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Section 18.2
Balancing Oxidation-Reduction Reactions
Warm-up
• Determine the oxidation state…• (1) Na
Na = 0
• (2) MgI2
Mg = +2, I= -1
• (3) (NH4)2O
N =-3, H = +1, O = -2• (4) NaOH
Na = +1, O = -2, H = +1
• (5) Cl2Cl = 0
• (6) Al(NO2)3
Al = +3, N = +3, O = -2
Section 18.2
Balancing Oxidation-Reduction Reactions
1. SWBAT identify oxidizing and reducing agents in a TOD.
Objectives
Section 18.2
Balancing Oxidation-Reduction Reactions
A. Oxidation-Reduction Reactions Between Nonmetals
• Na oxidized – Na is also called the reducing agent (electron donor).
• Cl2 reduced
– Cl2 is also called the oxidizing agent (electron acceptor).
2Na(s) + Cl2(g) 2NaCl(s)
Section 18.2
Balancing Oxidation-Reduction Reactions
Identifying Oxidation-Reduction Reactions Between Nonmetals
1. Find the oxidation state of each type of element
2. Compare the change in electrons
3. If electrons are lost -> Oxidation
i. The entire compound is the Reducing Agent
4. If electrons are gained -> Reduction
i. The entire compound is the Oxidizing agent
CH4(g) + 2O2(g) CO2(g) + 2H2O(g)
Steps:
Section 18.2
Balancing Oxidation-Reduction Reactions
Oxidation-Reduction Reactions Between Nonmetals
• C oxidized
– CH4 is the reducing agent.
• O2 reduced
– O2 is the oxidizing agent.
CH4(g) + 2O2(g) CO2(g) + 2H2O(g)
C=-4 O=0 C=+4 H=+1
H=+1 O=-2 O=-2
Section 18.2
Balancing Oxidation-Reduction Reactions
Practice Together
• 2Na(s) + Cl2(aq) --> 2NaCl(s)
Section 18.2
Balancing Oxidation-Reduction Reactions
Practice Together
• Zn(s) + Cu2+ (aq) --> Zn2+(aq) + Cu(s)
Section 18.2
Balancing Oxidation-Reduction Reactions
Practice Together
• 2Mg(s) + O2(g) --> 2MgO
Section 18.2
Balancing Oxidation-Reduction Reactions
You do it…#1
Fe = +3 C = +2 Fe = 0 C = +4
O = -2 O = -2 O = -2
• Fe reduced – CO is the reducing agent. • C oxidized
– Fe2O3 is the oxidizing agent.
Fe2O3(s) + 3CO(g) 2Fe(l) + 3CO2(g)
Section 18.2
Balancing Oxidation-Reduction Reactions
You do it…#2
Cl = 0 Na =+1 Na = +1 Br = 0
Br = -1 Cl = -1
• Cl reduced – NaBr is the reducing agent. • Br oxidized
– Cl2 is the oxidizing agent.
Cl2(g) + 2NaBr(aq) 2NaCl(aq) + Br2(l)
Section 18.2
Balancing Oxidation-Reduction Reactions
You do it…#3 (Reaction to make fertilizer)
N = 0 H =0 N = -3
H = +1
• N reduced
– H2 is the reducing agent.
• H oxidized
– N2 is the oxidizing agent.
N2(g) + 3H2(g) 2NH3(g)
Section 18.2
Balancing Oxidation-Reduction Reactions TOD
• 2PbS(s) + 3O2(g) 2PbO(s) + 2SO2(g)
– Oxidizing Agent = O2
– Reducing Agent = PbS
• 2PbO(s) + CO(g) Pb(s) + CO2(g)– Oxidizing Agent = PbO– Reducing Agent = CO
• 2Na(s) + Cl2(g) 2NaCl(aq)
– Oxidizing Agent = Cl2– Reducing Agent = Na
Section 18.2
Balancing Oxidation-Reduction Reactions
HW
• Page 651 #1-5
Section 18.2
Balancing Oxidation-Reduction Reactions Warm-upIdentify the following: 1) Oxidation States,
2) Oxidation/Reduction3) Oxidizing/Reducing Agents
• 2H2 (g) + O2 (g) 2H2O (l)– (0) (0) (+1) (-2)– Oxidized = H Reduced = O
– Oxidizing Agent = O2 Reducing Agent = H2
• 2PbS(s) + 3O2(g) 2PbO(s) + 2SO2(g) – (+2) (-2) (0) (+2) (-2) (+4) (-2)– Oxidized = S Reduced = O
– Oxidizing Agent = O2 Reducing Agent = PbS
Section 18.2
Balancing Oxidation-Reduction Reactions
Pop Quiz Time
HW
• Page 662: 9,10,18,19, 21
Section 18.2
Balancing Oxidation-Reduction Reactions Warm-upIdentify the following: 1) Oxidation States,
2) Oxidation/Reduction3) Oxidizing/Reducing Agents
• 6Na(s) + N2 (g) 2Na3N (s)– (0) (0) (+1) (-3)– Oxidized = Na Reduced = N
– Oxidizing Agent = N2 Reducing Agent = Na
• 2PbO(s) + CO(g) Pb(s) + CO2(g)– (+2) (-2) (+2) (-2) (0) (+4) (-2)– Oxidized = C Reduced = Pb
– Oxidizing Agent = PbO Reducing Agent = CO
Section 18.2
Balancing Oxidation-Reduction Reactions
SWBAT balance oxidation-reduction equations using half reactions in a Half-Reaction Practice
Objectives
Section 18.2
Balancing Oxidation-Reduction Reactions B. Balancing Oxidation-Reduction Reactions by the Half-Reaction Method
• Half reaction – equation which has electrons as products or reactants
Section 18.2
Balancing Oxidation-Reduction Reactions
Separate into half-reactions
• Sn + NO3¯ ---> SnO2 + NO2
• Sn ---> SnO2 and NO3¯ ---> NO2
Section 18.2
Balancing Oxidation-Reduction Reactions
• HClO + Co ---> Cl2 + Co2+
• HClO ---> Cl2 and Co ---> Co2+
Separate into half-reactions
Section 18.2
Balancing Oxidation-Reduction Reactions
• NO2 ---> NO3¯ + NO
• NO2 ---> NO3¯ and NO2 ---> NO
Separate into half-reactions
Section 18.2
Balancing Oxidation-Reduction Reactions
Practice balancing electrons
• Cl2 ---> Cl¯
• Cl2 + 2e¯ ---> 2Cl¯
Section 18.2
Balancing Oxidation-Reduction Reactions
• Sn ---> Sn2+
• Sn ---> Sn2+ + 2e¯
Practice balancing electrons
Section 18.2
Balancing Oxidation-Reduction Reactions
• Fe2+ ---> Fe3+
• Fe2+ ---> Fe3+ + e¯
Practice balancing electrons
Section 18.2
Balancing Oxidation-Reduction Reactions
• I3¯ ---> I¯
• I3¯ + 2e¯ ---> 3I¯
Practice balancing electrons
Section 18.2
Balancing Oxidation-Reduction Reactions
HW
• Page 663: #24(b,c,d), 25, 26
Section 18.2
Balancing Oxidation-Reduction Reactions
Warm-up
Complete the Handout…
Section 18.2
Balancing Oxidation-Reduction Reactions
SWBAT balance oxidation-reduction equations using half reactions in a Half-Reaction Practice
Objectives
Section 18.2
Balancing Oxidation-Reduction Reactions
Balancing Oxidation-Reduction Reactions by the Half-Reaction Method
• Half reaction – equation which has electrons as products or reactants
Section 18.2
Balancing Oxidation-Reduction Reactions
Balancing Oxidation-Reduction Reactions by the Half-Reaction Method
Step 1. Equalize the number of electrons gained and lost.
Ce4+ gained 1e-
Sn2+ lost 2e-
Step 2: Multiply reduction half-reaction by 2
2e- + 2Ce4+ 2Ce3+
Sn2+ Sn4+ + 2e-
2e- + 2Ce4+ + Sn2+ 2Ce3+ + Sn4+ + 2e-
2Ce4+ + Sn2+ 2Ce3+ + Sn4+
Section 18.2
Balancing Oxidation-Reduction Reactions
Example 1: Oxidation-Reduction using Half-Reaction
Pb (s) + PbO2(s) + H+ (aq) Pb2+ + H2O (l)
Pb Pb2+ PbO2 Pb2+
0 2+ 4+ 2+
Pb Pb2+ + 2e- 4H+ + PbO2 Pb2+ + 2H2O
0 2+ 2- 4+ 2+
2e- + 4H+ + PbO2 Pb2+ + 2H2O
Oxidation Half-reaction Reduction Half-reaction
Pb Pb2+ + 2e-
2e- + 4H+ + PbO2 Pb2+ + 2H2O
Pb (s) + 4H+ + (aq) + PbO2 (s) 2Pb2+ (aq) + 2H2O (l)
Section 18.2
Balancing Oxidation-Reduction Reactions
Example 2: Oxidation-Reduction using Half-ReactionBalance the equation for the reaction between permanganate and iron II ions in acidic (H+) solution. The net ionic equation for this reaction is
MnO4- (aq) + Fe2+ (aq) Fe3+ (aq) + Mn2+ (aq)
MnO4- Mn2+
Mn7+ (4O2-) Mn2+ Reduction Half-reaction
8H+ + MnO4- Mn2+ + 4H2O
8+ + 1- = 7+ 2+ + 0 = 2+
5e- + 8H+ + MnO4- Mn2+ + 4H2O
Fe2+ Fe3+ Oxidation Half-reaction
Fe2+ Fe3+
2+ 3+
Fe2+ Fe3+ + 1e-
Section 18.2
Balancing Oxidation-Reduction Reactions
Example 2: Oxidation-Reduction using Half-Reaction
5e- + 8H+ + MnO4- Mn2+ + 4H2O
Fe2+ Fe3+ + 1e-
Equalize number of electrons by multiplying by 5.
5e- + 8H+ + MnO4- Mn2+ + 4H2O
5Fe2+ 5Fe3+ + 5e-
8H+ + MnO4- + 5Fe2+ Mn2+ + 4H2O + 5Fe3+
8H+(aq) + MnO4- (aq) + 5Fe2+ (aq) Mn2+ (aq) + 4H2O(l) + 5Fe3+ (aq)
Section 18.2
Balancing Oxidation-Reduction Reactions
Balancing Oxidation-Reduction Reactions by the Half-Reaction Method
Section 18.2
Balancing Oxidation-Reduction Reactions
Section 18.2
Balancing Oxidation-Reduction Reactions
Section 18.2
Balancing Oxidation-Reduction Reactions
HW
• Page 651: #6, 7
Section 18.2
Balancing Oxidation-Reduction Reactions
Warm-up – Balance in Acid
1) ClO3¯ + SO2 ---> SO42¯ + Cl¯
2) H2S + NO3¯ ---> S8 + NO
Section 18.2
Balancing Oxidation-Reduction Reactions
HW solutions
• 6a) Br2 + 2e- --> 2Br-
• 6b) Zn -> Zn2+ + 2e-
• 7b) 4H+ + Ni + 2NO3- -> Ni2+ + 2NO2 + 2H2O
• 7c) 2H+ + CLO4- + 2Cl- -> CLO3 + Cl2 +H2O
Section 18.2
Balancing Oxidation-Reduction Reactions
SWBAT balance oxidation-reduction half- equations using half reactions in a Half-Reaction Practice
Objectives
Section 18.2
Balancing Oxidation-Reduction Reactions
HW
• Page 663: #26, 27
• Show all work!
Section 18.2
Balancing Oxidation-Reduction Reactions
the final answer: 40H2S + 48H+ + 16MnO4¯ ---> 5S8 + 16Mn2+ + 64H2O
Warm-up – Balance in Acid 1) Give the oxidation state of all atoms2) Identify the Ox/Red half-rxns3) Balance the reaction
1) MnO4¯ + H2S ---> Mn2+ + S8
Oxidation States Mn=+7, O=-2, H=+1, S=-2 Mn=+2, S=0
Oxidation Reduction
8H2S ---> S8 + 16H+ + 16e¯ 5e¯ + 8H+ + MnO4¯ ---> Mn2+ + 4H2O
make the number of electrons equal: 5[8H2S ---> S8 + 16H+ + 16e¯] 16[5e¯ + 8H+ + MnO4¯ ---> Mn2+ + 4H2O]
Section 18.2
Balancing Oxidation-Reduction Reactions
SWBAT balance oxidation-reduction half- equations using half reactions in a Half-Reaction Practice Quiz
Objectives
Section 18.2
Balancing Oxidation-Reduction Reactions
HW
• None, if you finish your practice quiz
Section 18.2
Balancing Oxidation-Reduction Reactions
1) Cu + SO42¯ ---> Cu2+ + SO2
Cu ---> Cu2+ + 2e¯ 2e¯ + 4H+ + SO42¯ --->
SO2 + 2H2O
the final answer: Cu + 4H+ + SO4
2¯ ---> Cu2+ + SO2 + 2H2O
Warm-up – Balance in Acid 1) Give the oxidation state of all atoms2) Identify the Ox/Red half-rxns3) Balance the reaction