th formers.

    I digress. Anyhow in the case of the iodide ions vs chloride ions reducing power it's all down to how tightly the stolen electron is bound to the ion. I'm sure you know about that with regards to the trends in the reactivity of groups 1, 2 and 7 (I'm not sure if they look at the other groups in A level chemistry). It's basically the same thing going on here. In iodide ions the stolen electron is much further from the nucleus than the stolen electron in chloride ions and it has 2 extra shells worth of shielding. So the attraction to the nucleus that the stolen electron experiences in iodide ions is far smaller than it experiences in chloride ions. This is why iodide ions are far less stable than chloride ions. The ion is pretty rubbish at holding onto that extra electron (compared to chloride at least), so oxidising agents can nick it more easily.

    It's like a tug of war game. Both the iron ion and iodide ion 'want' that electron, but iron's got muscle and the iodide is twiggy so iron ends up winning. However, when the iron ion goes against chloride, the chloride's the one with the most bulk and so it defeats the iron ion. The bulk each ion has represents the strength of the attraction between the extra electron and the nucleus of an ion.

    There's a bunch of other things that affect the stability, but for the chloride - iodide example this is the main thing.
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    (Original post by Peroxidation)
    Sorry for the late reply, I've had a busy one today. I see RMNDK's done a fantastic job of explaining it all to you. It's a good thing too, he's explained it in a far better way than I could've done.

    Electrode potentials are pretty handy. Instead of wasting brainpower trying to work out how unstable a molecule/ion/whatever is by studying the structure you can just grab your data book. Better still is that it gives you a numerical value instead of the qualitative analysis that you get from staring at the structure, it saves a lot of time too.

    The limitation of electrode potentials is that it gives you a measure of stability instead of the actual causes of a species' stability/instability. Normally this is just unnecessary information, but it is sometimes helpful to have an idea of what's causing instability. In these cases you basically just have to study someone else's work or go and start running experiments yourself. Obviously the latter isn't really doable for 6th formers.

    I digress. Anyhow in the case of the iodide ions vs chloride ions reducing power it's all down to how tightly the stolen electron is bound to the ion. I'm sure you know about that with regards to the trends in the reactivity of groups 1, 2 and 7 (I'm not sure if they look at the other groups in A level chemistry). It's basically the same thing going on here. In iodide ions the stolen electron is much further from the nucleus than the stolen electron in chloride ions and it has 2 extra shells worth of shielding. So the attraction to the nucleus that the stolen electron experiences in iodide ions is far smaller than it experiences in chloride ions. This is why iodide ions are far less stable than chloride ions. The ion is pretty rubbish at holding onto that extra electron (compared to chloride at least), so oxidising agents can nick it more easily.

    It's like a tug of war game. Both the iron ion and iodide ion 'want' that electron, but iron's got muscle and the iodide is twiggy so iron ends up winning. However, when the iron ion goes against chloride, the chloride's the one with the most bulk and so it defeats the iron ion. The bulk each ion has represents the strength of the attraction between the extra electron and the nucleus of an ion.

    There's a bunch of other things that affect the stability, but for the chloride - iodide example this is the main thing.
    You and RMNDK both have this natural flair to just explain things like magic, and I'm really not exaggerating when I say this because my Chemistry teacher is absolutely :poo: I'm sorry to say, and the difference like seriously...

    Thank you so much! - your analogy is so good and I know that I will remember this forever now.

    P.S: I'm sorry I wanted to give you two a rep but TSR didnt allow me to because I've to give someone else first.
 
 
 
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