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

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In summary, the conversation discusses balancing equations using the half-cell method, specifically focusing on the example of Cl2(g) + OH- <=> Cl- + ClO3^- + H2O(l). The oxidation and reduction reactions are shown and multiplied to cancel out the electrons. The equation is then balanced for a basic solution by swapping out the H+ with OH- and simplifying the coefficients. The conversation also mentions a previous mistake and confirms the correctness of the final answer.
  • #1
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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  • #2
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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  • #3
This thread is a double-post. The other copy is in Chemistry.

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

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

1. What is a balance half-cell equation?

A balanced half-cell equation is an equation that shows the transfer of electrons between two half-cells in a redox reaction. It is used to represent the oxidation and reduction reactions that occur in a redox reaction.

2. Why is it important to balance half-cell equations?

Balancing half-cell equations is important because it ensures that the number of electrons transferred in the oxidation and reduction reactions are equal. This is necessary for the overall reaction to be balanced and for the reaction to proceed correctly.

3. How do you balance a half-cell equation?

To balance a half-cell equation, you need to add coefficients to the reactants and products so that the number of atoms of each element and the overall charge are equal on both sides. The number of electrons transferred should also be equal in the oxidation and reduction reactions.

4. What are some examples of half-cell equations?

An example of a half-cell equation is the oxidation of zinc metal to zinc ions in a solution of copper sulfate. The half-cell equation for this reaction is:
Zn(s) → Zn2+(aq) + 2e-

Another example is the reduction of copper ions to copper metal in a solution of zinc sulfate. The half-cell equation for this reaction is:
Cu2+(aq) + 2e- → Cu(s)

5. How are half-cell equations used in electrochemistry?

Half-cell equations are used in electrochemistry to calculate the cell potential of a redox reaction. By balancing the half-cell equations and using the Nernst equation, the standard cell potential and the equilibrium constant of a reaction can be determined.

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