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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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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The answer to the problem is: Bi2S3 (s) + 8HNO3 (aq) -> 2Bi(NO3)3 (aq) + 2NO(g) + 3S (s) + 4H2O (l)
Thank you very much! I'll read it right now. :)
Wait, I know how to balance chemical equations but this problem makes us balance redox equations using the oxidation numbers. Not the moles..

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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
the S is going from -2 to 0 and then is get oxidized the N goes from +5 to +2 and then its reduce
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?
no it is also spectator there, doesn't matter
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?
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 _________________________________________________
Where did the 6e and 3e come from?
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-
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
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)
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.
So the oxidation half reaction would be S3(s) -> S(s) and the reduction half reaction HNO3(aq) -> NO(g) ?
S3-6(s) -> S(s)
\[S _{3}^{-6}{(s)} \rightarrow S _{(s)}\]
And it's -6 because of the \[S _{3}\] and the -2?
yes the oxidation state is -2 and you have 3 atoms (-2)x 3 = -6
Would I do the same for the HNO3 -> NO ?
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
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-
So I'm balancing the O in HNO3 on the left side with H2O on the right side?
yes
then you balance the H with H+
So so far it looks like 3HNO3 and 2H2O?
\[3e ^{-}+ 3H ^{+}+HNO _{3} \rightarrow NO+2H _{2}O\]
no with H+ only no with HNO3
@Jhannybean show them how it is done Ms. Chemist
lmao @Cuanchi has got this.
I think it can be made simpler tho
That one is so difficult... the other equation I have to do is \[SbCl _{5} + Kl -> SbCl _{3} + KCl + I _{2}\]
KI as in capital i not l.
\[6NO _{3}^{-}+Bi _{2}S _{3}\rightarrow 3S+2Bi(NO _{3})_{3}+6e ^{-}\] \[(3e ^{-}+3H ^{+}+HNO _{3}\rightarrow NO+2H _{2}O) 2\] ---------------------------------------------------------------------- \[6NO _{3}^{-}+ Bi _{2}S _{3}+6H ^{+}+2HNO _{3}\rightarrow 3S+2Bi(NO _{3})_{3}+2NO+4H _{2}O\] 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
the KI in the equation doesn't have any charge or you put a negative charge KI-?
Kl has no charge it was meant to be the arrow -> my apologies.
the Sb is +5 goes to +3 (reduce) the I goes from -1 to O (oxidize)
Okay. so now that I know it reduces -2 and oxidizes + 1, I balance the increase and decrease in oxidation numbers?
This is the example they show me.. I see that they multiply the reduction and oxidation numbers together?
1 Attachment
2KI -> I2 + 2 K+ +2e- 2e- + SbCl5 -> SbCl3 + 2Cl- ------------------------------------- 2KI+SbCl5 -> I2 + 2K+ + SbCl3 + 2Cl-
The -2 electrons came from the -2 times the +1 right? I think I'm understanding this now..
do you have a periodic table? what I can do is make the big picture a little clearer where those numbers come from
Yes. I have my periodic table right in front of me. :)
OK! you have -2 x 1 = -2 then 1SbCl5 +1 x (2)= 2 then 2 KI
perhaps we can back track a little on what oxidation states and ground states are
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
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.
@nincompoop do you have any easier way to balance redox reactions than the half reactions method?
reactions is all about the movement of electrons so I always start with that
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.
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.
whenever you're ready to go over with oxidation and ground states
@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!
http://www.bsqm.org.mx/PDFS/V2/N3/2-Ice%20B%20Risteski.pdf
@Cuanchi it is from the Prentice Hall California Chemistry book.. it's on Chapter 20.2 or Chapter 20.3
Prentice review is good, but any chemistry textbook is probably better just because the flow of information is much more systematically constructed.
Yeah, we always do our unit reviews from the book. Some are easier, rest are much more confusing!

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