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    My CGP As-Chemistry book says Oxidation is Exothermic, which is confusing me a little.

    It gives an example of the combustion of Methane CH4 + 2O2 = CO2 + 2H2O

    Firstly I can't work out what has been oxidised and what has been reduced in this reaction.

    I know that oxygen is more electronegative than carbon, so am I correct in thinking that the carbon is oxidised and the oxygen is reduced?

    Surely then according to CGP in this reaction the Oxygen has been reduced and is endothermic, and the carbon has been oxidised which is exothermic?

    So confused
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    you can look at the oxidation numbers to figure out which one has been reduced and which has been oxidised
    carbon in methane equals -4 and carbon in the RHS in co2 is +4 and so its oxidation number has increased therfore its been oxidised
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    You can't have oxidation without reduction (think about it). Are all redox reactions exothermic?

    All combustion reactions are exothermic.
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    (Original post by Pigster)
    You can't have oxidation without reduction (think about it). Are all redox reactions exothermic?

    All combustion reactions are exothermic.
    I understand your first point, but as for 'Are all redox reactions exothermic?', my guess would be they are both exothermic and endothermic right...?

    And yes all combustion reactions must be exothermic as heat is released from the breaking up of the bonds.

    This doesn't answer my question though, the book says Oxidation is exothermic, therefore is reduction endothermic? :s
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    (Original post by Sam00)
    I understand your first point, but as for 'Are all redox reactions exothermic?', my guess would be they are both exothermic and endothermic right...?

    And yes all combustion reactions must be exothermic as heat is released from the breaking up of the bonds.

    This doesn't answer my question though, the book says Oxidation is exothermic, therefore is reduction endothermic? :s
    It doesn't make sense to say oxidation is exothermic and reduction is endothermic as exo and endothermic refer to the total process, and a process which involves oxidation must also involve reduction. You could talk about the process of ionisation, that then would make sense. If oxidation is defined in the crudest way as the loss of electrons, then hell no, ionisation energies are always endothermic.

    Oxidation of hydrocarbons is always exothermic because the reduction of O2 in this case releases a lot of energy. Forming CO2 and Water is an extremely exothermic process due to the obvious stability of the two compounds. However, oxidation of many compounds can be endothermic. For example, the disproportionation of Vanadium 3 is endothermic even though Vanadium 3 is oxidised to Vanadium 4(there are other examples I can't remember).
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    (Original post by oShahpo)
    If oxidation is defined in the crudest way as the loss of electrons, then hell no, ionisation energies are always endothermic.
    What about the oxidation of e.g. O2-(g) to O-(g)?
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    (Original post by oShahpo)
    It doesn't make sense to say oxidation is exothermic and reduction is endothermic as exo and endothermic refer to the total process, and a process which involves oxidation must also involve reduction. You could talk about the process of ionisation, that then would make sense. If oxidation is defined in the crudest way as the loss of electrons, then hell no, ionisation energies are always endothermic.

    Oxidation of hydrocarbons is always exothermic because the reduction of O2 in this case releases a lot of energy. Forming CO2 and Water is an extremely exothermic process due to the obvious stability of the two compounds. However, oxidation of many compounds can be endothermic. For example, the disproportionation of Vanadium 3 is endothermic even though Vanadium 3 is oxidised to Vanadium 4(there are other examples I can't remember).
    Ok firstly I get that all reactions are oxidation/reduction, and that all combustion reactions are exothermic due to the breaking up of bonds right?

    Secondly I get that ionisation energies are always endothermic, would photosynthesis be an example of this whereby O2 and C6H12O6 is produced? Thus the CO2 and H2O have been oxidised and reduced to form bonds of the products?

    Could you please further explain why the forming and CO2 and Water is extremely exothermic? edit: or is this just because they're created from the combustion of the reactants?
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    (Original post by Pigster)
    What about the oxidation of e.g. O2-(g) to O-(g)?
    It doesn't make sense to specify an oxidation process without specifying what's reducing it. Are you taking about the energy required to go from O2- to O-? You might want to look up the "ionisation energy", not oxidation. Look up the ionisation energy from 0 to 2-, then look up that of 0 to -1 and take the second away from the first to get your total enthalpy.
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    (Original post by oShahpo)
    It doesn't make sense to specify an oxidation process without specifying what's reducing it. Are you taking about the energy required to go from O2- to O-? You might want to look up the "ionisation energy", not oxidation. Look up the ionisation energy from 0 to 2-, then look up that of 0 to -1 and take the second away from the first to get your total enthalpy.
    For example I know that the 'Oxidation' of glucose in respiration is Exothermic, bonds are broken and energy is released.....however this isn't what I am asking, it's the concept / bigger picture I am struggling to grasp!

    Is the Glucose Oxidised by O2? Therefore the O2 is Reduced?....
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    (Original post by oShahpo)
    It doesn't make sense to specify an oxidation process without specifying what's reducing it. Are you taking about the energy required to go from O2- to O-? You might want to look up the "ionisation energy", not oxidation. Look up the ionisation energy from 0 to 2-, then look up that of 0 to -1 and take the second away from the first to get your total enthalpy.
    Oxidation is the less of electrons
    Going from O2- to O- involves the loss of an electron, so it oxidation.
    The second electron gain enthalpy of oxygen (O-(g) + e- -> O2-(g) is endothermic.
    Therefore the removal of that e- (O2- -> O-(g) + e-) is exothermic
    Ergo, we have an example of an oxidation reaction that is exothermic

    For the record, ionisation enthalpy always involves the loss of e- so cannot be used to describe "0 to 2-"
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    (Original post by Sam00)
    Ok firstly I get that all reactions are oxidation/reduction, and that all combustion reactions are exothermic due to the breaking up of bonds right?

    Secondly I get that ionisation energies are always endothermic, would photosynthesis be an example of this whereby O2 and C6H12O6 is produced? Thus the CO2 and H2O have been oxidised and reduced to form bonds of the products?

    Could you please further explain why the forming and CO2 and Water is extremely exothermic? edit: or is this just because they're created from the combustion of the reactants?
    Let me try to explain to you a few things from the start.
    First of, there is a difference between redox and ionisation. Oxidation is the loss of electron from one species to another specified species. While ionisation is just the loss of electrons from one species. It always takes energy to remove the electron (ionisation is always endothermic), but when the species that takes the electron (the reduced species) releases more energy, the total enthalpy is negative and the reaction is exothermic. Thus, when you say oxidation of say, glucose, you must specify what's being reduced in order to determine the energy released.

    The reason we tend to say that oxidation of glucose or hydrocarbons releases energy is that in almost 100% of the cases, oxygen in the form of O2 is being reduced to O2- in water or carbon dioxide. Water and Carbon Dioxide are extremely stable species, and thus when oxygen (O2) and the hydrogen and carbon of the hydrocarbon combine to form the water and carbon dioxide, a lot of energy is released.

    It is also important to realise that in redox, electrons aren't really lost, rather shared unevenly.
    To make it simpler, just think of redox as an exchange of electrons that sometimes can release energy if the exchange is exo, or take energy if the exchange is endo. When you say oxidation is loss, just think that oxidation is when the element almost gives up its electrons to another specified species.
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    (Original post by Sam00)
    Ok firstly I get that all reactions are oxidation/reduction
    Na2CO3(aq) + CaCl2(aq) -> 2NaCl(aq) + CaCO3(s) involves no oxidation (and therefore no reduction).
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    In fact, let me one thing clear.

    Oxidation is NOT loss of electrons, if it were, ionic compounds would be formed. Oxidation is the giving away? of electrons such that the oxidised species gives up it's electron for most but not all the time. Therefore, considering ionisation energies in redox reactions is absurd as no ionisation takes place in most redox reactions (sometimes they do, as with transition metals). What you should consider is that, will the species that gives up it's electrons be more stable? and will the species that takes the electrons be stable? will the overall compound be stable?
    Again, the process of oxidation is always met with reduction as it's, again, not losing electrons, but unequally sharing electrons. Oxidation of hydrocarbons is exothermic for reasons I stated above. Sometimes, a compound would be less stable if it unequally shared it's electron, like the oxidation of copper (0) to copper (1).
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    (Original post by oShahpo)
    Let me try to explain to you a few things from the start.
    First of, there is a difference between redox and ionisation. Oxidation is the loss of electron from one species to another specified species. While ionisation is just the loss of electrons from one species. It always takes energy to remove the electron (ionisation is always endothermic), but when the species that takes the electron (the reduced species) releases more energy, the total enthalpy is negative and the reaction is exothermic. Thus, when you say oxidation of say, glucose, you must specify what's being reduced in order to determine the energy released.

    The reason we tend to say that oxidation of glucose or hydrocarbons releases energy is that in almost 100% of the cases, oxygen in the form of O2 is being reduced to O2- in water or carbon dioxide. Water and Carbon Dioxide are extremely stable species, and thus when oxygen (O2) and the hydrogen and carbon of the hydrocarbon combine to form the water and carbon dioxide, a lot of energy is released.

    It is also important to realise that in redox, electrons aren't really lost, rather shared unevenly.
    To make it simpler, just think of redox as an exchange of electrons that sometimes can release energy if the exchange is exo, or take energy if the exchange is endo. When you say oxidation is loss, just think that oxidation is when the element almost gives up its electrons to another specified species.
    Ok thanks this really helps, but I am not great at Chem so I need to make what you've said simpler for me to understand.

    In the simple case of the combustion of Ethane (C2H6 + O2 = CO2 + H2O) the ionisation would concern the 'energy put in to combust the Ethane (Endothermic)?

    Therefore the Ethane is Oxidised (i.e it is broken up, and electrons are lost to the Oxygen)

    However, the O2 gains electrons from the Reaction and so is Reduced?

    During this reaction, more energy is lost than is gained, and so the Enthalpy Change is Negative, therefore it's an overall Exothermic Reaction?

    So this is generally why Oxidation is exothermic (as the CGP book stated)

    I am just trying to make sense of this the best I can
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    (Original post by Pigster)
    Na2CO3(aq) + CaCl2(aq) -> 2NaCl(aq) + CaCO3(s) involves no oxidation (and therefore no reduction).
    I will say again that Chemistry is a weak subject for me.

    Could you explain rather than just show me an equation
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    (Original post by Pigster)
    Oxidation is the less of electrons
    Going from O2- to O- involves the loss of an electron, so it oxidation.
    The second electron gain enthalpy of oxygen (O-(g) + e- -> O2-(g) is endothermic.
    Therefore the removal of that e- (O2- -> O-(g) + e-) is exothermic
    Ergo, we have an example of an oxidation reaction that is exothermic

    For the record, ionisation enthalpy always involves the loss of e- so cannot be used to describe "0 to 2-"
    Oxidation is not always ionisatoin. Oxidation of alcohols to form aldehydes and ketones does not involve any ionisation at all.
    Even if it does involve ionisation, as you just stated with your example, you CAN'T just consider the oxidised species without considering the other species too. If the electron is taken in by something like Caesium, the total reaction might be endothermic.

    Btw, hydrogen peroxide, which has oxygen -1, will always want to form water, the oxygen will always want to be reduced. Thus it's irrelevant to say whether this oxidation process is endo or exo. I also doubt the process of going from oxygen 2 to oxygen 1 is endothermic, it's definitely exothermic. Oxygen 1 rarely exists in nature without being reduced immediately to oxygen 2.
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    (Original post by Sam00)
    Ok thanks this really helps, but I am not great at Chem so I need to make what you've said simpler for me to understand.

    In the simple case of the combustion of Ethane (C2H6 + O2 = CO2 + H2O) the ionisation would concern the 'energy put in to combust the Ethane (Endothermic)?

    Therefore the Ethane is Oxidised (i.e it is broken up, and electrons are lost to the Oxygen)

    However, the O2 gains electrons from the Reaction and so is Reduced?

    During this reaction, more energy is lost than is gained, and so the Enthalpy Change is Negative, therefore it's an overall Exothermic Reaction?

    So this is generally why Oxidation is exothermic (as the CGP book stated)

    I am just trying to make sense of this the best I can
    There is no ionisation when it comes to burning ethane, only unequal sharing of electrons. So the electron is not completely ripped off the atom, just taken away from it for most of the time. As long as you keep that in mind, everything else you said is correct. It takes a bit of energy to pull the electrons away from the carbon and the hydrogen, but once oxygen takes those electrons AND form the water molecule and the carbon dioxide molecule, a lot of energy is released. Remember, oxygen is a really tiny, highly electronegative species (it really, really wants some electrons) and thus when it gets them, she's happy as a mermaid eating a large pizza..
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    (Original post by oShahpo)
    Oxidation is not always ionisatoin. Oxidation of alcohols to form aldehydes and ketones does not involve any ionisation at all.
    Even if it does involve ionisation, as you just stated with your example, you CAN'T just consider the oxidised species without considering the other species too. If the electron is taken in by something like Caesium, the total reaction might be endothermic.

    Btw, hydrogen peroxide, which has oxygen -1, will always want to form water, the oxygen will always want to be reduced. Thus it's irrelevant to say whether this oxidation process is endo or exo. I also doubt the process of going from oxygen 2 to oxygen 1 is endothermic, it's definitely exothermic. Oxygen 1 rarely exists in nature without being reduced immediately to oxygen 2.
    I'd read it carefully before trying to use it against me: http://goldbook.iupac.org/O04362.html
    Why can't you consider oxidation in isolation?
    H2O2 can be reduced OR oxidised depending on whether you add a more powerful oxidising or reducing agent.
    You can doubt all you like about the second electron gain enthalpy. Or you could look it up. My data book says +844 kJ mol-1. The Law of conservation of energy states that the reverse process, i.e. O2- -> O- + e- must be -844 kJ mol-1, i.e. exothermic. Even if you quibble over the exact value, the sign is the important bit.
    Why pick Cs? The electron gain enthalpy for Li is much higher and therefore more likely to lead to an exothermic overall reaction!
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    (Original post by Sam00)
    My CGP As-Chemistry book says Oxidation is Exothermic, which is confusing me a little.

    It gives an example of the combustion of Methane CH4 + 2O2 = CO2 + 2H2O

    Firstly I can't work out what has been oxidised and what has been reduced in this reaction.

    I know that oxygen is more electronegative than carbon, so am I correct in thinking that the carbon is oxidised and the oxygen is reduced?

    Surely then according to CGP in this reaction the Oxygen has been reduced and is endothermic, and the carbon has been oxidised which is exothermic?

    So confused
    To try to be a little more helpful to the OP...

    I suspect that the book has a typo. Oxidation reactions aren't always exothermic, I think they meant to put combustion. These, as I said, are always exothermic.

    There are various ways to work out what is oxidised/reduced. Can you work out oxidation numbers? Oxidation involves the increase in o.n. The addition of O (unless F2 is involved) involves oxidation. Ditto the loss of H.

    In the combustion of CH4 example you quote, C is indeed oxidised and O is reduced (C loses e- to O in terms of o.n.).

    When you learn about Born Haber cycles, you'll see there is more to calculating enthalpy changes in reactions than enthalpy changes involved in the loss and gain of e-.
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    (Original post by Pigster)
    I'd read it carefully before trying to use it against me: http://goldbook.iupac.org/O04362.html
    Why can't you consider oxidation in isolation?
    H2O2 can be reduced OR oxidised depending on whether you add a more powerful oxidising or reducing agent.
    You can doubt all you like about the second electron gain enthalpy. Or you could look it up. My data book says +844 kJ mol-1. The Law of conservation of energy states that the reverse process, i.e. O2- -> O- + e- must be -844 kJ mol-1, i.e. exothermic. Even if you quibble over the exact value, the sign is the important bit.
    Why pick Cs? The electron gain enthalpy for Li is much higher and therefore more likely to lead to an exothermic overall reaction!
    There is a difference between ionisation and oxidation.There is a relationship between them, but they're two different processes. The equation you gave makes no sense when your considering redox between Hydrogen Peroxide and Ozone for example. There are no real ions in this reaction, O2- and O-1 for all intents do not exist in this reaction at all.
    If it was this easy, just considering ionisation enthalpies, then why did we need to establish electrode potential tables instead of just using ionisation enthalpies? http://hyperphysics.phy-astr.gsu.edu.../electpot.html
    Again, I repeat my point, oxidation =/ ionisation, thus you can't study a certain specific oxidation process by considering ionisation energies.
 
 
 
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