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) Cl[sub]2[/sub] (g) + S[sub]2[/sub]O[sub]3[/sub][sup]2-[/sup](aq) --> Cl[sup]-[/sup](aq)+SO[sub]4[/sub][sup]2-[/sup](aq)

My answer so far:
Half Reaction for chlorine: Cl[sub]2[/sub] + 2e[sup]-[/sup] --> Cl[sup]-[/sup]

This matched the back of the book

The problem:
The half reaction I wrote for S[sub]2[/sub]O[sub]3[/sub] was completely different from what was in the back of the book.

The answer from the back of the book:
S[sub]2[/sub]O[sub]3[/sub][sup]2-[/sup](aq) + 5H[sub]2[/sub]O(l) --> 2SO[sub]4[/sub][sup]2-[/sup](aq) + 10H[sup]+[/sup](aq) + 8 e[sup]-[/sup]

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

Thanks! ;D

Your post is almost unreadable. Better to stick with Cl2 and SO4 until you learn how to do th subscripts and superscripts.
The first one is
Cl2 + S2O3^-2 ==> Cl^- + SO4^-2

Here are the steps in acid solution.
1. Separate into half reactions.
Cl2 ==> 2Cl^-
S2O3^-2 ==> SO4^-2

I won't do Cl2 since you seem to have that one ok.

Step 2.
Identify the element changing oxidation state. That is S, THEN make the number of atoms th same. I can do that this way.
S2O3^-2 ==> 2SO4^-2

Step 3. Determine the total oxidation state of S on the left and right.
on the left total S is +4. On the right, the total oxidation state of 2 S atoms is +12.

Step 4. Add electrons to the appropriate side to balance the change in oxidation state. Going from +4 to +12 is loss of 8 electrons.
S2O3^-2 ==> 2SO4^-2 + 8e

Step 5. Count the charge on the left and right. I see a charge of -2 on the left and a charge of -12 on the right. Add H^+ (this is an acid solution) to the appropriate side to balance the charge.
S2O3^-2 ==> 2SO4^-2 + 8e + 10 H^+

Step 6. Add water to the appropriate side (often the other side) to balance the oxygens.
S2O3^-2 + 5H2O =>2SO4^-2 + 8e + 10 H^+

Step 7. Always check it for three things.
a. atoms.
I see 2 S on left and right.
I see 8 O on the left and right.
I see 10 H on left and right.

b. charge.
I see 2- on the left and (2x2- + 8- + 10+ = 2-)

c. change in oxidation state.
2S with a total charge of +4 goes to 2 S with a total charge of +12 for a loss of 8 electrons.

Everything balances.

If you wish to make subscripts and superscripts, write a < sign followed by sub or sup, followed by > sign. To turn sub or sup or, add a / before the sub or sup.
<.sub> and <./sub> without the . I must put that in to make it print so you can read it; otherwise, it doesn't show. I hope this helps.

If you wish to make subscripts and superscripts, write a < sign followed by sub or sup, followed by > sign. To turn sub or sup OFF, add a....

I made a typo there.

Here's the correct procedure:

To balance the equation for the reaction between Cl2 and S2O3^2-, follow these steps:

Step 1: Write the half-reactions for each species involved.

The half-reaction for Cl2 is:
Cl2 (g) --> 2Cl- (aq) + 2e-

The half-reaction for S2O3^2- is the one you provided:
S2O3^2- (aq) + 5H2O (l) --> 2SO4^2- (aq) + 10H+ (aq) + 8e-

Step 2: Balance the atoms other than hydrogen and oxygen in each half-reaction. In this case, the atoms are already balanced.

Step 3: Balance the oxygen atoms by adding water (H2O) molecules to the side that needs more oxygen. In this case, the S2O3^2- half-reaction needs 5 water molecules to balance the oxygen atoms.

Step 4: Balance the hydrogen atoms by adding H+ ions to the side that needs more hydrogen. In this case, the S2O3^2- half-reaction needs 10 H+ ions to balance the hydrogen atoms.

Step 5: Balance the charges by adding electrons (e-) to the side that needs more negative charge. In this case, the Cl2 half-reaction needs 2 electrons, and the S2O3^2- half-reaction needs 8 electrons.

Step 6: Multiply the half-reactions by appropriate integers to make the number of electrons the same in both reactions. In this case, we can multiply the Cl2 half-reaction by 4 and the S2O3^2- half-reaction by 1 to make the number of electrons equal.

After balancing all the steps, the final balanced equation is:

4Cl2 (g) + 5S2O3^2- (aq) + 20H2O (l) --> 8Cl- (aq) + 10SO4^2- (aq) + 40H+ (aq) + 16e-

Step 7: Identify the oxidizing agent and reducing agent in the reaction.

The oxidizing agent is the species that gets reduced, so in this case, Cl2 is the oxidizing agent.

The reducing agent is the species that gets oxidized, so in this case, S2O3^2- is the reducing agent.

I hope this helps clarify the process of balancing the equation and identifying the oxidizing and reducing agents!