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Where did oxygen come from?

  1. Feb 4, 2007 #1
    1. The problem statement, all variables and given/known data
    Consider the reaction between Sulfiric acid and zinc:
    H2SO4 + Zn ->
    What would be the ionic equation for this reaction?

    2. Relevant equations

    3. The attempt at a solution
    In solution it would be
    H2+ + (SO4)2- + Zn -> Zn2+ + 2e- + (SO4)2- + H2+ -> Zn2+ + H2 (in ionic form)

    but instead the answers gave Zn2+ + H2O
    where did the O come from?
  2. jcsd
  3. Feb 4, 2007 #2


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    From the O4? Why is this a problem?
  4. Feb 4, 2007 #3
    If that is the case than shouldn't they show (SO4)2- on the left and (SO3)2- on the right?

    ions can only be cancelled if they do not participate in the reaction.
    Last edited: Feb 4, 2007
  5. Feb 4, 2007 #4


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    But your answer to the equation seems to have LOST the SO4.

    Zn + (SO4) + H2 -> Zn + H2

    Wouldn't one of the products be SO2? leaving the rest of the O to combine with the H2?

    I guess I didn't study this. Nevermind me.
  6. Feb 5, 2007 #5


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    Are you sure the question wasn't:

    H2SO4 + ZnO --> Zn+2 + (SO4)-2 + H2O????

    If it isn't, you have found an error in your textbook!

    As you wrote it, Zn + H2SO4 --> H2(g) + ZnSO4
    is the correct answer. You might write the zinc sulfate as Zn+2 and (SO4)-2, but zinc will not react with sulfuric acid to produce water!

    Remember that bases react with acids to produce water (bases like ZnO) not zero valent metals!
  7. Feb 5, 2007 #6
    What about (SO4)2- turning to (SO3)2- which will donate the oxygen atom.

    The textbook worded zinc and not zincoxide in the reactants.
    Last edited: Feb 5, 2007
  8. Feb 6, 2007 #7


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    If that were true, zinc sulfate solutions would either spontaneously produce sulfurous acid and zinc oxide (H2SO3 and Zn+2) or excess zinc would react with sulfate solutions to produce sulfurous acid.

    Neither of these two happens.

    You can write all of the half reactions and see for yourself:

    Zn --> Zn+2 + 2e-
    2H+ + 2e- --> H2(g)

    (SO4)-2 doesn't participate in this reaction so,

    Zn + 2H+ + (SO4)-2 --> Zn+2 + H2(g) + (SO4)-2

    If you want to reduce the products further (SO4 to SO3 for example) you would have to apply a reducing potential to the products. Applying a reducing potential to the products would reduce Zn+2 to Zn metal first since it is the most easily reduced species in solution.

    ZnSO4 7H20 + 2e- --> Zn + (SO4)-2 + 7H20 (-0.7993V)
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