Section 18 2 Balancing OxidationReduction Reactions Warmup Determine

Section 18. 2 Balancing Oxidation-Reduction Reactions Warm-up • Determine the oxidation state… • (1) Na Na = 0 • (2) Mg. I 2 Mg = +2, I= -1 • (3) (NH 4)2 O N =-3, H = +1, O = -2 • (4) Na. OH 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 Objectives 1. SWBAT identify oxidizing and reducing agents in a TOD.

Section 18. 2 Balancing Oxidation-Reduction Reactions A. Oxidation-Reduction Reactions Between Nonmetals 2 Na(s) + Cl 2(g) 2 Na. Cl(s) • Na oxidized – Na is also called the reducing agent (electron donor). • Cl 2 reduced – Cl 2 is also called the oxidizing agent (electron acceptor).

Section 18. 2 Balancing Oxidation-Reduction Reactions Identifying Oxidation-Reduction Reactions Between Nonmetals CH 4(g) + 2 O 2(g) CO 2(g) + 2 H 2 O(g) Steps: 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

Section 18. 2 Balancing Oxidation-Reduction Reactions Between Nonmetals CH 4(g) + 2 O 2(g) CO 2(g) + 2 H 2 O(g) C=-4 O=0 C=+4 H=+1 O=-2 • C oxidized – CH 4 is the reducing agent. • O 2 reduced – O 2 is the oxidizing agent.

Section 18. 2 Balancing Oxidation-Reduction Reactions Practice Together • 2 Na(s) + Cl 2(aq) --> 2 Na. Cl(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 • 2 Mg(s) + O 2(g) --> 2 Mg. O

Section 18. 2 Balancing Oxidation-Reduction Reactions You do it…#1 Fe 2 O 3(s) + 3 CO(g) 2 Fe(l) + 3 CO 2(g) Fe = +3 O = -2 C = +2 O = -2 Fe = 0 C = +4 O = -2 • Fe reduced – CO is the reducing agent. • C oxidized – Fe 2 O 3 is the oxidizing agent.

Section 18. 2 Balancing Oxidation-Reduction Reactions You do it…#2 Cl 2(g) + 2 Na. Br(aq) 2 Na. Cl(aq) + Br 2(l) Cl = 0 • Cl – • Br – Na =+1 Br = -1 Na = +1 Cl = -1 reduced Na. Br is the reducing agent. oxidized Cl 2 is the oxidizing agent. Br = 0

Section 18. 2 Balancing Oxidation-Reduction Reactions You do it…#3 (Reaction to make fertilizer) N 2(g) + 3 H 2(g) 2 NH 3(g) N=0 • N – • H – H =0 N = -3 H = +1 reduced H 2 is the reducing agent. oxidized N 2 is the oxidizing agent.

Section 18. 2 Balancing Oxidation-Reduction Reactions TOD • 2 Pb. S(s) + 3 O 2(g) 2 Pb. O(s) + 2 SO 2(g) – Oxidizing Agent = O 2 – Reducing Agent = Pb. S • 2 Pb. O(s) + CO(g) Pb(s) + CO 2(g) – Oxidizing Agent = Pb. O – Reducing Agent = CO • 2 Na(s) + Cl 2(g) 2 Na. Cl(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 • 2 H 2 (g) + O 2 (g) 2 H 2 O (l) – (0) (+1) (-2) – Oxidized = H – Oxidizing Agent = O 2 Reduced = O Reducing Agent = H 2 • 2 Pb. S(s) + 3 O 2(g) 2 Pb. O(s) + 2 SO 2(g) – (+2) (-2) (0) – Oxidized = S – Oxidizing Agent = O 2 (+2) (-2) (+4) (-2) Reduced = O Reducing Agent = Pb. S

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 • 6 Na(s) + N 2 (g) 2 Na 3 N (s) – (0) (+1) (-3) – Oxidized = Na – Oxidizing Agent = N 2 Reduced = N Reducing Agent = Na • 2 Pb. O(s) + CO(g) Pb(s) + CO 2(g) – (+2) (-2) (0) – Oxidized = C – Oxidizing Agent = Pb. O (+4) (-2) Reduced = Pb Reducing Agent = CO

Section 18. 2 Balancing Oxidation-Reduction Reactions Objectives SWBAT balance oxidation-reduction equations using half reactions in a Half-Reaction Practice

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¯ ---> Sn. O 2 + NO 2 • Sn ---> Sn. O 2 and NO 3¯ ---> NO 2

Section 18. 2 Balancing Oxidation-Reduction Reactions Separate into half-reactions • HCl. O + Co ---> Cl 2 + Co 2+ • HCl. O ---> Cl 2 and Co ---> Co 2+

Section 18. 2 Balancing Oxidation-Reduction Reactions Separate into half-reactions • NO 2 ---> NO 3¯ + NO • NO 2 ---> NO 3¯ and NO 2 ---> NO

Section 18. 2 Balancing Oxidation-Reduction Reactions Practice balancing electrons • Cl 2 ---> Cl¯ • Cl 2 + 2 e¯ ---> 2 Cl¯

Section 18. 2 Balancing Oxidation-Reduction Reactions Practice balancing electrons • Sn ---> Sn 2+ + 2 e¯

Section 18. 2 Balancing Oxidation-Reduction Reactions Practice balancing electrons • Fe 2+ ---> Fe 3+ + e¯

Section 18. 2 Balancing Oxidation-Reduction Reactions Practice balancing electrons • I 3¯ ---> I¯ • I 3¯ + 2 e¯ ---> 3 I¯

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

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 Balancing Oxidation-Reduction Reactions by the Half-Reaction Method Step 1. Equalize the number of electrons gained and lost. Ce 4+ gained 1 e. Sn 2+ lost 2 e. Step 2: Multiply reduction half-reaction by 2 2 e- + 2 Ce 4+ 2 Ce 3+ Sn 2+ Sn 4+ + 2 e 2 e- + 2 Ce 4+ + Sn 2+ 2 Ce 3+ + Sn 4+ + 2 e 2 Ce 4+ + Sn 2+ 2 Ce 3+ + Sn 4+

Section 18. 2 Balancing Oxidation-Reduction Reactions Example 1: Oxidation-Reduction using Half-Reaction Pb (s) + Pb. O 2(s) + H+ (aq) Pb 2+ + H 2 O (l) Pb 2+ Pb. O 2 Pb 2+ 0 2+ 4+ 2+ Pb 2+ + 2 e 0 2+ 2 Oxidation Half-reaction 4 H+ + Pb. O 2 Pb 2+ + 2 H 2 O 4+ 2+ 2 e- + 4 H+ + Pb. O 2 Pb 2+ + 2 H 2 O Reduction Half-reaction Pb 2+ + 2 e 2 e- + 4 H+ + Pb. O 2 Pb 2+ + 2 H 2 O Pb (s) + 4 H+ + (aq) + Pb. O 2 (s) 2 Pb 2+ (aq) + 2 H 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 Mn. O 4 - (aq) + Fe 2+ (aq) Fe 3+ (aq) + Mn 2+ (aq) Mn. O 4 Mn 7+ (4 O 2 -) Mn 2+ Reduction Half-reaction 8 H+ + Mn. O 4 Mn 2+ + 4 H 2 O 8+ + 1 - = 7+ 2+ + 0 5 e- + 8 H+ + Mn. O 4 Mn 2+ + 4 H 2 O Fe 2+ Fe 3+ Oxidation Half-reaction Fe 2+ Fe 3+ 2+ 3+ Fe 2+ Fe 3+ + 1 e- = 2+

Section 18. 2 Balancing Oxidation-Reduction Reactions Example 2: Oxidation-Reduction using Half-Reaction 5 e- + 8 H+ + Mn. O 4 Mn 2+ + 4 H 2 O Fe 2+ Fe 3+ + 1 e. Equalize number of electrons by multiplying by 5. 5 e- + 8 H+ + Mn. O 4 Mn 2+ + 4 H 2 O 5 Fe 2+ 5 Fe 3+ + 5 e 8 H+ + Mn. O 4 - + 5 Fe 2+ Mn 2+ + 4 H 2 O + 5 Fe 3+ 8 H+(aq) + Mn. O 4 - (aq) + 5 Fe 2+ (aq) Mn 2+ (aq) + 4 H 2 O(l) + 5 Fe 3+ (aq)

Section 18. 2 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) Cl. O 3¯ + SO 2 ---> SO 42¯ + Cl¯ 2) H 2 S + NO 3¯ ---> S 8 + NO

Section 18. 2 Balancing Oxidation-Reduction Reactions HW solutions • 6 a) Br 2 + 2 e- --> 2 Br • 6 b) Zn -> Zn 2+ + 2 e • 7 b) 4 H+ + Ni + 2 NO 3 - -> Ni 2+ + 2 NO 2 + 2 H 2 O • 7 c) 2 H+ + CLO 4 - + 2 Cl- -> CLO 3 + Cl 2 +H 2 O

Section 18. 2 Balancing Oxidation-Reduction Reactions Objectives SWBAT balance oxidation-reduction half- equations using half reactions in a Half-Reaction Practice

Section 18. 2 Balancing Oxidation-Reduction Reactions HW • Page 663: #26, 27 • Show all work!

Section 18. 2 Balancing Oxidation-Reduction Reactions 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) Mn. O 4¯ + H 2 S ---> Mn 2+ + S 8 Oxidation States Mn=+7, O=-2, H=+1, S=-2 Mn=+2, S=0 Oxidation 8 H 2 S ---> S 8 + 16 H+ + 16 e¯ Reduction 5 e¯ + 8 H+ + Mn. O 4¯ ---> Mn 2+ + 4 H 2 O make the number of electrons equal: 5[8 H 2 S ---> S 8 + 16 H+ + 16 e¯] 16[5 e¯ + 8 H+ + Mn. O 4¯ ---> Mn 2+ + 4 H 2 O] the final answer: 40 H 2 S + 48 H+ + 16 Mn. O 4¯ ---> 5 S 8 + 16 Mn 2+ + 64 H 2 O

Section 18. 2 Balancing Oxidation-Reduction Reactions Objectives SWBAT balance oxidation-reduction half- equations using half reactions in a Half-Reaction Practice Quiz

Section 18. 2 Balancing Oxidation-Reduction Reactions HW • None, if you finish your practice quiz

Section 18. 2 Balancing Oxidation-Reduction Reactions 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) Cu + SO 42¯ ---> Cu 2+ + SO 2 Cu ---> Cu 2+ + 2 e¯ SO 2 + 2 H 2 O 2 e¯ + 4 H+ + SO 42¯ ---> the final answer: Cu + 4 H+ + SO 42¯ ---> Cu 2+ + SO 2 + 2 H 2 O