Section 18.2 Balancing Oxidation-Reduction Reactions Warm-up Determine the oxidation state… (1) Na Na = 0 (2) MgI 2 Mg = +2, I= -1 (3) (NH 4 ) 2 O N =-3, H = +1, O = -2 (4) NaOH Na = +1, O = -2, H = +1 (5) Cl 2 Cl = 0 (6) Al(NO 2 ) 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). Cl 2  reduced –Cl 2 is also called the oxidizing agent (electron acceptor). 2Na(s) + Cl 2 (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 CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g) Steps:

Section 18.2 Balancing Oxidation-Reduction Reactions Oxidation-Reduction Reactions Between Nonmetals C  oxidized –CH 4 is the reducing agent. O 2  reduced –O 2 is the oxidizing agent. CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g) C=-4 O=0 C=+4H=+1 H=+1 O=-2O=-2

Section 18.2 Balancing Oxidation-Reduction Reactions Practice Together 2Na(s) + Cl 2 (aq) --> 2NaCl(s)

Section 18.2 Balancing Oxidation-Reduction Reactions Practice Together Zn(s) + Cu 2+ (aq) --> Zn 2+ (aq) + Cu(s)

Section 18.2 Balancing Oxidation-Reduction Reactions Practice Together 2Mg(s) + O 2 (g) --> 2MgO

Section 18.2 Balancing Oxidation-Reduction Reactions You do it…#1 Fe = +3 C = +2 Fe = 0C = +4 O = -2 O = -2 O = -2 Fe  reduced –CO is the reducing agent. C  oxidized –Fe 2 O 3 is the oxidizing agent. Fe 2 O 3 (s) + 3CO(g)  2Fe(l) + 3CO 2 (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 –Cl 2 is the oxidizing agent. Cl 2 (g) + 2NaBr(aq)  2NaCl(aq) + Br 2 (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 –H 2 is the reducing agent. H  oxidized –N 2 is the oxidizing agent. N 2 (g) + 3H 2 (g)  2NH 3 (g)

Section 18.2 Balancing Oxidation-Reduction Reactions TOD 2PbS(s) + 3O 2 (g)  2PbO(s) + 2SO 2 (g) –Oxidizing Agent = O 2 –Reducing Agent = PbS 2PbO(s) + CO(g)  Pb(s) + CO 2 (g) –Oxidizing Agent = PbO –Reducing Agent = CO 2Na(s) + Cl 2 (g)  2NaCl(aq) –Oxidizing Agent = Cl 2 –Reducing Agent = Na

Section 18.2 Balancing Oxidation-Reduction Reactions HW Page 651 #1-5

Section 18.2 Balancing Oxidation-Reduction Reactions Warm-up Identify the following: 1) Oxidation States, 2) Oxidation/Reduction 3) Oxidizing/Reducing Agents 2H 2 (g) + O 2 (g)  2H 2 O (l) –(0) (0) (+1) (-2) –Oxidized = HReduced = O –Oxidizing Agent = O 2 Reducing Agent = H 2 2PbS(s) + 3O 2 (g)  2PbO(s) + 2SO 2 (g) –(+2) (-2) (0) (+2) (-2) (+4) (-2) –Oxidized = SReduced = O –Oxidizing Agent = O 2 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-up Identify the following: 1) Oxidation States, 2) Oxidation/Reduction 3) Oxidizing/Reducing Agents 6Na(s) + N 2 (g)  2Na 3 N (s) –(0) (0) (+1) (-3) –Oxidized = NaReduced = N –Oxidizing Agent = N 2 Reducing Agent = Na 2PbO(s) + CO(g)  Pb(s) + CO 2 (g) –(+2) (-2) (+2) (-2) (0) (+4) (-2) –Oxidized = CReduced = 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 + NO 3 ¯ ---> SnO 2 + NO 2 Sn ---> SnO 2 and NO 3 ¯ ---> NO 2

Section 18.2 Balancing Oxidation-Reduction Reactions HClO + Co ---> Cl 2 + Co 2+ HClO ---> Cl 2 and Co ---> Co 2+ Separate into half-reactions

Section 18.2 Balancing Oxidation-Reduction Reactions NO 2 ---> NO 3 ¯ + NO NO 2 ---> NO 3 ¯ and NO 2 ---> NO Separate into half-reactions

Section 18.2 Balancing Oxidation-Reduction Reactions Practice balancing electrons Cl 2 ---> Cl¯ Cl 2 + 2e¯ ---> 2Cl¯

Section 18.2 Balancing Oxidation-Reduction Reactions Sn ---> Sn 2+ Sn ---> Sn e¯ Practice balancing electrons

Section 18.2 Balancing Oxidation-Reduction Reactions Fe > Fe 3+ Fe > Fe 3+ + e¯ Practice balancing electrons

Section 18.2 Balancing Oxidation-Reduction Reactions I 3 ¯ ---> I¯ I 3 ¯ + 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. Ce 4+ gained 1e- Sn 2+ lost 2e- Step 2: Multiply reduction half-reaction by 2 2e - + 2Ce 4+  2Ce 3+ Sn 2+  Sn e - 2e- + 2Ce 4+ + Sn 2+  2Ce 3+ + Sn e- 2Ce 4+ + Sn 2+  2Ce 3+ + Sn 4+

Section 18.2 Balancing Oxidation-Reduction Reactions Example 1: Oxidation-Reduction using Half-Reaction Pb (s) + PbO 2 (s) + H + (aq)  Pb 2+ + H 2 O (l) Pb  Pb 2+ PbO 2  Pb Pb  Pb e- 4H+ + PbO 2  Pb H 2 O e- + 4H + + PbO 2  Pb H 2 O Oxidation Half-reaction Reduction Half-reaction Pb  Pb e- 2e- + 4H + + PbO 2  Pb H 2 O Pb (s) + 4H + + (aq) + PbO 2 (s)  2Pb 2+ (aq) + 2H 2 O (l)

Section 18.2 Balancing Oxidation-Reduction Reactions Example 2: Oxidation-Reduction using Half-Reaction Balance the equation for the reaction between permanganate and iron II ions in acidic (H+) solution. The net ionic equation for this reaction is MnO 4 - (aq) + Fe 2+ (aq)  Fe 3+ (aq) + Mn 2+ (aq) MnO 4 -  Mn 2+ Mn 7+ (4O 2- ) Mn 2+ Reduction Half-reaction 8H + + MnO 4 -  Mn H 2 O = = 2+ 5e- + 8H+ + MnO 4 -  Mn H 2 O Fe 2+  Fe 3+ Oxidation Half-reaction Fe 2+  Fe Fe 2+  Fe e-

Section 18.2 Balancing Oxidation-Reduction Reactions Example 2: Oxidation-Reduction using Half-Reaction 5e- + 8H+ + MnO 4 -  Mn H 2 O Fe 2+  Fe e- Equalize number of electrons by multiplying by 5. 5e- + 8H+ + MnO 4 -  Mn H 2 O 5Fe 2+  5Fe e- 8H+ + MnO Fe 2+  Mn H 2 O + 5Fe 3+ 8H+(aq) + MnO 4 - (aq) + 5Fe 2+ (aq)  Mn 2+ (aq) + 4H 2 O(l) + 5Fe 3+ (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) ClO 3 ¯ + SO 2 ---> SO 4 2 ¯ + Cl¯ 2) H 2 S + NO 3 ¯ ---> S 8 + NO

Section 18.2 Balancing Oxidation-Reduction Reactions HW solutions 6a) Br 2 + 2e- --> 2Br- 6b) Zn -> Zn e- 7b) 4H+ + Ni + 2NO 3 - -> Ni NO2 + 2H 2 O 7c) 2H+ + CLO Cl- -> CLO 3 + Cl 2 +H 2 O

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: 40H 2 S + 48H MnO 4 ¯ ---> 5S Mn H 2 O Warm-up – Balance in Acid 1)Give the oxidation state of all atoms 2)Identify the Ox/Red half-rxns 3)Balance the reaction 1) MnO 4 ¯ + H 2 S ---> Mn 2+ + S 8 Oxidation StatesMn=+7, O=-2, H=+1, S=-2  Mn=+2, S=0 OxidationReduction 8H 2 S ---> S H e¯ 5e¯ + 8H + + MnO 4 ¯ ---> Mn H 2 O make the number of electrons equal: 5[8H 2 S ---> S H e¯] 16[5e¯ + 8H + + MnO 4 ¯ ---> Mn H 2 O]

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 + SO 4 2 ¯ ---> Cu 2+ + SO 2 Cu ---> Cu e¯ 2e¯ + 4H + + SO 4 2 ¯ ---> SO 2 + 2H 2 O the final answer: Cu + 4H + + SO 4 2 ¯ ---> Cu 2+ + SO 2 + 2H 2 O Warm-up – Balance in Acid 1)Give the oxidation state of all atoms 2)Identify the Ox/Red half-rxns 3)Balance the reaction