Unit 28
Electrochemistry (Chapter 8)
•Electrochemistry as Applied to Batteries (8.3)
•Corrosion (8.4)
•Important Oxidizing Agents (8.6 – 8.7)
•Important Reducing Agents (8.8 – 8.9)
Electrochemistry (8.3)
•Electricity is due to the motion of electrons. Since oxidation-reduction
reactions involve an exchange of electrons, such reactions can be used to
generate electricity through applications such as batteries.
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Electrochemistry Cont. (8.3)
•In the reaction from the previous page, the
copper goes from being in the elemental form to
dissolving in solution as Cu2+ ions which cause
the blue color in solution.
•The Ag+ ions previously dissolved in solution
become elemental silver and form the “hangings” seen in the second
beaker.
•In equation form:
Cu (s) → Cu2+ (aq) + 2 eAg+ (aq) +
e-
→
Ag (s)
•An important aspect in understanding electrochemistry is to understand
how these two equations may be combined to form the overall reaction.
Electrochemistry Cont. (8.3)
•Each of the reactions below is called a half-reaction. One represents an
oxidation and the other a reduction.
Cu (s) →
Ag+ (aq) +
Cu2+ (aq)
e-
→
+
2 e-
Ag (s)
•Since the electrons donated by the copper are the ones accepted by the
silver, the number of electrons being accepted and donated must match.
•In order for the electrons to balance, each half-reaction is multiplied by an
integer as necessary to ensure that the number of electrons donated matches
those accepted. In this example, the Cu equation involves two electrons and
the Ag equation involves only one so multiplying the Ag equation by 2 will give
two electrons accepted to go along with the two electrons donated by copper
(continued on next page).
Electrochemistry Cont. (8.3)
•Multiplying the first equation by “1” and the second by “2” gives:
1 × (Cu (s) →
Cu2+ (aq)
2 × (Ag+ (aq) +
e-
+
→
2 e- )
Ag (s) )
•These simplify to:
Cu (s) →
Cu2+ (aq)
2 Ag+ (aq) +
2 e-
→
+
2 e-
2 Ag (s)
•Adding these two equations gives:
Cu (s) +
2 Ag+ (aq) → Cu2+ (aq) + 2 Ag (s)