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moongazer

  • 3 years ago

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

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  1. Callisto
    • 3 years ago
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    Oxidation number of ..?

  2. moongazer
    • 3 years ago
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    all elements there

  3. Callisto
    • 3 years ago
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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?

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

  5. moongazer
    • 3 years ago
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    yes, there is a typo :)

  6. Callisto
    • 3 years ago
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    Okay. What have you got for K?

  7. moongazer
    • 3 years ago
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    +1

  8. moongazer
    • 3 years ago
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    could you check my answers?

  9. moongazer
    • 3 years ago
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    I'll just post it :)

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

  11. moongazer
    • 3 years ago
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    I'll just answer it :)

  12. moongazer
    • 3 years ago
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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 ?

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

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

  15. Callisto
    • 3 years ago
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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.

  16. moongazer
    • 3 years ago
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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.

  17. moongazer
    • 3 years ago
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    Is it correct ?

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

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

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

  21. Callisto
    • 3 years ago
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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!

  22. moongazer
    • 3 years ago
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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 :)

  23. Callisto
    • 3 years ago
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    Nice :)

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

  25. moongazer
    • 3 years ago
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    please wait :)

  26. Callisto
    • 3 years ago
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    Take your time!

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

  28. moongazer
    • 3 years ago
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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.

  29. moongazer
    • 3 years ago
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    you will solve for the unknown oxidation number

  30. Callisto
    • 3 years ago
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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...

  31. Callisto
    • 3 years ago
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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.

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

  33. Callisto
    • 3 years ago
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    Yup :D

  34. moongazer
    • 3 years ago
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    I hope it is correct. :)

  35. moongazer
    • 3 years ago
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    Yey! thanks!

  36. Callisto
    • 3 years ago
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    Welcome :)

  37. Callisto
    • 3 years ago
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    There are three definitions....

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

  39. moongazer
    • 3 years ago
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    transfer of electrons?

  40. Callisto
    • 3 years ago
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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.

  41. moongazer
    • 3 years ago
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    and reduction is the opposite of that?

  42. Callisto
    • 3 years ago
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    Yup...

  43. moongazer
    • 3 years ago
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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 ?

  44. Callisto
    • 3 years ago
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    -H?!

  45. moongazer
    • 3 years ago
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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

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

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

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

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