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    Hey

    I know orders can only be obtained from experimental data but I was wondering if anyone knows the orders of reactants or rate equation for the reaction between potassium manganate and ethanedioic acid
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    I know ethanedioate is the same as oxalate so we're talking about H2C2O4. If you react this with KMnO4, all the metals will be left ionic and you'll get CO2 and H2O. As follows:

    Acidic solution:
    KMnO4 + H2C2O4 + 6H+ + 3e- -> 2CO2 + 4H2O + K+ + Mn2+

    Alkaline solution:
    KMnO4 + H2C2O4 + 2H2O + 3e- -> 2CO2 + 6OH- + K+ + Mn2+

    Note: I'm just guessing that in both cases this is just a simple reaction which can be balanced as an ionic equation straight away, without needing to come up with two separate reactions and then combine them. This assumption could well be wrong.

    Anyway, if you're really looking for the orders of this reaction you're best bet would be an Internet search rather than asking people. This would be decently well documented but not that well known.
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    (Original post by Big-Daddy)
    I know ethanedioate is the same as oxalate so we're talking about H2C2O4. If you react this with KMnO4, all the metals will be left ionic and you'll get CO2 and H2O. As follows:

    Acidic solution:
    KMnO4 + H2C2O4 + 6H+ + 3e- -> 2CO2 + 4H2O + K+ + Mn2+

    Alkaline solution:
    KMnO4 + H2C2O4 + 2H2O + 3e- -> 2CO2 + 6OH- + K+ + Mn2+

    Note: I'm just guessing that in both cases this is just a simple reaction which can be balanced as an ionic equation straight away, without needing to come up with two separate reactions and then combine them. This assumption could well be wrong.

    Anyway, if you're really looking for the orders of this reaction you're best bet would be an Internet search rather than asking people. This would be decently well documented but not that well known.
    Trust me, I've tried searching. Found countless methods but never the actual final rate equation
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    (Original post by NiceToMeetYou)
    Trust me, I've tried searching. Found countless methods but never the actual final rate equation
    I see. Try asking on a chemistry-focused website instead: http://www.chemicalforums.com/index.php?board=9.0. Though I personally cannot say it's likely anyone has the specific answer to this. You may need some reference-book of rate laws.
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    (Original post by Big-Daddy)
    I know ethanedioate is the same as oxalate so we're talking about H2C2O4. If you react this with KMnO4, all the metals will be left ionic and you'll get CO2 and H2O. As follows:

    Acidic solution:
    KMnO4 + H2C2O4 + 6H+ + 3e- -> 2CO2 + 4H2O + K+ + Mn2+

    Alkaline solution:
    KMnO4 + H2C2O4 + 2H2O + 3e- -> 2CO2 + 6OH- + K+ + Mn2+

    Note: I'm just guessing that in both cases this is just a simple reaction which can be balanced as an ionic equation straight away, without needing to come up with two separate reactions and then combine them. This assumption could well be wrong.

    Anyway, if you're really looking for the orders of this reaction you're best bet would be an Internet search rather than asking people. This would be decently well documented but not that well known.
    Sorry BD, you can't have a reaction equation with electrons appearing from nowhere.

    The manganate ion gets reduced:

    MnO4- + 8H+ + 5e --> Mn2+ + 4H2O

    and the ethandioate gets oxidised:

    (COO)22- - 2e --> CO2

    Then as usual balance the electrons and add:

    2MnO4- + 16H+ + 5(COO)22- --> 2Mn2+ + 8H2O + 5CO2
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    (Original post by charco)
    Sorry BD, you can't have a reaction equation with electrons appearing from nowhere.

    The manganate ion gets reduced:

    MnO4- + 8H+ + 5e --> Mn2+ + 4H2O

    and the ethandioate gets oxidised:

    (COO)22- - 2e --> CO2

    Then as usual balance the electrons and add:

    2MnO4- + 16H+ + 5(COO)22- --> 2Mn2+ + 8H2O + 5CO2
    Thanks for helping.

    Any idea how can I do this similar problem by the way:

    In the presence of ammonia, O2 oxidizes [Cu(S2O3)3]5- back to [Cu(NH3)4]2+. Write a balanced equation for this oxidation-reduction reaction in basic solution.

    Clearly [Cu(S2O3)3]5- + 4 NH3 -> [Cu(NH3)4]2+ + 3S2O32- + e- (and indeed from the gain of charge we can see copper has been oxidized). But if O2 gets reduced, what does it get reduced to?
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    This reaction will have a highly complex reaction mechanism, probably unknown. You may find some studies in some journals, but I doubt you will find anything. Lots of complex oxidation reactions have no universally accepted mechanism or even any good suggestions. For the oxidation course at my uni half the mechanisms were rather contrived examples 'made up' by the lecturer that are only reasonable in the sense that they rationalise the product. They had no grounding in hard data or studies.

    Your best bet is to find some data someone has for the kinetics of the reaction and work out the orders from that. Failing that, do it yourself although it would be way beyond the capabilities of a standard A-level lab.
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    (Original post by JMaydom)
    This reaction will have a highly complex reaction mechanism, probably unknown. You may find some studies in some journals, but I doubt you will find anything. Lots of complex oxidation reactions have no universally accepted mechanism or even any good suggestions. For the oxidation course at my uni half the mechanisms were rather contrived examples 'made up' by the lecturer that are only reasonable in the sense that they rationalise the product. They had no grounding in hard data or studies.

    Your best bet is to find some data someone has for the kinetics of the reaction and work out the orders from that. Failing that, do it yourself although it would be way beyond the capabilities of a standard A-level lab.
    Would it be impossible to deduce a few mechanisms based on the ionic reactions going on, and then come up with some candidate rate laws?
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    (Original post by Big-Daddy)
    Would it be impossible to deduce a few mechanisms based on the ionic reactions going on, and then come up with some candidate rate laws?
    A lot of oxidations are radical mechanisms. I don't think you grasp how complex these are in comparison to your standard Sn2 reaction.
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    (Original post by Big-Daddy)
    Thanks for helping.

    Any idea how can I do this similar problem by the way:

    In the presence of ammonia, O2 oxidizes [Cu(S2O3)3]5- back to [Cu(NH3)4]2+. Write a balanced equation for this oxidation-reduction reaction in basic solution.

    Clearly [Cu(S2O3)3]5- + 4 NH3 -> [Cu(NH3)4]2+ + 3S2O32- + e- (and indeed from the gain of charge we can see copper has been oxidized). But if O2 gets reduced, what does it get reduced to?
    The methodology is the same.

    The ligands are a bit distracting but don't affect the outcome:

    Cu+ --> Cu2+ + 1e

    O2 + 2H2O + 4e --> 4OH-

    MTB 4 in the first equation to equalise electrons and add:

    4Cu+ --> 4Cu2+ + 4e
    O2 + 2H2O + 4e --> 4OH-
    ------------------------------------------------------------ add
    4Cu+ + O2 + 2H2O --> 4Cu2+ + 4OH-

    now if you're bothered you can put in the ligands and any balancing ions ...
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    (Original post by charco)
    The methodology is the same.

    The ligands are a bit distracting but don't affect the outcome:

    Cu+ --> Cu2+ + 1e

    O2 + 2H2O + 4e --> 4OH-

    MTB 4 in the first equation to equalise electrons and add:

    4Cu+ --> 4Cu2+ + 4e
    O2 + 2H2O + 4e --> 4OH-
    ------------------------------------------------------------ add
    4Cu+ + O2 + 2H2O --> 4Cu2+ + 4OH-

    now if you're bothered you can put in the ligands and any balancing ions ...
    Ah I see, thanks. The reduction of O2 was providing the difficulty.

    Is it a general rule that once you have a balanced ionic equation (i.e. not for oxidation or reduction individually but overall) it can no longer contain any lone electrons (e.g. 4e-)? I presume that's what you meant when you said "we can't just have electrons coming from nowhere".
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    (Original post by Big-Daddy)
    Ah I see, thanks. The reduction of O2 was providing the difficulty.

    Is it a general rule that once you have a balanced ionic equation (i.e. not for oxidation or reduction individually but overall) it can no longer contain any lone electrons (e.g. 4e-)? I presume that's what you meant when you said "we can't just have electrons coming from nowhere".
    yes, electrons have to be transferred from the reducing agent to the oxidising agent.
 
 
 
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