Transcript Section_10-2

Section 10.2—Redox
Reactions
How do we know if a reaction will work?
Definition
Single Replacement Reactions – A
single element reacts with a compound
and replaces one of the elements in the
compound
Generic Form: A + BX  B + AX
Example: Mg + CuSO4  Cu + MgSO4
Magnesium replaces copper
Single Replacement & Redox
Single atom has
no charge
That atom
becomes part
of a
compound &
therefore has
a charge
A change in
charge on an
atom is a
Redox
reaction
All Single Replacement Reactions are Redox Reaction
(But not all redox reactions are single replacement!)
Activity Series
 An activity series is a list of metals from most reactive to
least reactive
 The more reactive an element is, the more it wants to be
in a compound
 A more reactive element will “kick out” a less reactive
element that is in a compound
 If the more reactive element is the “lone atom” in a single
replacement reaction, the reaction will occur
 If the more reactive element is already in the compound,
the reaction will not occur
 There is an activity series at the back of Chapter 10 & in
the Appendix
Example #1
Example:
If the following reaction will occur, write
the products. If the reaction will not
occur, write “no reaction”
FeCl3 + Zn 
Example #1
Example:
If the following reaction will occur, write
the products. If the reaction will not
occur, write “no reaction”
2 FeCl3 +3 Zn 3 ZnCl2 +2 Fe
Zn is higher on the activity series than Fe
Zn is more reactive and wants to be in the compound more than Fe
Let’s Practice #1
Example:
If the following reaction will occur, write
the products. If the reaction will not
occur, write “no reaction”
Al + NaCl 
Let’s Practice #1
Example:
If the following reaction will occur, write
the products. If the reaction will not
occur, write “no reaction”
Al + NaCl 
No reaction
Na is higher on the activity series than Al.
Na wants to be in the compound more than Al.
Lets Practice #2
Cu + 2 AgNO3
Cu(s) + 2 AgNO3 (aq) → Cu(NO3)2 (aq) + 2 Ag(s)
Lets Practice #3
Li+1 + Na 
Li+1 + Na  Li + Na+1
Half Reactions
Definition
Half-Reaction method – Method of
balancing even the most complex
Redox equations.
Redox reactions are split into a reduction halfreaction and an oxidation half-reaction
Redox Reactions in Acid or Base
The half-reaction method occurs in either an acidic
or basic environment.
This method uses the species produced by acids
or bases to balance the equation:
Acidic Redox
balances with
Basic Redox
balances with
H
H+
H 2O
O
H2O
OH-
Balancing Redox in Acid:
Determine oxidation numbers & use to decide
what’s being reduced and oxidized.
Mn O4- + H2 C2 O4  Mn2+ + CO2
+7 -2
+1 +3 -2
+2
+4 -2
Reduction
Oxidation
Balancing Redox in Acid:
1
Split the reaction into half reaction—one for
oxidation and one for reduction
Mn O4- +
Mn O4- 
H2 C2 O4  Mn2+ +
CO2
Mn2+
H2 C2 O4 
CO2
Balancing Redox in Acid:
Balance all atoms except H and O
2
Perform each of the following steps for both half-reactions
Mn O4- +
Mn O4- 
Reactants
H2 C2 O4  Mn2+ +
CO2
Mn2+
Products
Mn
1
1
O
4
0
H2 C2 O4  2 CO2
Reactants
Products
C
2
1
2
O
4
2
4
H
2
0
Balancing Redox in Acid:
3
Balance O’s by adding H2O to the side needing
more O
Mn O4- +
Mn O4- 
Reactants
H2 C2 O4  Mn2+ +
CO2
Mn2+ + 4 H2O
Products
Mn
1
1
O
4
0
H
0
8
H2 C2 O4  2 CO2
4
Reactants
Products
C
2
1
2
O
4
2
4
H
2
0
Balancing Redox in Acid:
4
Balance H’s by adding H+ to the side needing more
H
Mn O4- +
8 H+ + Mn O4- 
Reactants
CO2
Mn2+ + 4 H2O
Products
Mn
1
1
O
4
0
H
0
8
H2 C2 O4  Mn2+ +
H2 C2 O4  2 CO2 + 2 H+
4
Reactants
8
Products
C
2
1
2
O
4
2
4
H
2
0
2
Balancing Redox in Acid:
Find the total charge on the reactants and products
side. Add electrons (e-) to balance the charges
5
Electrons should need to be added to the reactants in the reduction
half-reaction and to the products in the oxidation half-reaction
Mn O4- +
5 e- + 8 H+ + Mn O4- 
Reactants
1
1
O
4
0
H
0
Charge
+7 +2
CO2
Mn2+ + 4 H2O
Products
Mn
8
H2 C2 O4  Mn2+ +
H2 C2 O4  2 CO2 + 2 H+ + 2 e4
Reactants
8
+2
Products
C
2
1
2
O
4
2
4
H
2
0
2
Charge
0
+2 0
Balancing Redox in Acid:
6
Multiply the half-reactions by a factor to make the
electrons lost in the oxidation equal to the electrons
gained in the reduction.
Mn O4- +
2 × ( 5 e- + 8 H+ + Mn O4- 
H2 C2 O4  Mn2+ +
CO2
Mn2+ + 4 H2O )
5 × ( H2 C2 O4  2 CO2 + 2 H+ + 2 e- )
Balancing Redox in Acid:
7
Add the two half-reactions back together.
Mn O4- +
2 × ( 5 e- + 8 H+ + Mn O4- 
H2 C2 O4  Mn2+ +
CO2
Mn2+ + 4 H2O )
5 × ( H2 C2 O4  2 CO2 + 2 H+ + 2 e- )
10 e- + 16H+ + 2 MnO4- + 5 H2C2O4  2 Mn2+ + 8 H2O + 10 CO2 + 10 H+ + 10 e-
Balancing Redox in Acid:
8
Cancel out anything that appears on both sides.
Mn O4- +
2 × ( 5 e- + 8 H+ + Mn O4- 
H2 C2 O4  Mn2+ +
CO2
Mn2+ + 4 H2O )
5 × ( H2 C2 O4  2 CO2 + 2 H+ + 2 e- )
6
10 e- + 16H+ + 2 MnO4- + 5 H2C2O4  2 Mn2+ + 8 H2O + 10 CO2 + 10 H+ + 10 e6H+ + 2 MnO4- + 5 H2C2O4  2 Mn2+ + 8 H2O + 10 CO2
Balancing Redox in Base:
Determine oxidation numbers & use to decide
what’s being reduced and oxidized.
Mn O4- +
+7 -2
Br1- 
-1
Reduction
MnO2
+4 -2
Oxidation
+
BrO31+5 -2
Balancing Redox in Base:
1
Split the reaction into half reaction—one for
oxidation and one for reduction
Mn O4- +
Br1- 
MnO2
+
BrO31-
Mn O4-  MnO2
Br1-  BrO31-
Balancing Redox in Base:
Balance all atoms except H and O
2
Perform each of the following steps for both half-reactions
Mn O4- +
Br1- 
MnO2
+
BrO31-
Mn O4-  MnO2
Reactants
Products
Mn
1
1
O
4
2
Br1-  BrO31Reactants
Products
Br
1
1
O
0
3
Balancing Redox in Base:
Balance O’s by adding twice as many OH- to the
side needing more O.
3
It’s OK that the O will not be balanced after this step, they will become
balanced in the next step when balancing the H’s.
Mn O4- +
Br1- 
MnO2
+
BrO31-
Mn O4-  MnO2 + 4 OH1Reactants
Products
Mn
1
1
O
4
2
H
0
4
6 OH1- + Br1-  BrO316
Reactants
Products
Br
1
1
O
0 6
3
H
6
0
Balancing Redox in Base:
Balance H’s by adding H2O to the side needing more
H
4
Remember that there are 2 H’s in H2O, so add ½ the number of H’s that
you need.
Mn O4- +
Br1- 
MnO2
+
BrO31-
2 H2O + Mn O4-  MnO2 + 4 OH1Reactants
Products
Mn
1
O
4
6
2
H
0
4
4
6 OH1- + Br1-  BrO31- + 3 H2O
1
6
Reactants
Products
Br
1
1
O
0 6
3
6
H
6
0
6
Balancing Redox in Base:
Find the total charge on the reactants and products
side. Add electrons (e-) to balance the charges
5
Electrons should need to be added to the reactants in the reduction halfreaction and to the products in the oxidation half-reaction
Mn O4- +
Br1- 
MnO2
+
BrO31-
3 e- + 2 H2O + Mn O4-  MnO2 + 4 OH1Reactants
Products
Mn
1
O
4
6
2
H
0
4
4
Charge
6 OH1- + Br1-  BrO31- + 3 H2O + 6 e-
1
-1 -4
-4
6
Reactants
Products
Br
1
1
O
0 6
3
6
H
6
0
6
Charge
-7
-1 -7
Balancing Redox in Base:
6
Multiply the half-reactions by a factor to make the
electrons lost in the oxidation equal to the electrons
gained in the reduction.
Mn O4- +
Br1- 
MnO2
+
BrO31-
2 × (3 e- + 2 H2O + Mn O4-  MnO2 + 4 OH1- )
1 × ( 6 OH1- + Br1-  BrO31- + 3 H2O + 6 e- )
Balancing Redox in Base:
7
Add the two half-reactions back together.
Mn O4- +
Br1- 
MnO2
+
BrO31-
2 × (3 e- + 2 H2O + Mn O4-  MnO2 + 4 OH1- )
1 × ( 6 OH1- + Br1-  BrO31- + 3 H2O + 6 e- )
6 e- + 4 H2O + 2 MnO4- + 6 OH1- + Br1-  2 MnO2 + 8 OH1- + BrO31- + 3 H2O + 6 e-
Balancing Redox in Base:
8
Cancel out anything that appears on both sides.
Mn O4- +
Br1- 
MnO2
+
BrO31-
2 × (3 e- + 2 H2O + Mn O4-  MnO2 + 4 OH1- )
1 × ( 6 OH1- + Br1-  BrO31- + 3 H2O + 6 e- )
2
6 e- + 4 H2O + 2 MnO4- + 6 OH1- + Br1-  2 MnO2 + 8 OH1- + BrO31- + 3 H2O + 6 eH2O + 2 MnO4- + Br1-  2 MnO2 + 2 OH1- + BrO31-