Oxidation-Reduction Reactions Chemistry Worksheet Page 16
ADVERTISEMENT
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41222
Chapter 6
Oxidation-Reduction Reactions
Single-Displacement Reactions
Now let’s consider another type of chemical reaction. In single‑displacement reactions,
O
7
bjeCtive
atoms of one element in a compound are displaced (or replaced) by atoms from a
pure element. These reactions are also called single‑replacement reactions. All of the
following are single‑displacement reactions.
Pure element displaces element in compound
Zn(s) + CuSO
(aq)
ZnSO
(aq) + Cu(s)
4
4
Cd(s) +
H
SO
(aq)
CdSO
(aq) +
H
( g )
2
4
4
2
Cl
( g ) + 2NaI(aq)
2NaCl(aq) +
I
(s)
2
2
Our model of particle behavior enables us to visualize the movements of atoms
O
9
bjeCtive
and ions participating in single‑displacement reactions. For example, consider the
first equation above, a reaction in which atoms of zinc replace ions of copper in a
copper sulfate solution. Because copper(II) sulfate is a water‑soluble ionic compound,
2+
the CuSO
solution consists of free Cu
ions surrounded by the negatively charged
4
2‒
oxygen ends of water molecules and free SO
ions surrounded by the positively
4
charged hydrogen ends of water molecules. These ions move throughout the solution,
colliding with each other, with water molecules, and with the walls of their container.
Now imagine that a lump of solid zinc is added to the solution. Copper ions begin to
2+
collide with the surface of the zinc. When the Cu
ions collide with the uncharged
zinc atoms, two electrons are transferred from the zinc atoms to the copper(II) ions.
The resulting zinc ions move into solution, where they become surrounded by the
negatively charged ends of water molecules, and the uncharged copper solid forms on
the surface of the zinc (Figure 6.4).
Because the zinc atoms lose electrons in this reaction and change their oxidation
2+
number from 0 to +2, they are oxidized, and zinc is the reducing agent. The Cu
ions
gain electrons and decrease their oxidation number from +2 to 0, so they are reduced
and act as the oxidizing agent. The half reaction equations and the net ionic equation
for this reaction are below.
Zn(s) → Zn
2+
(aq) + 2e
‒
oxidation:
2+
(aq) + 2e
‒
→ Cu(s)
reduction:
Cu
2+
2+
Zn(s) + Cu
(aq) → Zn
(aq) + Cu(s)
Net Ionic Equation:
You can see an animation that illustrates the reaction above at the textbook’s Web
site.
ADVERTISEMENT
0 votes
Related Articles
Related forms
Related Categories
Parent category: Education