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

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Discussion Overview

The discussion focuses on balancing chemical equations using the half-cell method, specifically addressing a given reaction involving chlorine and hydroxide ions. Participants explore the steps involved in identifying oxidation and reduction reactions, as well as the manipulation of coefficients in the final balanced equation.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a balanced equation using the half-cell method, identifying oxidation and reduction reactions involving Cl2, Cl-, and ClO3-.
  • Another participant suggests simplifying the final equation by canceling out water molecules and dividing coefficients, noting a common occurrence with diatomic molecules.
  • A third participant points out that the thread is a double-post, indicating a potential issue with the organization of the discussion.

Areas of Agreement / Disagreement

Participants express differing views on the final balanced equation, with one proposing a specific simplification while another has not confirmed this approach. The discussion remains unresolved regarding the correctness of the proposed simplifications.

Contextual Notes

Some assumptions about the balancing process and the treatment of diatomic molecules may not be explicitly stated. The discussion does not resolve the mathematical steps involved in the half-cell method.

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.
 
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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: .
 
Last edited:
This thread is a double-post. The other copy is in Chemistry.

Yellowduck - please do not post multiple copies of the same thing!
 

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