Download Slide 1

Figure 4.11
A summary of terminology for oxidation-reduction (redox)
reactions.
e-
X
transfer
Y
or shift of
electrons
X loses electron(s)
Y gains electron(s)
X is oxidized
Y is reduced
X is the reducing agent
Y is the oxidizing agent
X increases its
oxidation number
Y decreases its
oxidation number
Oxidation-Reduction Reactions
(electron transfer reactions)
2Mg
O2 + 4e-
2Mg2+ + 4e- Oxidation half-reaction (lose e-)
2O2Reduction half-reaction (gain e-)
2Mg + O2 + 4e2Mg2+ + 2O2- + 4e2Mg + O2
2MgO
2
3
Table 4.3 Rules for Assigning an Oxidation Number (O.N.)
General rules
1. For an atom in its elemental form (Na, O2, Cl2, etc.): O.N. = 0
2. For a monoatomic ion: O.N. = ion charge
3. The sum of O.N. values for the atoms in a compound equals zero. The sum of
O.N. values for the atoms in a polyatomic ion equals the ion’s charge.
Rules for specific atoms or periodic table groups
1. For Group 1A(1):
O.N. = +1 in all compounds
2. For Group 2A(2):
O.N. = +2 in all compounds
3. For hydrogen:
O.N. = +1 in combination with nonmetals
4. For fluorine:
O.N. = -1 in combination with metals and boron
5. For oxygen:
O.N. = -1 in peroxides
O.N. = -2 in all other compounds(except with F)
O.N. = -1 in combination with metals, nonmetals (except
O), and other halogens lower in the group
6. For Group 7A(17):
Sample Problem 4.6
PROBLEM:
Determining the Oxidation Number of an Element
Determine the oxidation number (O.N.) of each element in these
compounds:
(a) zinc chloride (b) sulfur trioxide
(c) nitric acid
The O.N.s of the ions in a polyatomic ion add up to the charge of the ion
PLAN:
and the O.N.s of the ions in the compound add up to zero.
SOLUTION:
(a) ZnCl2. The O.N. for zinc is +2 and that for chloride is -1.
(b) SO3. Each oxygen is an oxide with an O.N. of -2. Therefore the O.N. of
sulfur must be +6.
(c) HNO3. H has an O.N. of +1 and each oxygen is -2. Therefore the N
must have an O.N. of +5.
Figure 4.10
Highest and lowest
oxidation numbers of
reactive main-group
elements.
Zn (s) + CuSO4 (aq)
Zn2+ + 2e-
Zn
Cu2+ + 2e-
Cu
ZnSO4 (aq) + Cu (s)
Zn is oxidized
Zn is the reducing agent
Cu2+ is reduced
Cu2+ is the oxidizing agent
Copper wire reacts with silver nitrate to form silver metal.
What is the oxidizing agent in the reaction?
Cu (s) + 2AgNO3 (aq)
Cu
Ag+ + 1e-
Cu(NO3)2 (aq) + 2Ag (s)
Cu2+ + 2eAg
Ag+ is reduced
Ag+ is the oxidizing agent
7
What are the oxidation numbers of all the
elements in each of these compounds?
NaIO3
IF7
K2Cr2O7
IF7
F = -1
7x(-1) + ? = 0
NaIO3
I = +7
Na = +1 O = -2
3x(-2) + 1 + ? = 0
I = +5
K2Cr2O7
O = -2
K = +1
7x(-2) + 2x(+1) + 2x(?) = 0
Cr = +6
8
Types of Oxidation-Reduction Reactions
Combination Reaction
A+B
0
C
+3 -1
0
2Al + 3Br2
2AlBr3
Decomposition Reaction
C
+1 +5 -2
2KClO3
A+B
+1 -1
0
2KCl + 3O2
9
Types of Oxidation-Reduction Reactions
Combustion Reaction
A + O2
B
0
0
S + O2
0
0
2Mg + O2
+4 -2
SO2
+2 -2
2MgO
10
Types of Oxidation-Reduction Reactions
Displacement Reaction
A + BC
0
+1
+2
Sr + 2H2O
+4
0
TiCl4 + 2Mg
0
AC + B
-1
Cl2 + 2KBr
0
Sr(OH)2 + H2
0
Hydrogen Displacement
+2
Ti + 2MgCl2
-1
Metal Displacement
0
2KCl + Br2
Halogen Displacement
11
The Activity Series for Halogens
F2 > Cl2 > Br2 > I2
Halogen Displacement Reaction
0
-1
Cl2 + 2KBr
I2 + 2KBr
-1
0
2KCl + Br2
2KI + Br2
12
Types of Oxidation-Reduction Reactions
Disproportionation Reaction
The same element is simultaneously oxidized
and reduced.
Example:
reduced
+1
0
Cl2 + 2OH-
-1
ClO- + Cl- + H2O
oxidized
13
Classify each of the following reactions.
Ca2+ + CO32NH3 + H+
Zn + 2HCl
Ca + F2
CaCO3
NH4+
ZnCl2 + H2
CaF2
Precipitation
Acid-Base
Redox (H2 Displacement)
Redox (Combination)
14
Solution Stoichiometry
The concentration of a solution is the amount of solute
present in a given quantity of solvent or solution.
M = molarity =
moles of solute
liters of solution
What mass of KI is required to make 500. mL of a
2.80 M KI solution?
volume of KI solution
500. mL x
1L
1000 mL
M KI
x
moles KI
2.80 mol KI
1 L soln
x
M KI
166 g KI
1 mol KI
grams KI
= 232 g KI
15
Preparing a Solution of Known Concentration
16
Dilution is the procedure for preparing a less concentrated
solution from a more concentrated solution.
Dilution
Add Solvent
Moles of solute
before dilution (i)
MiVi
=
=
Moles of solute
after dilution (f)
MfVf
17
How would you prepare 60.0 mL of 0.200 M HNO3
from a stock solution of 4.00 M HNO3?
MiVi = MfVf
Mi = 4.00 M Mf = 0.200 M Vf = 0.0600 L
Vi =
MfVf
Mi
= 0.200 M x 0.0600 L
4.00 M
Vi = ? L
= 0.00300 L = 3.00 mL
Dilute 3.00 mL of acid with water to a total volume
of 60.0 mL.
18