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anonymous

  • one year ago

Hey guys! I'm doing unit review in Chemistry and I'm stuck on how to balance redox equations using the oxidation-number change method. It's from section 20.3 Balancing Redox Equations in the California Prentice Hall Chemistry book. The equation is: Bi2S3 (s) + HNO3 (aq) -> Bi(NO3)3 (aq) + NO (g) + S (s) + H2O (l). I know the oxidation numbers for most of them: S3 (s): -2 H: +1 / N: +5 / O3: -2 N: +5 / O3: -2 N: +2 / O: -2 S: 0 H2: +1 / O: -2 I guess my confusion is on what the oxidation numbers for Bi are and then the steps to balance it once I get all the oxidation numbers.

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  1. anonymous
    • one year ago
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    The answer to the problem is: Bi2S3 (s) + 8HNO3 (aq) -> 2Bi(NO3)3 (aq) + 2NO(g) + 3S (s) + 4H2O (l)

  2. anonymous
    • one year ago
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    Thank you very much! I'll read it right now. :)

  3. anonymous
    • one year ago
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    Wait, I know how to balance chemical equations but this problem makes us balance redox equations using the oxidation numbers. Not the moles..

  4. cuanchi
    • one year ago
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    Hi @rheaanderson341 the oxidation state of the Bi is +3 but doesnt matter because it is an spectator ion, it doesnt get reduce neither oxidize

  5. cuanchi
    • one year ago
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    the S is going from -2 to 0 and then is get oxidized the N goes from +5 to +2 and then its reduce

  6. anonymous
    • one year ago
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    The N is from the HNO3 and the NO right? So the oxidation number on the other N on the product side Bi(NO3)3 doesn't matter?

  7. cuanchi
    • one year ago
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    no it is also spectator there, doesn't matter

  8. anonymous
    • one year ago
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    Oh okay. Then it says the next step to balancing redox equations is to balance the increase and decrease in oxidation numbers. I don't understand what they mean?

  9. cuanchi
    • one year ago
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    What I do is to use the method of the half reactions. I separate the oxidation and reduction reactions and balance them Bi2S3 -> 3S+2Bi+3 + 6e (3e + 3H+ + HNO3 ->HO + 2H2O) x2 _________________________________________________

  10. anonymous
    • one year ago
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    Where did the 6e and 3e come from?

  11. cuanchi
    • one year ago
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    you have 2Bi+3 it hast 3 positive charge and you have 2 of them = 6+ charge, to balance you have to put electrons so you put 6 e- you have 3H+ then you add 3 e-

  12. anonymous
    • one year ago
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    So what side would the 6 electrons go on? Also the 3H+ comes from the HNO3 correct? Sorry this is all so confusing I'm trying to understand bit by bit lol. The book made it so confusing and since I'm homeschooled I'm stuck trying to teach myself. ;S

  13. cuanchi
    • one year ago
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    you should start with more simple reactions, this is a complicated one look at this tutorial http://chemwiki.ucdavis.edu/Analytical_Chemistry/Electrochemistry/Redox_Chemistry/Balancing_Redox_reactions they start with a Cu(s)+2Ag+(aq)→Cu2+(aq)+2Ag(s) then Cu+(aq)+Fe(s)→Fe3+(aq)+Cu(s) then Cr2O2−7(aq)+HNO2(aq)→Cr3+(aq)+NO−3(aq)

  14. cuanchi
    • one year ago
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    this are the rules for balancing the half reactions Each equation is balanced by adjusting coefficients and adding H2O, H+, and e- in this order: Balance elements in the equation other than O and H. Balance the oxygen atoms by adding the appropriate number of water (H2O) molecules to the opposite side of the equation. Balance the hydrogen atoms (including those added in step 2 to balance the oxygen atom) by adding H+ ions to the opposite side of the equation. Add up the charges on each side. Make them equal by adding enough electrons (e-) to the more positive side. (Rule of thumb: e- and H+ are almost always on the same side.) The e- on each side must be made equal; if they are not equal, they must be multiplied by appropriate integers (the lowest common multiple) to be made the same. The half-equations are added together, canceling out the electrons to form one balanced equation. Common terms should also be canceled out. (If the equation is being balanced in a basic solution, through the addition of one more step, the appropriate number of OH- must be added to turn the remaining H+ into water molecules.) The equation can now be checked to make sure that it is balanced.

  15. anonymous
    • one year ago
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    So the oxidation half reaction would be S3(s) -> S(s) and the reduction half reaction HNO3(aq) -> NO(g) ?

  16. cuanchi
    • one year ago
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    S3-6(s) -> S(s)

  17. cuanchi
    • one year ago
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    \[S _{3}^{-6}{(s)} \rightarrow S _{(s)}\]

  18. anonymous
    • one year ago
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    And it's -6 because of the \[S _{3}\] and the -2?

  19. cuanchi
    • one year ago
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    yes the oxidation state is -2 and you have 3 atoms (-2)x 3 = -6

  20. anonymous
    • one year ago
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    Would I do the same for the HNO3 -> NO ?

  21. cuanchi
    • one year ago
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    I found this in yahoo answers that may help you to follow your procedure, in this case they dont use the two half reactions https://answers.yahoo.com/question/index?qid=20080504150552AAFHraO

  22. cuanchi
    • one year ago
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    balance the O with H2O put 2H2O in the right side, then the H with H+, you have 4H in the water and one in the HNO3 then add 3H+ in the left side finally balance the charge with e-

  23. anonymous
    • one year ago
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    So I'm balancing the O in HNO3 on the left side with H2O on the right side?

  24. cuanchi
    • one year ago
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    yes

  25. cuanchi
    • one year ago
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    then you balance the H with H+

  26. anonymous
    • one year ago
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    So so far it looks like 3HNO3 and 2H2O?

  27. cuanchi
    • one year ago
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    \[3e ^{-}+ 3H ^{+}+HNO _{3} \rightarrow NO+2H _{2}O\]

  28. cuanchi
    • one year ago
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    no with H+ only no with HNO3

  29. nincompoop
    • one year ago
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    @Jhannybean show them how it is done Ms. Chemist

  30. Jhannybean
    • one year ago
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    lmao @Cuanchi has got this.

  31. nincompoop
    • one year ago
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    I think it can be made simpler tho

  32. anonymous
    • one year ago
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    That one is so difficult... the other equation I have to do is \[SbCl _{5} + Kl -> SbCl _{3} + KCl + I _{2}\]

  33. anonymous
    • one year ago
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    KI as in capital i not l.

  34. cuanchi
    • one year ago
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    \[6NO _{3}^{-}+Bi _{2}S _{3}\rightarrow 3S+2Bi(NO _{3})_{3}+6e ^{-}\] \[(3e ^{-}+3H ^{+}+HNO _{3}\rightarrow NO+2H _{2}O) 2\] ---------------------------------------------------------------------- \[6NO _{3}^{-}+ Bi _{2}S _{3}+6H ^{+}+2HNO _{3}\rightarrow 3S+2Bi(NO _{3})_{3}+2NO+4H _{2}O\] you can combine the 6H+ with the 6NO3- and get 6 HNO3 add the 6 HNO3 plus the 2 HNO3 and you have the 8 HNO3

  35. cuanchi
    • one year ago
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    the KI in the equation doesn't have any charge or you put a negative charge KI-?

  36. anonymous
    • one year ago
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    Kl has no charge it was meant to be the arrow -> my apologies.

  37. cuanchi
    • one year ago
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    the Sb is +5 goes to +3 (reduce) the I goes from -1 to O (oxidize)

  38. anonymous
    • one year ago
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    Okay. so now that I know it reduces -2 and oxidizes + 1, I balance the increase and decrease in oxidation numbers?

  39. anonymous
    • one year ago
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    This is the example they show me.. I see that they multiply the reduction and oxidation numbers together?

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  40. cuanchi
    • one year ago
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    2KI -> I2 + 2 K+ +2e- 2e- + SbCl5 -> SbCl3 + 2Cl- ------------------------------------- 2KI+SbCl5 -> I2 + 2K+ + SbCl3 + 2Cl-

  41. anonymous
    • one year ago
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    The -2 electrons came from the -2 times the +1 right? I think I'm understanding this now..

  42. nincompoop
    • one year ago
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    do you have a periodic table? what I can do is make the big picture a little clearer where those numbers come from

  43. anonymous
    • one year ago
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    Yes. I have my periodic table right in front of me. :)

  44. cuanchi
    • one year ago
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    OK! you have -2 x 1 = -2 then 1SbCl5 +1 x (2)= 2 then 2 KI

  45. nincompoop
    • one year ago
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    perhaps we can back track a little on what oxidation states and ground states are

  46. cuanchi
    • one year ago
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    you multiply the number of electron that loose the oxidized by the number of electrons that gain the reduced and vice-verse to get the same number of electrons. Then you put the numbers as coeficients

  47. anonymous
    • one year ago
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    Yes please @nincompoop lol. And oh.. okay I kept seeing the whole multiplying thing but wasn't sure what they were talking about.. let me look back at what you said and see fi I can understand this better now.

  48. cuanchi
    • one year ago
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    @nincompoop do you have any easier way to balance redox reactions than the half reactions method?

  49. nincompoop
    • one year ago
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    reactions is all about the movement of electrons so I always start with that

  50. anonymous
    • one year ago
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    Okay, that will work I believe. I mean, I understand the whole oxidation numbers and where they get them from because of where they are in the periodic table. For example, o is -2 because it's two away from the 8A group. Na is +1 and so on so forth. It's mainly the part of using the oxidation numbers and the reduce and oxidation numbers, multiplying them, then using the electrons to balance the equation. The SbCl equation seemed easier to understand than the previous equation.

  51. anonymous
    • one year ago
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    My teacher is going to go through it with me before I take the chapter test. Thank you guys so much though! I definitely learned a thing or two speaking with you all. I understand it much more than I did when I wrote this question, lol.

  52. nincompoop
    • one year ago
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    whenever you're ready to go over with oxidation and ground states

  53. cuanchi
    • one year ago
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    @rheaanderson341 can you tell me from what study guide are you getting this method? The page that you post it before what is the Author or name of the guide? Thank you!

  54. nincompoop
    • one year ago
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    http://www.bsqm.org.mx/PDFS/V2/N3/2-Ice%20B%20Risteski.pdf

  55. anonymous
    • one year ago
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    @Cuanchi it is from the Prentice Hall California Chemistry book.. it's on Chapter 20.2 or Chapter 20.3

  56. nincompoop
    • one year ago
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    Prentice review is good, but any chemistry textbook is probably better just because the flow of information is much more systematically constructed.

  57. anonymous
    • one year ago
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    Yeah, we always do our unit reviews from the book. Some are easier, rest are much more confusing!

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