How Do You Balance Equations Using the Half-Cell Method?

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yellowduck
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Ok, here is another one I have, 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-

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.
 
on Phys.org
Ha, I was just reviewing these today.
yellowduck said:
6Cl2 + 6H2O + 12OH- <-> 10Cl- + 2ClO3- + 12HOH
You can cancel out the H2O since you have it on both sides.

6Cl2 + 12OH- --> 10Cl- + 2ClO3- + 6H2O

You can divide all of the coefficients by 2. It seems that sometimes when you work with diatomic molecules like Cl2 that your final answer will be doubled. I made the same mistake here: https://www.physicsforums.com/showthread.php?t=96870

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

Copy and paste that line into google and the results show that you're good :smile: .
 
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