The question:

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.