9.2 Balancing Redox Reactions

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Here, we’ll go through an example of balancing a redox equation in acid solution using the half-reaction method.

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Presentation transcript:

9.2 Balancing Redox Reactions The Half Cell Method

“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.

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)

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.

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!

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)