## 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.

1. anonymous

The answer to the problem is: Bi2S3 (s) + 8HNO3 (aq) -> 2Bi(NO3)3 (aq) + 2NO(g) + 3S (s) + 4H2O (l)

2. anonymous

Thank you very much! I'll read it right now. :)

3. anonymous

Wait, I know how to balance chemical equations but this problem makes us balance redox equations using the oxidation numbers. Not the moles..

4. anonymous

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. anonymous

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

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. anonymous

no it is also spectator there, doesn't matter

8. anonymous

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. anonymous

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

Where did the 6e and 3e come from?

11. anonymous

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

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. anonymous

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. anonymous

15. anonymous

So the oxidation half reaction would be S3(s) -> S(s) and the reduction half reaction HNO3(aq) -> NO(g) ?

16. anonymous

S3-6(s) -> S(s)

17. anonymous

$S _{3}^{-6}{(s)} \rightarrow S _{(s)}$

18. anonymous

And it's -6 because of the $S _{3}$ and the -2?

19. anonymous

yes the oxidation state is -2 and you have 3 atoms (-2)x 3 = -6

20. anonymous

Would I do the same for the HNO3 -> NO ?

21. anonymous

22. anonymous

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

So I'm balancing the O in HNO3 on the left side with H2O on the right side?

24. anonymous

yes

25. anonymous

then you balance the H with H+

26. anonymous

So so far it looks like 3HNO3 and 2H2O?

27. anonymous

$3e ^{-}+ 3H ^{+}+HNO _{3} \rightarrow NO+2H _{2}O$

28. anonymous

no with H+ only no with HNO3

29. nincompoop

@Jhannybean show them how it is done Ms. Chemist

30. anonymous

lmao @Cuanchi has got this.

31. nincompoop

I think it can be made simpler tho

32. anonymous

That one is so difficult... the other equation I have to do is $SbCl _{5} + Kl -> SbCl _{3} + KCl + I _{2}$

33. anonymous

KI as in capital i not l.

34. anonymous

$6NO _{3}^{-}+Bi _{2}S _{3}\rightarrow 3S+2Bi(NO _{3})_{3}+6e ^{-}$ $(3e ^{-}+3H ^{+}+HNO _{3}\rightarrow NO+2H _{2}O) 2$ ---------------------------------------------------------------------- $6NO _{3}^{-}+ Bi _{2}S _{3}+6H ^{+}+2HNO _{3}\rightarrow 3S+2Bi(NO _{3})_{3}+2NO+4H _{2}O$ 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. anonymous

the KI in the equation doesn't have any charge or you put a negative charge KI-?

36. anonymous

Kl has no charge it was meant to be the arrow -> my apologies.

37. anonymous

the Sb is +5 goes to +3 (reduce) the I goes from -1 to O (oxidize)

38. anonymous

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

39. anonymous

This is the example they show me.. I see that they multiply the reduction and oxidation numbers together?

40. anonymous

2KI -> I2 + 2 K+ +2e- 2e- + SbCl5 -> SbCl3 + 2Cl- ------------------------------------- 2KI+SbCl5 -> I2 + 2K+ + SbCl3 + 2Cl-

41. anonymous

The -2 electrons came from the -2 times the +1 right? I think I'm understanding this now..

42. nincompoop

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

Yes. I have my periodic table right in front of me. :)

44. anonymous

OK! you have -2 x 1 = -2 then 1SbCl5 +1 x (2)= 2 then 2 KI

45. nincompoop

perhaps we can back track a little on what oxidation states and ground states are

46. anonymous

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

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. anonymous

@nincompoop do you have any easier way to balance redox reactions than the half reactions method?

49. nincompoop

50. anonymous

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

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

whenever you're ready to go over with oxidation and ground states

53. anonymous

@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
55. anonymous

@Cuanchi it is from the Prentice Hall California Chemistry book.. it's on Chapter 20.2 or Chapter 20.3

56. nincompoop

Prentice review is good, but any chemistry textbook is probably better just because the flow of information is much more systematically constructed.

57. anonymous

Yeah, we always do our unit reviews from the book. Some are easier, rest are much more confusing!