Electrochemistry and Half Reactions

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  • #1
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The question:
Balance each of the following skeletal equations by using oxidation and reduction half reactions. All the reactions take place in acidic solution. Identify the oxidizing agent and reducing agent in each reaction.
a) Cl2 (g) + S2O32-(aq) --> Cl-(aq)+SO42-(aq)

My answer so far:
Half Reaction for chlorine: Cl2 + 2e- --> Cl-

This matched the back of the book

The problem:
The half reaction I wrote for S2O3 was completely different from what was in the back of the book.

The answer from the back of the book:
S2O32-(aq) + 5H2O(l) --> 2SO42-(aq) + 10H+(aq) + 8 e-

Would someone please walk me through how to come to this answer? I would very much appreciate it!

Thanks! ;D
 

Answers and Replies

  • #2
Gokul43201
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The question:
Balance each of the following skeletal equations by using oxidation and reduction half reactions. All the reactions take place in acidic solution. Identify the oxidizing agent and reducing agent in each reaction.
a) Cl2 (g) + S2O32-(aq) --> Cl-(aq)+SO42-(aq)

My answer so far:
Half Reaction for chlorine: Cl2 + 2e- --> Cl-

This matched the back of the book
That should really be 2Cl- on the right, but you got the basic idea right.

The problem:
The half reaction I wrote for S2O3 was completely different from what was in the back of the book.

The answer from the back of the book:
S2O32-(aq) + 5H2O(l) --> 2SO42-(aq) + 10H+(aq) + 8 e-

Would someone please walk me through how to come to this answer? I would very much appreciate it!

Thanks! ;D
Okay, this step is clearly a little more tricky than the first one.

You start with:

1. S2O32-(aq) + unknowns --> SO42-(aq) + unknowns

The first thing to recognize is that you want to fix the number of S atoms on the right (don't worry about O atoms as there's always H2O that can be used to fix that). 2 S atoms on the left, so you want 2 S atoms on the right.

2. S2O32-(aq) + unknowns --> 2SO42-(aq) + unknowns

Next, you fix the number of O atoms by adding H2O to the required side. In this case, we've got 3 O atoms on the left and 8 on the right. So we need 5 more on the left, and the way to do that is by adding 5 H2O to the left.

3. S2O32-(aq) + 5H2O(l) + unknowns --> 2SO42-(aq) + unknowns

Now you see that you've got 10 H atoms on the left and none on the right. You fix this by adding 10 H+ ions (you're allowed to do this, if the reaction takes place in acidic medium) to the right.

4. S2O32-(aq) + 5H2O(l) + unknowns --> 2SO42-(aq) + 10H+(aq) + unknowns

Now all elements are balanced, but the charge is not. On the left, you have -2. On the right you have -4 + 10 = +6. So you need to add 8 electrons on the right to balance the charge.

5. S2O32-(aq) + 5H2O(l) --> 2SO42-(aq) + 10H+(aq) + 8 e-

Now you've got both charge and chemical elements balanced, and you're done.
 
  • #3
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Thanks so much!

So you balance everything except hydrogen and oxygen,
balance oxygen by adding water to the other side,
balance hydrogen by adding hydrogen ions to the other side,
and balance charge by adding electrons on the side that needs it?

Would you say that those would be the proper steps?
 
  • #4
Borek
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  • #5
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thanks! (and thanks for the link, too!)
 
  • #6
chemisttree
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Balanced?

I don't think it is balanced yet...
 
Last edited:

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