Register to reply 
Oxidation numbers 
Share this thread: 
#1
Jul605, 01:08 PM

P: 63

Hello Im confused with the halfcell method. Thought i had it. Did two questions. now I am stuck on the third?? PLEASE HELP!
It says: Balance the following equations by the halfcell method. Show both halfcell reactions and identify them ad oxidation or reduction. 3) Cl2(g) + OH <=> Cl + ClO3^ + H2O(l) I have been able to balance them and follow the steps just for this one im confused on finding the oxidation numbers???? For ClO3^ I got O as 2 and therefore Cl as +5. in theH2O i know H is 1+ and O is 2. the Cl2 i believe is 1 and same for the Cl. Im not sure when it comes to the OH. The Cl will be the oxidation since it increases from 1 to 5+. But im not sure with the OH so i cant tell which other one changes. PLEASE HELP!! THANKS 


#2
Jul605, 04:01 PM

Emeritus
Sci Advisor
PF Gold
P: 11,155

In Cl2, the oxidation state of Cl is 0. This is because the "net oxidation state" must be zero for the molecule.



#3
Jul605, 10:13 PM

P: 63

Wow this question got extra confusing from the others.
Cl2(g) + OH <=> Cl + ClO3^ + H2O(l) So if the Cl2 is 0. and the Cl is 1 and the other Cl is +5.. then they all differ, are they going to be my oxidation and reduction. And is OH just H going to be 1+ and O 2???Thanks for the help 


#4
Jul705, 12:30 AM

Emeritus
Sci Advisor
PF Gold
P: 11,155

Oxidation numbers
Yes, the oxidation states of H and O in OH^{} are indeed +1 and 2 respectively.
Your other statement : You have to write down (and balance) the halfcell reactions and say that "this one" is the oxidation reaction (because the reactant loses electrons) and "that one" is the reduction reaction (because the reactant gains electrons). 


#5
Jul1005, 09:04 PM

P: 63

Cl2(g) + OH <=> Cl + ClO3^ + H2O(l)
So the O's balance on both sides it's 2. The H's are equal on both sides they are +1. The Cl2 is 0 and the Cl is 1 and for ClO3^. 2*3=6+x=1 so x (Cl = +5) So the Cl's in the equation are unbalanced. So is it: with Cl2 and Cl 1e is lost therefore Cl2 is the oxidation. But then with Cl2 and ClO3^, 5e is gained therefore Cl2 is the reduction. Like can Cl2 be both.. Im sooo confused or is it i dont use the Cl2 and I use the Cl = oxidation and ClO3^ =reduction? Thanks for your help though so far I appreciate it 


#6
Jul1005, 09:53 PM

Emeritus
Sci Advisor
PF Gold
P: 11,155

But you are not reading carefully ! You CAN NOT say "x = oxidation, y = reduction" where x and y are molecules or ions.
Oxidation/reduction are (halfcell) reactions. You must write down the reaction and say that "this reaction is the oxidation/reduction reaction". What you can say is that "x is getting oxidized" and "y is getting reduced" (but this is not what is being asked). So, if your question is "can the same species get reduced as well as oxidized ?" the answer "yes." In the same reaction some of x can get reduced to y and some can get oxidized to z. But this is not what your question is asking for. It is asking for the oxidation and reduction halfcell reactions, which are used to balance the overall reaction. But if you have already done this and only want to identify the species that get oxidized/reduced, then keep in mind that you should be looking for species from among the reactants. And in this case, some of the Cl2 gets oxidized (to Cl^5+ in the ClO3^)and some of it gets reduced (to Cl^). Your reasoning is mostly okay, but you need to use the right language to describe things in science. 


#7
Jul1305, 05:39 AM

P: 47

The reaction isnt quite so simple as you've put it m0286,
the first thing that happens to the [tex]Cl_2[/tex] in alkaline solutions is [tex]Cl_2 + 2OH^ \xrightarrow{} Cl^ + ClO^[/tex] Notice how [tex]Cl^[/tex] has a charge of 1 and [tex]ClO^[/tex] has a charge of +1. Then this happens: [tex]3ClO^ \xrightarrow{high~temperature} ClO_3^ + 2Cl^[/tex] Now 2[tex]ClO^[/tex] have oxidized the Cl in 1 other [tex]ClO^[/tex] to yield [tex]2Cl^[/tex] with a formal charge of 1 and [tex]ClO_3^[/tex] where the Cl atom has a formal charge of +5. So you see the oxidation and reduction occurs only to the Cl atoms and not to any other atoms in this particular reaction. 


#8
Jul1305, 12:45 PM

P: 63

Thanks for all your help guys I understand now



#9
Jul506, 06:19 PM

P: 16

Ok, I have this question now and the explanation here is not clear to me.
this is one of a series of halfcell questions... this one is confusing. 79. Balance the following equations by the halfcell method. Show both halfcell reactions and identify them as oxidation or reduction. b) Cl2(g) + OH <=> Cl + ClO3^ + H2O(l) Oxidation Reaction: Cl2 (g) <> Cl 2e + Cl2 <> 2Cl  Balanced Cl molecules and electrons Reduction Reaction: Cl2 (g) <> ClO3 Cl2 + 6H2O <> 2ClO3 + 12H+ 10e Multiply the oxidation reaction by a factor of 5 to cross out the e 5Cl2 <> 10Cl Add them together: 6Cl2 + 6H2O <> 10Cl + 2ClO3 + 12H+ Since this is a basic solution we must swap out the H+ with OH... in this case adding 6OH to each side 6Cl2 + 6H2O + 12OH <> 10Cl + 2ClO3 + 12HOH This seems kinda insane but it may just be right. Can anyone give me some input on this? Thanks. 


#10
Jul506, 11:24 PM

Emeritus
Sci Advisor
PF Gold
P: 11,155

It's almost correct  you've got the identification of oxidation and reduction wrong. But other than that all you've left to do is cancel off some of the H2O from both sides and reduce coefficients to the lowest integer ratio.



#11
Jul606, 05:16 PM

P: 16

Very, Very Sorry for the double post. Had some computer issues and thought it did not post in this old thread so I stated a new post.
Thank you for taking the time to look at this for me... I have updated the answer a bit and if you have time to look at it I would be thankful. 79. Balance the following equations by the halfcell method. Show both halfcell reactions and identify them as oxidation or reduction. b) Cl2(g) + OH <=> Cl + ClO3^ + H2O(l) Reduction Reaction: Cl2 (g) <> Cl 2e + Cl2 <> 2Cl  Balanced Cl molecules and electrons Oxidation Reaction: Cl2 (g) <> ClO3 Cl2 + 6H2O <> 2ClO3 + 12H+ 10e Multiply the oxidation reaction by a factor of 5 to cross out the e 5Cl2 <> 10Cl Add them together: 6Cl2 + 6H2O <> 10Cl + 2ClO3 + 12H+ Since this is a basic solution we must swap out the H+ with OH... in this case adding 6OH to each side 6Cl2 + 6H2O + 12OH <> 10Cl + 2ClO3 + 12HOH 3Cl2 + 3H2O + 6OH <> 5Cl + ClO3 + 6H2O not shure the rules for canceling H2O... I need the H2O on the left side to balance the ClO3 on the right... If I take any away from the right side and not the left I would unbalance the equation... Thanks Again. 


#12
Jul706, 08:25 AM

Emeritus
Sci Advisor
PF Gold
P: 11,155




#13
Jul706, 10:53 PM

P: 16

Thank you, that makes perfect sense now.
your help is very much appreciated. 


#14
Jul1006, 05:57 AM

P: 1

Interesting question. I balanced it a different way, please tell me why this is not correct:
recognizing that the Cl2 is beibg both oxidized and reduced: 2e + 2Cl > 2Cl, and 2Cl > 2Cl(+5) + 10e the ration being 1:5, so 5Cl2 + 6OH > 9Cl + ClO3 + 3H2O Any input is appreciated. Thanks. 


#15
Jul1006, 05:59 AM

Emeritus
Sci Advisor
PF Gold
P: 9,781




#16
Jul1006, 08:38 AM

Emeritus
Sci Advisor
PF Gold
P: 11,155

2. You final equation is unbalanced in charge. LHS has 6 and RHS has 9 + 1 = 10 


#17
May2907, 03:16 PM

P: 83

[tex]SO_4^2^ + I^ +H^+ \xrightarrow{} S^2^ + H_2O[/tex]
I have a simular question and I just wanted to go over it to make sure I have it right. Reduction Reaction: [tex]SO_4^2^ \xrightarrow{} S^2^[/tex] [tex]8e^ + 8H^+ +SO_4^2^ \xrightarrow{} S^2^ +4H_2O[/tex] Oxidation Reaction: [tex]I^ \xrightarrow{} I_2[/tex] [tex]2I^ \xrightarrow{} I_2 + 2e^[/tex] Multiply the oxidation reaction by a factor of 4 to cross out the e^ [tex]8I^ \xrightarrow{} 4I_2 +8e^[/tex] Add them together: [tex]SO_4^2^ +8H^+ +8I^ \xrightarrow{} S^2^ +4I_2 +4H_2O[/tex] Is there any else I need to do to this question? 


Register to reply 
Related Discussions  
What are oxidation numbers and how do they work?  Chemistry  2  
Oxidation numbers  Chemistry  1  
Balancing Oxidation Numbers  Chemistry  3  
Irrational numbers depends on rational numbers existence  General Math  0 