section 18.2 balancing oxidation-reduction reactions warm-up determine the oxidation state… (1) na...

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


1. SWBAT identify oxidizing and reducing agents in a TOD.

Objectives

Section 18.2


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


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


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


Practice Together

• 2Na(s) + Cl2(aq) --> 2NaCl(s)

Section 18.2


Practice Together

• Zn(s) + Cu2+ (aq) --> Zn2+(aq) + Cu(s)

Section 18.2


Practice Together

• 2Mg(s) + O2(g) --> 2MgO

Section 18.2


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


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


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


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


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


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


Separate into half-reactions

• Sn + NO3¯ ---> SnO2 + NO2

• Sn ---> SnO2 and NO3¯ ---> NO2

Section 18.2


• HClO + Co ---> Cl2 + Co2+

• HClO ---> Cl2 and Co ---> Co2+


Section 18.2


• NO2 ---> NO3¯ + NO

• NO2 ---> NO3¯ and NO2 ---> NO


Section 18.2


Practice balancing electrons

• Cl2 ---> Cl¯

• Cl2 + 2e¯ ---> 2Cl¯

Section 18.2


• Sn ---> Sn2+

• Sn ---> Sn2+ + 2e¯


Section 18.2


• Fe2+ ---> Fe3+

• Fe2+ ---> Fe3+ + e¯


Section 18.2


• I3¯ ---> I¯

• I3¯ + 2e¯ ---> 3I¯


Section 18.2


HW

• Page 663: #24(b,c,d), 25, 26

Section 18.2


Warm-up

Complete the Handout…

Section 18.2


SWBAT balance oxidation-reduction equations using half reactions in a Half-Reaction Practice

Objectives

Section 18.2


Balancing Oxidation-Reduction Reactions by the Half-Reaction Method

• Half reaction – equation which has electrons as products or reactants

Section 18.2



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


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


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


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



Section 18.2


HW

• Page 651: #6, 7

Section 18.2


Warm-up – Balance in Acid

1) ClO3¯ + SO2 ---> SO42¯ + Cl¯

2) H2S + NO3¯ ---> S8 + NO

Section 18.2


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


SWBAT balance oxidation-reduction half- equations using half reactions in a Half-Reaction Practice

Objectives

Section 18.2


HW

• Page 663: #26, 27

• Show all work!

Section 18.2


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


SWBAT balance oxidation-reduction half- equations using half reactions in a Half-Reaction Practice Quiz

Objectives

Section 18.2


HW

• None, if you finish your practice quiz

Section 18.2


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

section 18.2 balancing oxidation-reduction reactions warm-up determine the oxidation state… (1) na...

Documents