How to Balance Redox Reactions (Acidic Solution)
Since the total number of electrons gained by one atom must be the same as the number of electrons lost by another, we can use this knowledge to balance complicated chemical equations. Here's a challenge: Balance the following equation. Remember that the number of each type of atom has to be the same on each side, and so does the total charge!
This is how it's done, folks:
Cl's oxidation number changes from +5 to -1 (a change of 6)
I's oxidation number changes from 0 to +5 (a change of 5)
The lowest common multiple between these two is 30. So, 6 I atoms must give up 5 electrons each, and 5 Cl atoms must gain 6 each.
Cl's oxidation number changes from +5 to -1 (a change of 6)
I's oxidation number changes from 0 to +5 (a change of 5)
The lowest common multiple between these two is 30. So, 6 I atoms must give up 5 electrons each, and 5 Cl atoms must gain 6 each.
Notice that the coefficient on I2 is “3” since this makes for 6 I atoms total.
Unfortunately, O is not balanced. Add H2O to whichever side is short of O, and add H+ to the other side to counteract the H that you just added.
Unfortunately, O is not balanced. Add H2O to whichever side is short of O, and add H+ to the other side to counteract the H that you just added.
Note: After all this work, the charges balance too! WHAT UP.
Also Note: Don't be afraid to add H2O to either side, especially if any of the reactants or products are aqueous. You may end up added OH- to one side, in lieu of adding H+, if the question explicitly tells you that the reaction is occurring in basic solution. I've done an example of that HERE.
Also Note: Don't be afraid to add H2O to either side, especially if any of the reactants or products are aqueous. You may end up added OH- to one side, in lieu of adding H+, if the question explicitly tells you that the reaction is occurring in basic solution. I've done an example of that HERE.