# Oxidation numbers

## Main Question or Discussion Point

Hello Im confused with the half-cell 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 half-cell method. Show both half-cell 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 Gokul43201
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In Cl2, the oxidation state of Cl is 0. This is because the "net oxidation state" must be zero for the molecule.

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 Gokul43201
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Yes, the oxidation states of H and O in OH- are indeed +1 and -2 respectively.

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.
...makes no sense. "Oxidation" and "reduction" are half-cell reactions; they are not ions or molecules.

You have to write down (and balance) the half-cell 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).

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 Gokul43201
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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 (half-cell) 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 half-cell 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.

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The reaction isnt quite so simple as you've put it m0286,

the first thing that happens to the $$Cl_2$$ in alkaline solutions is
$$Cl_2 + 2OH^- \xrightarrow{} Cl^- + ClO^-$$
Notice how $$Cl^-$$ has a charge of -1 and $$ClO^-$$ has a charge of +1.

Then this happens:
$$3ClO^- \xrightarrow{high~temperature} ClO_3^- + 2Cl^-$$
Now 2$$ClO^-$$ have oxidized the Cl in 1 other $$ClO^-$$ to yield $$2Cl^-$$ with a formal charge of -1 and $$ClO_3^-$$ 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.

gotcha!

Thanks for all your help guys I understand now Ok, I have this question now and the explanation here is not clear to me.
this is one of a series of half-cell questions... this one is confusing.

79. Balance the following equations by the half-cell method. Show both half-cell 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-

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.

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Gokul43201
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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.

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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 half-cell method. Show both half-cell 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-

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.

Gokul43201
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yellowduck said:
Multiply the oxidation reaction by a factor of 5 to cross out the e-
5Cl2 <-> 10Cl-
Probably just a typo, but that should be reduction.

3Cl2 + 3H2O + 6OH- <-> 5Cl- + ClO3- + 6H2O

not shure the rules for canceling H2O...
With any equation, you can add or subtract equal amounts (moles) of the same quantity to both sides without affecting it. In this case, you can subtract 3H2O from both sides.

I need the H2O on the left side to balance the ClO3 on the right...
Why do you need that? And what do you mean by "to balance ClO3"? You do not balance compounds or radicals in a chemical equation; you only balance atomic species and charge. The O-atoms to balance the O in ClO3- comes from OH- on the left.

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

balancing redox

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.

Hootenanny
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t_falle said:
2Cl ---> 2Cl(+5) + 10e-
. When was the last time you saw a chlorine atom losing five electrons? Why would it want to do this?

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Gokul43201
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t_falle said:
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
1. How exactly have you utilized the fact that the "ratio" (which specific ratio?) is 1:5?
2. You final equation is unbalanced in charge. LHS has 6- and RHS has 9- + 1- = 10-

$$SO_4^2^- + I^- +H^+ \xrightarrow{} S^2^- + H_2O$$

I have a simular question and I just wanted to go over it to make sure I have it right.

Reduction Reaction:
$$SO_4^2^- \xrightarrow{} S^2^-$$

$$8e^- + 8H^+ +SO_4^2^- \xrightarrow{} S^2^- +4H_2O$$

Oxidation Reaction:
$$I^- \xrightarrow{} I_2$$

$$2I^- \xrightarrow{} I_2 + 2e^-$$

Multiply the oxidation reaction by a factor of 4 to cross out the e^-
$$8I^- \xrightarrow{} 4I_2 +8e^-$$

$$SO_4^2^- +8H^+ +8I^- \xrightarrow{} S^2^- +4I_2 +4H_2O$$

Is there any else I need to do to this question?

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chemisttree