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moongazer Group Title

K2Cr2O7 + S8 ----> K2O + Cr2O3 + SO2 (the numbers are subscripts) find the oxidation number

  • 2 years ago
  • 2 years ago

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  1. Callisto Group Title
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    Oxidation number of ..?

    • 2 years ago
  2. moongazer Group Title
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    all elements there

    • 2 years ago
  3. Callisto Group Title
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    K -> Group I metal. Either 0, or +1. Here, it's an ion. So, what is the oxidation number of K?

    • 2 years ago
  4. Callisto Group Title
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    Btw, I think it's K2Cr2O7 on the left?

    • 2 years ago
  5. moongazer Group Title
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    yes, there is a typo :)

    • 2 years ago
  6. Callisto Group Title
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    Okay. What have you got for K?

    • 2 years ago
  7. moongazer Group Title
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    +1

    • 2 years ago
  8. moongazer Group Title
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    could you check my answers?

    • 2 years ago
  9. moongazer Group Title
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    I'll just post it :)

    • 2 years ago
  10. Callisto Group Title
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    Yup. For both sides, there is no oxidation number change for K. So, K on both sides are +1. Sure :)

    • 2 years ago
  11. moongazer Group Title
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    I'll just answer it :)

    • 2 years ago
  12. moongazer Group Title
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    Isn't it that if the subscript of O is greater than 2, it will have an oxidation number of -2 ?

    • 2 years ago
  13. Callisto Group Title
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    O has an oxidation number of -2 - You're right! :)

    • 2 years ago
  14. moongazer Group Title
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    but if "O" only it will have -1 and O2 it will have 0?

    • 2 years ago
  15. Callisto Group Title
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    No, not really. O is in group 6, it need 2 more electrons. So, usually, its oxidation number is -2 when it is bonded to other atoms. If it is an element (i.e. O2), its oxidation number is 0 as element in its own form. But remember, it's just 'usually' when its oxidation number is of -2. There are some exceptions. For instance, if I remember correctly, O in F2O is +2 since F is a more electronegative element and it has an oxidation number of -1. Since there are two F, oxidation number of O has to be +2 in order to get a net oxidation number of the compound equal to zero.

    • 2 years ago
  16. moongazer Group Title
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    ohh, thanks for the info. Here are my answers :) K2(+1)Cr2(+6)O7(-2) + S8(0) ----> K2(+1)O(-2) + Cr2(+3)O3(-2) + S(0)O2(0) The oxidation number is enclosed by parenthesis.

    • 2 years ago
  17. moongazer Group Title
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    Is it correct ?

    • 2 years ago
  18. Callisto Group Title
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    Almost. S(0)O2(0) <- not correct (for both)

    • 2 years ago
  19. moongazer Group Title
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    K2(+1)Cr2(+6)O7(-2) + S8(0) ----> K2(+1)O(-2) + Cr2(+3)O3(-2) + S(-2)O2(+1)

    • 2 years ago
  20. moongazer Group Title
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    I made the oxidation number of O2 the unknown. Is it correct now?

    • 2 years ago
  21. Callisto Group Title
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    Not really. For SO2, O is -2 (you can take a look at O first as it is more electronegative than S!) O2 is NOT an element in this case, so O doesn't have a oxidation number of 0!

    • 2 years ago
  22. moongazer Group Title
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    K2(+1)Cr2(+6)O7(-2) + S8(0) ----> K2(+1)O(-2) + Cr2(+3)O3(-2) + S(+4)O2(-2) I think this is correct already :)

    • 2 years ago
  23. Callisto Group Title
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    Nice :)

    • 2 years ago
  24. moongazer Group Title
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    I'll say something. I'll just fix the laptop :)

    • 2 years ago
  25. moongazer Group Title
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    please wait :)

    • 2 years ago
  26. Callisto Group Title
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    Take your time!

    • 2 years ago
  27. moongazer Group Title
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    Whew! The battery almost got dead. It keeps on saying plugged in, not charging.

    • 2 years ago
  28. moongazer Group Title
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    So it means, If you have a molecule with an oxygen,to find the oxidation number of the elements in it. You will identify first the more electronegative element and made the charge of it equal to its oxidation number. Then make the oxidation number of the other element the unknown.

    • 2 years ago
  29. moongazer Group Title
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    you will solve for the unknown oxidation number

    • 2 years ago
  30. Callisto Group Title
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    Hmm.. you can say so. But when you consider electronegativity of the element, that would be talking about the group V - group VII For metal, it usually has positive oxidation number. If I remember correctly, for group I metal, when it does not exist in element state, it should be of an oxidation number of +1, and looks like it works for most of the group II metal too. But for transition metals, that's another case...

    • 2 years ago
  31. Callisto Group Title
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    One little practice for you (if you want one lol) Find the oxidation number of H, O and Cl in the compound HOCl.

    • 2 years ago
  32. moongazer Group Title
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    H(+1)O(-2)Cl(+1)

    • 2 years ago
  33. Callisto Group Title
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    Yup :D

    • 2 years ago
  34. moongazer Group Title
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    I hope it is correct. :)

    • 2 years ago
  35. moongazer Group Title
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    Yey! thanks!

    • 2 years ago
  36. Callisto Group Title
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    Welcome :)

    • 2 years ago
  37. Callisto Group Title
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    There are three definitions....

    • 2 years ago
  38. moongazer Group Title
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    is it the loss of electrons and reactions directly with O2 ?

    • 2 years ago
  39. moongazer Group Title
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    transfer of electrons?

    • 2 years ago
  40. Callisto Group Title
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    Definition of oxidation: 1. addition of oxygen to a substance 2. a process in which a substance loses electrons 3. a process in which the oxidation number of an element in a substance increases.

    • 2 years ago
  41. moongazer Group Title
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    and reduction is the opposite of that?

    • 2 years ago
  42. Callisto Group Title
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    Yup...

    • 2 years ago
  43. moongazer Group Title
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    In my notes it says below oxidation it says: +O -H What does "H" have to do with oxidation and reduction ?

    • 2 years ago
  44. Callisto Group Title
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    -H?!

    • 2 years ago
  45. moongazer Group Title
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    idk I just saw in my notes. I was late during our class so I didn't know what is H there

    • 2 years ago
  46. moongazer Group Title
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    Anyway, thank you very much in your time and effort helping me. :)

    • 2 years ago
  47. Callisto Group Title
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    You're welcome.. Sorry that I couldn't help :(

    • 2 years ago
  48. moongazer Group Title
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    It's alright you helped me a lot. :)

    • 2 years ago
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