Balancing Redox Reactions using half reactions?

[kshah93]

Homework Statement

Use half reactions to balance the following redox reactions and underline the oxidizing agent.
a) Cl2 + ClO3{-} -> ClO{-} (acidic)

{} is the charge
e{-} is electrons

Homework Equations

Not applicable

The Attempt at a Solution

Well I tried to separate and write the two half reactions:

1) Cl2 + 2e{-} -> 2Cl{-} (I took this directly from my standard reduction potentials table)

2) ClO3{-} -> ClO{-} (I attempted to balance this half-reaction as it didn't appear on my table)
4H{+} + ClO3{-} + 4e{-} -> ClO{-} + 2H2O (I added 2H2O to the right side to balance the oxygen and then added 4H{+} on the left side to balance the hydrogen, then added 4 electrons (4e{-}) to the left side to balance the charges)

Once determining both half reactions, I am stuck, and am not sure exactly how to proceed.
The answer key to this question states:

2Cl2 + ClO{3-} + 2H2O -> 5ClO{-} + 4H{+} with ClO{3-} as the oxidizing agent.

Did I approach this question incorrectly and how am I supposed to balance this redox equation using half reactions?

 

[symbolipoint]

This might help.

Using standard reduction potential table from Wikipedia, you could find this half:

Cl₂ + 2H₂O <------> 2e + 2H⁺ + 2HClO

and other half, based on your telling from message:

4H⁺ + ClO₃^- +4e <--------> ClO^- + 2H2O

 

[kshah93]

This might help.

Using standard reduction potential table from Wikipedia, you could find this half:

Cl₂ + 2H₂O <------> 2e + 2H⁺ + 2HClO

and other half, based on your telling from message:

4H⁺ + ClO₃^- +4e <--------> ClO^- + 2H2O

Okay, thanks for responding. I have one question. For your first half reaction, why is there an HClO?

 

[symbolipoint]

Okay, thanks for responding. I have one question. For your first half reaction, why is there an HClO?

You stated that the conditions are acidic solution. HClO is a weak acid.

 

[kshah93]

You stated that the conditions are acidic solution. HClO is a weak acid.

Yeah, but the answer key doesn't have an HClO in it at all.