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A half-reaction is simply one which shows either reduction OR oxidation, but not both. Here is the example redox reaction used in a different file:
Ag+ + Cu ---> Ag + Cu2+
It has BOTH a reduction and an oxidation in it. That is why we call it a redox reaction, from REDuction and OXidation.
What you must be able to do is look at a redox reaction and separate out the two half-reactions in it. To do that, identify the atoms which get reduced and get oxidized. Here are the two half-reactions from the above example:
Ag+ ---> Ag
Cu ---> Cu2+
The silver is being reduced, its oxidation number going from +1 to zero. The copper's oxidation number went from zero to +2, so it was oxidized in the reaction. In order to figure out the half-reactions, you MUST be able to calculate the oxidation number of an atom.
Keep in mind that a half-reaction shows only one of the two behaviors we are studying. A single half-reaction will show ONLY reduction or ONLY oxidation, never both in the same equation.
Also, notice that the reaction is read from left to right to determine if it is reduction or oxidation. If you read the reaction in the opposite direction (from right to left) it then becomes the other of our two choices (reduction or oxidation). For example, the silver half-reaction above is a reduction, but in the reverse direction it is an oxidation, going from zero on the right to +1 on the left.
There will be times when you want to switch a half-reaction from one of the two types to the other. In that case, rewrite the entire equation and swap sides for everything involved. If I needed the silver half-reaction to be oxidation, I'd write Ag ---> Ag+ rather than just doing it mentally.
The next step is that both half-reactions must be balanced. However, there is a twist. When you learned about balancing equation, you made equal the number of atoms of each element on each side of the arrow. That still applies, but there is one more thing: the total amount of charge on each side of the half-reaction MUST be the same.
When you look at the two half-reactions above, you will see they are already balanced for atoms with one Ag on each side and one Cu on each side. So, all we need to do is balance the charge. To do this you add electrons to the more positive side. You add enough to make the total charge on each side become EQUAL.
To the silver half-reaction, we add one electron:
Ag+ + e¯ ---> Ag
To the copper half-reaction, we add two electrons:
Cu ---> Cu2+ + 2e¯
One point of concern: notice that each half-reaction wound up with a total charge of zero on each side. This is not always the case. You need to strive to get the total charge on each side EQUAL, not zero.
One more point to make before wrapping this up. A half-reaction is a "fake" chemical reaction. It's just a bookkeeping exercise. Half-reactions NEVER occur alone. If a reduction half-reaction is actually happening (say in a beaker in front of you), then an oxidation reaction is also occuring. The two half-reactions can be in separate containers, but they do have to have some type of "chemical connection" between them. The nature of this connection is the subject of another tutorial.
Click the sentence for the last step: recombine the two half-reactions to obtain the balanced redox equation.
Half-Reactions Practice Problems
Balnce each half-reaction for atoms and charge:
1) Cl2 ---> Cl¯
2) Sn ---> Sn2+
3) Fe2+ ---> Fe3+
4) I3¯ ---> I¯
5) ICl2¯ ---> I¯ (I'm being mean on this one. Hint: the iodine is the only thing reduced or oxidized.)
Separate each of these redox reactions into their two half-reactions (but do not balance):
6) Sn + NO3¯ ---> SnO2 + NO2
7) HClO + Co ---> Cl2 + Co2+
8) NO2 ---> NO3¯ + NO
Answers to the problems
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