Skyler Wild - Final Chemistry Notebook

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 occurring. The two halfreactions
can be in separate containers, but they do have to have some type of "chemical
connection" between them.
Half-Reaction Practice Problems
Balance each half-reaction for atoms and charge:
1) Cl2 → Cl¯
2) Sn → Sn 2+
Reduced
Oxidized
3) Fe 2+ → Fe 3+
Oxidized
4) I3¯ → I¯
Reduced
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 + Co 2+
8) NO2 →NO3¯ + NO
Here are the two half-reactions to be combined:
Here is the rule to follow:
Ag+ + e¯ → Ag
Cu → Cu 2+ + 2e¯
the total electrons MUST cancel when the two half-reactions are added.
Another way to say it:
the number of electrons in each half-reaction MUST be equal when the two half-reactions are
added.
What that means is that one (or both) equations must be multiplied through by a factor. The value of
the factor is selected so as to make the number of electrons equal.
In our example problem, the top reaction (the one with silver) must be multiplied by two, like this:
2Ag+ + 2e¯ → 2Ag