Chemistry Reference
In-Depth Information
a charge of (-1 × 2) = -2 on the negative side. The tin (Sn) on the reac-
tant side must have a charge of +2 in order to ensure that the compound
has a net charge of 0. Following this pattern, we are able to assign oxida-
tion numbers to all of the elements in the equation, as shown here:
Sn
2+
Cl
2
1-
+ 2Fe
3+
Cl
3
1-
2Fe
2+
Cl
2
1-
+ Sn
4+
Cl
4
1-
B.
The substance that is being oxidized is the one that appears to lose
electrons, as its oxidation number is algebraically increased. The oxi-
dation number of tin (Sn) changes from +2 to +4. Tin (Sn
2+
) is the
substance being oxidized.
C.
The oxidizing agent is the substance that is taking the extra electrons,
allowing the tin to be oxidized. Iron (Fe
3+
) is our oxidizing agent.
D.
The oxidizing agent is the substance that is reduced. The oxidation num-
ber of iron changes from +3 to +2. Fe
3+
is the substance that is being
reduced.
E.
We also mentioned that the substance that is being oxidized is also the
reducing agent. Tin provides the electrons for the iron to become re-
duced. Tin (Sn
2+
) is the reducing agent.
Example 2 summary:
SnCl
2
+ 2FeCl
3
2FeCl
2
+ SnCl
4
A.
The oxidation numbers for each substance involved
Sn
2+
Cl
2
1-
+ 2Fe
3+
Cl
3
1-
2Fe
2+
Cl
2
1-
+ Sn
4+
Cl
4
1-
B.
The substance being oxidized:
tin (Sn
2+
)
C.
The oxidizing agent:
iron (Fe
3+
)
D.
The substance being reduced:
iron (Fe
3+
)
E.
The reducing agent:
tin (Sn
2+
)
Now, try the following review problems. Be sure to check your answers
at the end of the chapter, and we will continue our discussion of redox
reactions in our next lesson.
