Download PP 27: Oxidation/Reduction Reactions

PP 27: Oxidation/Reduction Reactions
Drill: Calculate the solubility of PbSO4:
(Ksp = 2.0 x 10-14)
REDOX Reactions: Chemical reactions in which electrons are transferred
Chemical Charge: A positive, neutral, or negative number assigned to a particle & determined by
the difference between the # of protons & electrons
Oxidation Number: The charge of a single atom
Oxidation: The act of losing electrons (When oxidation occurs, the oxidation number will increase)
Reduction: The act of gaining electrons (When reduction occurs, the oxidation number will decrease)
REDOX Fact: You cannot have oxidation without reduction or reduction without oxidation
Oxidizing Agent:
• A substance that oxidizes another
• The oxidizing agent will be reduced
Reducing Agent:
• A substance that reduces another
• The reducing agent will be oxidized
Demonstrate how to determining oxidation numbers of elements in compounds or polyatomic ions:

MnO4-1 + NH3  MnO2 + NO2-1
The oxidation number of neutral elements is zero.
The oxidation number of an element with a charge is equal the charge.
The oxidation number of negative elements within compounds or polyatomic ions is equal to its
column number – eight.
 O is -2 & N is -3
o The oxidation number of positive elements within compounds or polyatomic ions must be
determined by balancing the charges.
 In NH3, N = -3: thus, each of 3 Hs must be 1 to balance the charge of the compound to 0.
Rxn:
o
o
o

In MnO2, each O is -2, adding up to -4. The Mn must be +4 to balance the charge.

In NO2-1, each O is -2, adding up to -4. The charge must add up to the charge of the
polyatomic ion which is -1. The N must be +3 because +3 & -4 add up to -1.
In MnO7-1, each O is -2, adding up to -8. The charge must add up to the charge of the
polyatomic ion which is -1. The Mn must be +7 because +7 & -8 add up to -1.

Drill: determine ox # of each element in the following rxn:
MnO4-1 + Fe  Mn2O3 + FeO
Balancing REDOX Reactions:
1. Determine the oxidation numbers of all elements in the reaction.
2. Determine the element being oxidized & the element being reduced.
3. Divide the reaction into ½ reactions. (one oxidation rxn & one reduction rxn)
4. Balance each ½ reaction.
a. Balance the elements being oxidized or reduced
b. Balance the oxidation numbers with electrons.
c. Balance the electrons between the two ½ reactions.
d. Balance the charge with either H+ (in acid) or OH- (in base).
e. Balance H s with H2O.
f. Check the balance of each ½ rxn with O s.
5. Add the two ½ reactions
a. Species on opposite sides of the reaction can be cancelled additively.
b. Species on the same side of the reaction must be added.
c. Check your work
Demonstration: Rxn:
(1)
MnO4-1 + Fe  Mn2O3 + FeO (in acid)
+7
(2)
0
+3
-2
+2 -2
MnO4-1 + Fe  Mn2O3 + FeO
+7
-2
0
+3
-2
+2 -2
(3)
Fe  FeO
MnO4-1  Mn2O3
(4a)
Fe  FeO
2 MnO4-1  Mn2O3
(4b)
Fe  FeO + 2 e-1
8 e-1 + 2 MnO4-1  Mn2O3
4( Fe  FeO + 2 e-1)
8 e-1 + 2 MnO4-1  Mn2O3
(4c)
10 H+1
4 Fe  4 FeO + 8 e-1 + 8 H+1
+ 8 e-1 + 2 MnO4-1  Mn2O3
10 H+1
4 H2O + 4 Fe  4 FeO + 8 e-1 + 8 H+1
+ 8 e-1 + 2 MnO4-1  Mn2O3 + 5 H2O
(4d)
(4e)
10 H
4 H2O + 4 Fe  4 FeO + 8 e-1 + 8 H+1
+ 8 e-1 + 2 MnO4-1  Mn2O3 + 5 H2O
2 10 H+1
4 H2O + 4 Fe  4 FeO + 8 e-1 + 8 H+1
+ 8 e-1 + 2 MnO4-1  Mn2O3 + 1 5 H2O
(4f)
+1
(5)
(5)
-2
4 O s on each side
8 O s on each side
4 Fe + 2 MnO4-1 + 2 H+1  4 FeO + Mn2O3 + 1 H2O
Balance the following rxns:
 MnO4-1 + NH4+  Mn+2 + NO2-1
(in base)
Drill: Determine the oxidation number of each element:
MnO2 + Cl-1  Mn+2 + Cl2
Determine the oxidation number of each element: MnO4-1 + N2O3  Mn+2 + NO3-1
Balance each or the following reactions:


MnO4-1 + N2O3  Mn+2 + NO3-1
Cr2O7-2 + NH3  Cr+3 + NO3-1
(in acid)
(in base)
Drill: Determine the oxidation number of each element:
SO4-2
N2O3
Balance each of the two ½ reactions:


MnO2  Mn+2
Cr2O7-2  Cr+3
Balance each or the following reactions:



MnO3 + Cl-1  Mn+2 + Cl2
(in acid)
FeCl2 + KNO3  FeCl3 + KNO2
(in base)
SO2 + Pb(OH)2  SO3 + Pb
(in base)
Drill: Balance the following half-reaction: H2O  O2
(in acid)
Balance each or the following reactions:





HNO3 + Cu  N2O + Cu+2
Fe(NO3)3 + Cr(OH)4-1  Fe(NO3)2 + CrO4-2
MnO2 + P4  Mn2O7 + PH3
(in acid)
+
PbBr2 + O2 + K  PbBr4 + K2O
(in acid)
-1
-1
-1
IO3 + I  I3
(in acid)
Drill: Balance the following half-reaction:
KIO4 
Balance each or the following reactions:







SnF2 + KIO4  KI + SnF4
Ca + VO4-3  CaO + V+2
Ni(OH)2 + N2H4  Ni + N2
P2O3 + K2CrO3  K3PO4 + Cr+2
SO + H2Cr2O7  H2SO4 + Cr+2
PH3 + KMnO4  P2O5 + Mn2O3
(in acid)
(in acid)
(in base)
(in acid)
(in base)
KI (in acid)
K2CO3
N2O5