1. 9.2 Balancing Redox Reactions
The Half Cell Method

2. “Chlorate ions and iodine react in acidic solution to produce chloride ions and iodate ions”
Important - redox reactions nearly always react in water.
So to balance them you can add H2O to add extra O atoms as needed.
You can then add H+ atoms to balance the H as needed. But that means you have acidic protons in your equation. So…
If the reaction is in basic conditions, add enough OH− ions to turn all the H+ ions into water.

3. First step – write the overall skeleton reaction:
“Chlorate ions and iodine react in acidic solution to produce chloride ions and iodate ions”
First step – write the overall skeleton reaction:
ClO3−(aq) + I2(aq)  Cl−(aq) + IO3−(aq)
2nd step – write two half cell reactions, one for the oxidation, the other for the reduction:
ClO3−(aq)  Cl−(aq) I2(aq)  IO3−(aq)

4. 4th step – balance O by adding water as needed.
“Chlorate ions and iodine react in acidic solution to produce chloride ions and iodate ions”
ClO3−(aq)  Cl−(aq)
+ 6H+
+ 3H2O(l)
I2(aq)  IO3−(aq)
+ 6H2O(l) 2
+ 12H+
+ 6e−
+ 10e−
3rd step – deal with each half reaction separately; balance any atoms other than O and H (for now).
4th step – balance O by adding water as needed.
5th step – balance H by adding H+ as needed.
6th step – balance charge by adding e− as needed.

5. Multiply the 2nd reaction by 3, to get 30 e−
ClO3−(aq) + 6H+(aq) + 6e−  Cl−(aq) + 3H2O(l) I2(aq) + 6H2O(l) 2 IO3−(aq) + 12H+(aq)+10e−
5( )
3( )
Nearly done! Examine the two half cell reactions … we need to balance the electrons. Multiply the 1st reaction by 5, to get 30 e−
Multiply the 2nd reaction by 3, to get 30 e−
5ClO3−(aq) + 30 H+(aq) + 30e−  5Cl−(aq) + 15H2O(l) 3I2(aq) + 18H2O(l) 6 IO3−(aq) + 36H+(aq)+ 30e− 3 6
Now we combine the two half cells, cancelling out any species that are the same. Watch and be amazed!

6. 5ClO3−(aq) + 3I2(aq) + 3H2O(l)  5Cl−(aq) + 6 IO3−(aq) + 6H+(aq)
“Chlorate ions and iodine react in acidic solution to produce chloride ions and iodate ions”
5ClO3−(aq) + 3I2(aq) + 3H2O(l)  5Cl−(aq) + 6 IO3−(aq) + 6H+(aq)