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One point of concern: notice that each half-reaction wound up with a total charge of zero on each<br />
side. This is not always the case. You need to strive to get the total charge on each side EQUAL, not<br />
zero.<br />
One more point to make before wrapping this up. A half-reaction is a "fake" chemical reaction. It's just<br />
a bookkeeping exercise. Half-reactions NEVER occur alone. If a reduction half-reaction is actually<br />
happening (say in a beaker in front of you), then an oxidation reaction is also occurring. The two halfreactions<br />
can be in separate containers, but they do have to have some type of "chemical<br />
connection" between them.<br />
Half-Reaction Practice Problems<br />
Balance each half-reaction for atoms and charge:<br />
0<br />
-1<br />
1) Cl2 → Cl¯<br />
0 2+<br />
2) Sn → Sn 2+<br />
up oxidation<br />
down reduction<br />
The numbers are going own so its a reduction because its reducing.<br />
The reaction is going up so its oxidation.<br />
3) Fe 2+ → Fe 3+<br />
The reaction is going up so its oxidation.<br />
4) I3¯ → I¯<br />
5) ICl2¯ → I¯ (I'm being mean on this one. Hint: the iodine is the only thing reduced or oxidized.)<br />
Separate each of these redox reactions into their two half-reactions (but do not balance):<br />
6) Sn + NO3¯ →SnO2 + NO2<br />
7) HClO + Co →Cl2 + Co 2+<br />
8) NO2 →NO3¯ + NO<br />
Here are the two half-reactions to be combined:<br />
Here is the rule to follow:<br />
Ag+ + e¯ → Ag<br />
Cu → Cu 2+ + 2e¯<br />
the total electrons MUST cancel when the two half-reactions are added.<br />
Another way to say it:<br />
the number of electrons in each half-reaction MUST be equal when the two half-reactions are<br />
added.<br />
What that means is that one (or both) equations must be multiplied through by a factor. The value of<br />
the factor is selected so as to make the number of electrons equal.<br />
In our example problem, the top reaction (the one with silver) must be multiplied by two, like this:<br />
2Ag+ + 2e¯ → 2Ag<br />
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