Question

How do we figure this out? NaOCl + Na2S2O3 ® ???

Answer 1

it is a redox reaction?

Answer 2

Yes @cuanchi I'm pretty sure :/

Answer 3

ok do you know how to find the oxidation state of the elements?

Answer 4

@cuanchi I have notes on how :)

Answer 5

http://www.webqc.org/balance.php?reaction=NaOCl+%2B+Na2S2O3+%3D+NaCl+%2B+Na2S2O6

Answer 6

@welshfella that's a pretty cool website but I don't think I have to balance?

Answer 7

OK but it gives you the products

Answer 8

I still don't know how to figure it out though :(

Answer 9

What exactly is the question? I'm just seeing half of an equation, are you supposed to be figuring out the other half? What's that little symbol next to the question marks for?

Answer 10

@SapphireMoon Yes, I'm supposed to swap ions. The real question is this: Laundry bleach such as Clorox is a dilute solution of sodium hypochlorite, NaOCl. Write a balanced net ionic equation for the reaction of NaOCl with Na2S2O3. The OCl- is reduced to chloride ion and the S2O3-2 is oxidized to sulfate ion.

Answer 11

I just made it easier on the eyes.

Answer 12

Ah, see that straightens it out a lot... That bit of extra information is what makes it a complete question. You'll want to balance it so that:\[NaOCl ^{-} + Na_{2}SO_{3} \rightarrow Cl^{-} + (SO_{4})^{2-}\]. That equation obviously looks fishy, where did the sodium go? Well you know that sodium turns into an Na+ ion, that disappears in a net ionic equation. You have three of those floating around now. Generally I'd think one of those would take the chlorine and make a salt, and the other two would make sodium sulfate but I haven't had this type of problem before.

Answer 13

My teacher gave me solutions

Answer 14

NO il looks like a REDOX reaction

Answer 15

NaOCl + Na2S2O3 ® NaCl + Na2SO4
Net ionic: OCl- + S2O3-2 ® Cl- + SO4-2 This is the first

Answer 16

Yes, It is a redox reaction

Answer 17

you have to figure out what element is going to oxidize and which one is going to reduce

Answer 18

The \[S _{2}O _{3^{-2}}\]

Answer 19

S2O3 ^2- -> SO4^2-

Answer 20

S (+2) -> S (+6)

Answer 21

Yup

Answer 22

S (+2) -> S (+6) + 4 e-

Answer 23

Wait it would be -2. On the S2O3

Answer 24

Cl (+1) + 2 e- -> Cl (-1)

Answer 25

S2O3 ^2-

2 x + (-2) 3 = -2

calculate x

Answer 26

Wait why would it be +2 and +6? :(

Answer 27

Hmm, maybe redox was too long ago, I thought it was different from the net ionic stuff.

Answer 28

And didn't the problem statement say what would oxidize and what would reduce?

Answer 29

SO4-2

x + (-2) 4 = -2

calculate x

Answer 30

OCl- + S2O3-2 -> Cl- + SO4-2

Answer 31

@cuanchi You're not answering questions, you're just dumping math... Are we trying to figure out by what amount chlorine and sulfur change their oxidation numbers? I'm still quite unclear on the question, and Melissa asked you a different question.

Answer 32

@SapphireMoon Yeah the math confuses the heck outta me

Answer 33

@Melissa_Something So the question you're asking is how to get from the reaction to the balanced ionic equation?

Answer 34

@SapphireMoon Yes, I have to balance the half reactions and get the balanced equation.

Answer 35

Okay, half-reactions might require a refresher, let me look at my chem notes from last year... (There's a reason I didn't take the AP last year lol... memory needs two years of chem first)

Answer 36

@SapphireMoon What chem class are you in?

Answer 37

@Melissa_Something I'm taking AP Chem this year, after Honors last year. Sorry, the power went out here and the lack of internet freaked OS out and I have to start over...
Just for clarity, the net ionic reaction is\[OCl^{-} + S_{2}O_{3}^{2-} \rightarrow Cl^{-} + SO_{4}^{2-}\] right?

Answer 38

Yes @SapphireMoon

Answer 39

@Melissa_Something After a lot of hard work for the first month of school, I finally have it! My notes from last year are pretty good (though I did use the interwebs a bit lol). Here's what I have:
My teacher gave us six steps to balance redox reactions.
1. Assign oxidation numbers to all elements.
2. Determine which are reduced & which are oxidized.
3. Split the equation into two half-reactions, one oxidation and one reduction. Include electrons.
4. Balance half reactions.
5. Balance electrons (trickiest bit)
6. Add half-reactions back together (deceptively easy-looking)

Step 2 is done fore us already, so let's see what we can do with step 3. Here is the oxidation half-reaction before balancing:
\[S_{2}O_{3}^{2-} \rightarrow SO_{4}^{2-}\]
That's nice, but it obviously isn't right, where did the extra oxygen come from, and where did the other sulfur go? Let's add another sulfate to balance the sulfurs:
\[S_{2}O_{3}^{2-} \rightarrow 2SO_{4}^{2-}\]
Lovely, the sulfur is balanced! Unfortunately, there are a lot more oxygens that come out than go in. We add waters to balance oxygens in oxidation reactions. The extra hydrogens will come out as hydride ions on the other side. It looks like we have five extra oxygens, so let's add five waters. Here's the equation with waters to balance the oxygens:
\[S_{2}O_{3}^{2-} + 5H_{2}O\rightarrow 2SO_{4}^{2-} + 10H^{+}\]
I've also put the extra hydrogen on the other side. Now we have a lovely balanced equation, right? Wrong! There's something wrong with the charges here. Originally, the sulfurs each had a charge of 2+ to balance out their companion oxygens. Now, however, they're in sulfate ions, which have more oxygens, so their oxidation numbers have each changed to 6+! Oh, boy, they must have lost their electrons. Let's scoop them up and put them on the right side for them:
\[S_{2}O_{3}^{2-} + 5H_{2}O\rightarrow 2SO_{4}^{2-} + 10H^{+} + 8e^{-}\]
Hooray! We have a brilliantly balanced oxidation half-reaction! Let's go to intermission so I can save and hope I don't lose all of this...

Answer 40

Okay, @Melissa_Something , we're back and ready to reduce! Here's our original half-reaction:
\[ClO^{-} \rightarrow Cl^{-}\]
OCl- has been changed to the proper format for hypochlorite, just to be easier on me. :) Now this equation is just as ridiculous as the first. It's got a rogue oxygen, just up and left the solution! We can't have that, let's get a water on the other side of that...
\[ClO^{-} + 2H^{+} \rightarrow Cl^{-} + H_{2}O\]
I've also balanced the hydrogens with a couple of hydrides. Brilliant! A balanced chemical reaction. Unfortunately, That chlorine seems to have found a couple of electrons changing it's oxygen-balancing +1 charge to a -1 charge. Let's put those on the left side.
\[2e^{-} + ClO^{-} + 2H^{+} \rightarrow Cl^{-} + H_{2}O\]
Yay! Flawlessly chemically and electrically balanced. Now onto step five, in which things get a bit hairy..

Alrighty, we have two half-reactions wanting to be whole, so let's put them together real quick:
\[2e^{-} + ClO^{-} + 2H^{+} S_{2}O_{3}^{2-} + 5H_{2}O \rightarrow Cl^{-} + H_{2}O + 2SO_{4}^{2-} + 10H^{+} + 8e^{-} \]
Ahahaha. Those charges are not balanced at all. The only way to fix this is to multiply both half-reactions to the least common multiple of the numbers of electrons in both reactions. These ones conveniently meet up at 8.
\[8e^{-} + 4ClO^{-} + 8H^{+} + S_{2}O_{3}^{2-} + 5H_{2}O \rightarrow 4Cl^{-} + 4H_{2}O + 2SO_{4}^{2-} + 10H^{+} + 8e^{-}\]
Okay, now that may have been confusing. I really should have done that before I put them together, but it's easier to see what you need to do if you combine it (IMO). What I did was multiply everything involved in the reduction reaction by 4 to get the number of electrons to 8. (I can leave the oxidation side alone because it's already at 8, sometimes you may have to mess with it.) Phew. Finally, chemically and electrically balanced. But it's not very concise. Let's cancel out everything that happens twice, namely, the electrons, four of the waters, and eight of the hydrides. That should leave one water on the left side and two hydrides on the right (balanced!). Here we go:
\[4ClO^{-} + S_{2}O_{3}^{2-} + H_{2}O \rightarrow 4Cl^{-} + 2SO_{4}^{2-} + 2H^{+} \]

Congratulations! You now have a balanced redox reaction! Sorry for the wall of text, feel free to ask loads of questions!

Answer 41

@SapphireMoon the mathematical formulas that I posted were to calculate the oxidation state of the elements in the formula.
It is the first step in the series of step that you listed
1. Assign oxidation numbers to all elements.
2. Determine which are reduced & which are oxidized.

To me still not clear how did you figure out the oxidation states of the elements and how did you find out which one is oxidized and which one it is reduced.

All the rest of the procedure looks fine (I havent go in the details)

http://nobel.scas.bcit.ca/chem0010/unit5/oxrule.htm

the formulas that I used it is a way to calculate the oxidation statte

What is the oxidation state of chromium in the dichromate ion, Cr2O7^2-?

The oxidation state of the oxygen is -2, and the sum of the oxidation states is equal to the charge on the ion. Don't forget that there are 2 chromium atoms present.

2n + 7(-2) = -2

n = +6

Answer 42

@cuanchi I understood that, but Melissa had no idea what you were doing, it was just numbers and she couldn't understand. I based my answer off of the fact that she said she had a way to find the oxidation numbers, I felt it would shorten my response (which is still very very long) to skip the part she could figure out on her own (and I could still explain it afterward if she had a question.)

Answer 43

@SapphireMoon im confused on why you put 5 H2O when you only needed one more oxygen? You said you had 5 more oxygens so you add 5 more waters but there is no 5 in the equarion. Also how did you guys find out the charges changed? (the 2 and 6) How did you know chlorine went from -1 to +1? I got the rest of it thank you sooooo much!!!!!!!!!!

Answer 44

@Melissa_Something There is no 5 in the equation, but we went from three oxygens in the thiosulfate to four in the sulfate, which is one, then remember I doubled the number of sulfates to make up for the 2 sulfurs in thiosulfate. That's another four oxygens to give us a total of 5.
In the hypochlorite ion (ClO-) it has a single negative charge. Oxygen carries a charge of -2 (except in certain scenarios which I don't remember and this isn't one of them, lol) so to get -1 there has to be a +1 charge somewhere, the only possible suspect would be Mr. Chlorine over there. Then in the chloride ion, shifty Mr. Chlorine changes his charge to -1 (cause it's Cl-). He's reduced his oxidation number, which is the reduction in the redox reaction. Does all of that make sense? (I should do a tutorial on this, lol).