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Oxidation and Reduction

  1. May 30, 2006 #1
    At this moment, I'm on balancing redox equations. I'm stuck on step one though...:redface:

    1. Assign oxidation numbers to all atoms in the equation...

    S + HNO3 --> SO2 + NO + H2O

    This particular problem is an example in my Chem book so I already know the answers. I just don't understand why they are what they are. I understand this part: All free, uncombined elements have an oxidation number of zero. ...so that's simple...it's the rest I'm not getting.

    The first part that throws me is the HNO3 part. Why is N +5?
     
  2. jcsd
  3. May 30, 2006 #2

    Hootenanny

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    Your chemistry book should list a series of 'rules' for assigning oxidation states. One of them is Oxygen when not in the O2 or O22- or OF2 form always has an oxidation state of -2. The next one is hydrogen always has an oxidation state of +1 unless in the H2 form or in the H- form. An easy way to remember this is usually the oxidation state is the same as the standard elementry ion.

    Do you follow?

    ~H
     
  4. May 30, 2006 #3
    Ah, found the "rules list". Stupid...:rolleyes:

    Okay, on to an actual question then...:rofl: When first learning to balance equations, we learned that the number of atoms of each element in the products and reactants must be equivalent. What are some additional factors that must be taken into account when balancing equations for redox reactions?

    The question says "What are some additional factors..." but I'm only seeing one factor listed here and that's "the number of electrons lost in an oxidation process must equal the number of electrons gained in the reduction process" and that's the end of the chapter, basically.
     
  5. May 30, 2006 #4

    Hootenanny

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    Well, they are the only two rules I have used, but then again I'm not a chemistry 'specialist', so you might want to wait for GCT or a chemistry Homework Helper to come online before you fill your answer sheet in.

    Sorry I couldn't have been more of a help.

    ~H
     
  6. May 30, 2006 #5
    That's alright. :smile:

    While I'm waiting for help with that question... :biggrin:

    What two aspects of the half-reaction equations must be balanced?

    The number of atoms and the charge?
     
  7. May 30, 2006 #6

    Hootenanny

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    I'd go with that but I would also add number of electrons, but that will probably be covered with charge and atoms.

    ~H
     
  8. May 30, 2006 #7

    Borek

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    balancing by oxidation numbers method

    Charge must be balanced, and that's covered by the electrons gained=electrons lost rule.

    However, it is hardly a rule specific for redox reactions - in the net ionic reactions charge must be balanced as well, if it is not balanced reaction equation is not balanced. Although if there is no redox (charge transfer) process present once the atoms are balanced charge is usually balanced too.
     
  9. Jan 2, 2008 #8
    May I offer an answer?

    Starting with the original equation:

    S + HNO3 --> SO2 + NO + H2O

    Let's start with HNO3. If you already get this part, sorry....

    Lists of priorities for redox equations can vary, however, if your list does not have this, it might be helpful to learn this:there are 3exceptions: F,H,O
    Think of the exceptions of F, H, O like this:
    1. F=-1
    2. H= +1
    3. O= -2
    If you have to break any of these rules, break the lower down ones.
    When dealing with an equation that has one or more exception molecule (H,O,F), plug the oxidation value for the higher one in first, then solve the rest

    Here, plug in H first, since it will take precedence over O and N. It takes precedence over N because if your need to choose between breaking F/O/H and anything else, say N for example, choose the something else.
    SO, H NO3 will have:
    +1 ? ?
    Then, since O is the next exception, get O. since O is near always-2, total the Os to be -6, with a -2 charge each
    Since the charges have to balance, H +N will equal +6
    If H is already +1, then N has to be +5, since 6-1=5

    While this question didn't directly break the charges of O or H, if you try to give N precedence over H, the charge for H goes crazy.

    Ok, the rest:
    S + H N O3 --> S O2 + N O + H2 O
    N +1+5-2 +4-2 +2-2 +1 -2
    S will be N(neutral, or 0) since it is unaccompanied
    HNO3 already covered, SO2: plug in -2 for O, S has to equal -2(2) so as to cancel out O
    NO: plug in -2 for O, N has to be +2 to equal O
    H2O: Plug in +1 for H first, then since its H2, O has to cancel out +1(2), so is -2

    OH, and sorry if the explanation was a little detail heavy. It's not that I think anyone really needs this much detail, it's just that this is something I struggled with since my teachers didn't give me the detail, and I don't want that to happen to anyone else. ^_^
     
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