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Redox reactions

  1. Oct 26, 2005 #1
    I had some trouble balancing these using the "ion electron method", ordinarily I'd balance the half reactions (excluding H and O), add H20 to compensate for O, H+ to compensate for H, and electrons to compensate for charge.

    P4(s) + NO3-(aq) ---> H2PO4-(aq) + NO(g) (in an acidic solution)

    2CuS(s) + HNO3 --> Cu(NO3)2 + H20 + NO (g) + S (in an acidic solution)

    ClO2(aq) + OH-(aq) + ---> ClO3-(aq) + Cl-(aq) + H2O(l) (in a basic solution)

    1) how should the half reactions of these chemical equations look ?

    2) aren't all types of chemical reactions also redox-reactions(chemical reactions in which a transfer of electrons occur) ?
    Last edited by a moderator: Oct 26, 2005
  2. jcsd
  3. Oct 26, 2005 #2
    1) a) P4 --> H2PO4
    NO3- --> NO

    b) CuS --> Cu+2 + S
    HNO3 --> NO

    c) ClO2 basically oxides itself:
    ClO2 --> ClO3-
    ClO2 --> Cl-

    2) No e.g. In double displacement reactions oxidation numbers will always stay the same.

    But lots of reactions are like the combustion of hydrogen and fluorine

    H2 + F2 --> 2HF

    H2 → 2H+ + 2e-
    + 2e- + F2 → 2F-
    H2 + F2 → 2H+ + 2F-

    Even though you would never review this equation, when learning about redox.
    Last edited: Oct 26, 2005
  4. Oct 27, 2005 #3
    for b) what happened to the NO3- and H2O on the right ?

    for c) what happened to the OH- on the left and the H2O on the right ?

    also what if you are given a half reaction deffieicent of an element other than H and O, would you add ions of that element like for ex if you where defficient in Na on one side do you add Na+ to the other ?
    Last edited by a moderator: Oct 27, 2005
  5. Oct 27, 2005 #4


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    Staff: Mentor

    All three equations can be balanced without using redox approach.

    In the first case you must add water on the left and H+ on the right.

    Second and third are ready to be balanced without further modifications.
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