QUICK QUIZ
Match the substance with its correct oxidation number below.
Answers:
Example: Assign oxidation numbers to the atoms in the following reaction in order to
determine the oxidized and reduced species and the oxidizing and reducing agents.
SnCl 2 + PbCl 4 → SnCl 4 + PbCl 2
Solution: All these species are neutral, so the oxidation numbers of all the atoms in each
compound must add up to zero. In SnCl 2 , since there are two chlorines present, and
chlorine has an oxidation number of −1, Sn must have an oxidation number of +2.
Similarly, the oxidation number of Sn in SnCl 4 is +4; the oxidation number of Pb is
+4 in PbCl 4 and +2 in PbCl 2 . Notice that the oxidation number of Sn goes from +2 to
+4; i.e., it loses electrons and thus is oxidized, making it the reducing agent. Since
the oxidation number of Pb has decreased from +4 to +2, it has gained electrons and
been reduced. Pb is the oxidizing agent. The sum of the charges on both sides of the
1. Na+
2. Cu2+
3. Cl−
4. Fe3+^
(A) −1
(B) +3
(C) +2
(D) +1
1. = (D)
2. = (C)
3. = (A)
4. = (B)
