:-) Emoticons Are Born :-( (1982) Reading for Thursday: Chapter 5 section 3 HOMEWORK – DUE Thursday 9/21/17 BW 4b: CH 4 #'s 9 - 23 (odd), 37, 39, 41 WS 7: (from course website) Lab Wednesday/Thursday EXP 6 – wet lab Prelab Lab Monday/Tuesday EXP 6 continued My office has moved LIB 023

Rules for Assigning Oxidation Numbers 5) Oxidation numbers for nonmetals are typically found based on their group typically?!?! Fluorine = –1 always! OF2 (F = –1, O = +2) Hydrogen = +1 unless paired with a metal NaH (Na = +1, H = –1) Oxygen = –2 unless peroxide BaO2 (Ba = +2, O = –1) Other nonmetals: the element closest to fluorine on the PT gets to keep its “usual” O.S. SCl4 ClO3–1 P2S5 O.S. chlorine = –1 O.S. oxygen = –2 O.S. sulfur = –2 O.S. sulfur = +4 O.S. chlorine = +5 O.S. phosphorus = +5

Practice!! Assign oxidation numbers to all atoms in each of the following: a) H2CO b) S2O32– c) NH4+ d) NO3– e) Br2 f) Ca3(PO4)2 g) 2 K(s) + 2 H2O(l)  2 KOH(aq) + H2(g)

Oxidation and Reduction Oxidation is the process that occurs when the oxidation number increases (gets more positive) an element loses electrons a compound adds bonds to oxygen a compound loses bonds to hydrogen a half–reaction has electrons as products Reduction is the process that occurs when the oxidation number of an element decreases (gets more negative) an element gains electrons a compound loses bonds to oxygen a compound gains bonds to hydrogen a half–reaction has electrons as reactants

Redox Reactions Cu(s) + 2 Ag+(aq)  Cu2+(aq) + 2 Ag(s) +1 +2 Lost Electrons OXIDIZED reducing agent Cu(s) + 2 Ag+(aq)  Cu2+(aq) + 2 Ag(s) Gain Electrons REDUCED oxidizing agent

Redox Reactions 2 I–(aq) + Br2(aq)  I2(aq) + 2 Br–(aq) –1 –1 Lost Electrons OXIDIZED reducing agent 2 I–(aq) + Br2(aq)  I2(aq) + 2 Br–(aq) Gain Electrons REDUCED oxidizing agent

Redox Reactions 3 Cl2(g) + 2 Al(s)  2 AlCl3(s) +3 –1 Gain Electrons REDUCED oxidizing agent 3 Cl2(g) + 2 Al(s)  2 AlCl3(s) reducing agent Lost Electrons OXIDIZED

Redox Reactions Cr2O72–(aq) + IO–(aq)  IO3–(aq) + Cr3+(aq) +6 +1 +5 +3 Lost Electrons OXIDIZED reducing agent Cr2O72–(aq) + IO–(aq)  IO3–(aq) + Cr3+(aq) Gain Electrons REDUCED oxidizing agent

Half–Reactions 3 Cl2 + I− + 3H2O → 6 Cl− + IO3− + 6 H+ We can split a RedOx reaction into two separate half–reactions (or half–cell), each involving only an oxidation OR a reduction. Oxidation half–cells have electrons as products (OIL) Reduction half–cells have electrons as reactants (RIG) 3 Cl2 + I− + 3H2O → 6 Cl− + IO3− + 6 H+ –1 +1 –2 –1 +5 –2 +1 Reduction: Cl2 + 2 e− → 2 Cl− Reduction: Cl2 + 2 e− → 2 Cl− Reduction: Cl2 + 2 e− → 2 Cl− Oxidation: I− → IO3− + 6 e− Oxidation: I− → IO3− + 6 e− Oxidation: I− → IO3− + 6 e−

Balancing Redox Reactions – Half–Cells 1. Write oxidation and reduction half–cells a) Assign oxidation numbers to determine what was oxidized and what was reduced.

Fe2+(aq) + MnO4−(aq) → Fe3+(aq) + Mn2+(aq) Lost Electrons OXIDIZED Fe2+(aq) + MnO4−(aq) → Fe3+(aq) + Mn2+(aq) +2 +7 +3 +2 Gain Electrons REDUCED ox.: 5x ( 5 Fe2+(aq) → 5 Fe3+(aq) + 5 e− ) red.: MnO4−(aq) + 5 e− + 8 H+(aq) → Mn2+(aq) + 4 H2O(l)

Balancing Redox Reactions – Half–Cells 1. Write oxidation and reduction half–cells a) Assign oxidation numbers to determine what was oxidized and what was reduced. 2. Balance half–cells by mass. a) First balance elements other than H and O.

Fe2+(aq) + MnO4−(aq) → Fe3+(aq) + Mn2+(aq) +2 +7 +3 +2 ox.: 5x ( 5 Fe2+(aq) → 5 Fe3+(aq) + 5 e− ) red.: MnO4−(aq) + 5 e− + 8 H+(aq) → Mn2+(aq) + 4 H2O(l)

Balancing Redox Reactions – Half–Cells 1. Write oxidation and reduction half–cells a) Assign oxidation numbers to determine what was oxidized and what was reduced. 2. Balance half–cells by mass. a) First balance elements other than H and O. b) Add H2O where O is needed.

Fe2+(aq) + MnO4−(aq) → Fe3+(aq) + Mn2+(aq) +2 +7 +3 +2 ox.: 5x ( 5 Fe2+(aq) → 5 Fe3+(aq) + 5 e− ) red.: MnO4−(aq) + 5 e− + 8 H+(aq) → Mn2+(aq) + 4 H2O(l) red.: MnO4−(aq) + 5 e− + 8 H+(aq) → Mn2+(aq) + 4 H2O(l) red.: MnO4−(aq) + 5 e− + 8 H+(aq) → Mn2+(aq) + 4 H2O(l)

Balancing Redox Reactions – Half–Cells 1. Write oxidation and reduction half–cells a) Assign oxidation numbers to determine what was oxidized and what was reduced. 2. Balance half–cells by mass. a) First balance elements other than H and O. b) Add H2O where O is needed. c) Add H+ where H is needed.

Fe2+(aq) + MnO4−(aq) → Fe3+(aq) + Mn2+(aq) +2 +7 +3 +2 ox.: 5x ( 5 Fe2+(aq) → 5 Fe3+(aq) + 5 e− ) red.: MnO4−(aq) + 5 e− + 8 H+(aq) → Mn2+(aq) + 4 H2O(l) red.: MnO4−(aq) + 5 e− + 8 H+(aq) → Mn2+(aq) + 4 H2O(l) red.: MnO4−(aq) + 5 e− + 8 H+(aq) → Mn2+(aq) + 4 H2O(l)

Balancing Redox Reactions – Half–Cells 1. Write oxidation and reduction half–cells a) Assign oxidation numbers to determine what was oxidized and what was reduced. 2. Balance half–cells by mass. a) First balance elements other than H and O. b) Add H2O where O is needed. c) Add H+ where H is needed. 3. Balance half–cells by charge. a) Balance charge by adding electrons

Fe2+(aq) + MnO4−(aq) → Fe3+(aq) + Mn2+(aq) +2 +7 +3 +2 ox.: 5x ( 5 Fe2+(aq) → 5 Fe3+(aq) + 5 e− ) ox.: 5x ( 5 Fe2+(aq) → 5 Fe3+(aq) + 5 e− ) red.: MnO4−(aq) + 5 e− + 8 H+(aq) → Mn2+(aq) + 4 H2O(l) red.: MnO4−(aq) + 5 e− + 8 H+(aq) → Mn2+(aq) + 4 H2O(l) red.: MnO4−(aq) + 5 e− + 8 H+(aq) → Mn2+(aq) + 4 H2O(l)

Balancing Redox Reactions – Half–Cells 1. Write oxidation and reduction half–cells a) Assign oxidation numbers to determine what was oxidized and what was reduced. 2. Balance half–cells by mass. a) First balance elements other than H and O. b) Add H2O where O is needed. c) Add H+ where H is needed. 3. Balance half–cells by charge. a) Balance charge by adding electrons 4. Balance electrons between half–cells.

Fe2+(aq) + MnO4−(aq) → Fe3+(aq) + Mn2+(aq) +2 +7 +3 +2 ox.: 5x ( 5 Fe2+(aq) → 5 Fe3+(aq) + 5 e− ) ox.: 5x ( 5 Fe2+(aq) → 5 Fe3+(aq) + 5 e− ) ox.: 5x ( 5 Fe2+(aq) → 5 Fe3+(aq) + 5 e− ) red.: MnO4−(aq) + 5 e− + 8 H+(aq) → Mn2+(aq) + 4 H2O(l)

Balancing Redox Reactions – Half–Cells 1. Write oxidation and reduction half–cells a) Assign oxidation numbers to determine what was oxidized and what was reduced. 2. Balance half–cells by mass. a) First balance elements other than H and O. b) Add H2O where O is needed. c) Add H+ where H is needed. 3. Balance half–cells by charge. a) Balance charge by adding electrons 4. Balance electrons between half–cells. 5. Add half–cells

Fe2+(aq) + MnO4−(aq) → Fe3+(aq) + Mn2+(aq) +2 +7 +3 +2 ox.: 5x ( 5 Fe2+(aq) → 5 Fe3+(aq) + 5 e− ) red.: MnO4−(aq) + 5 e− + 8 H+(aq) → Mn2+(aq) + 4 H2O(l) 5 Fe2+(aq) + MnO4−(aq) + 5 e− + 8 H+(aq) → 5 Fe3+(aq) + 5 e− + Mn2+(aq) + 4 H2O(l) 5 Fe2+(aq) + MnO4−(aq) + 5 e− + 8 H+(aq) → 5 Fe3+(aq) + 5 e− + Mn2+(aq) + 4 H2O(l) 5 Fe2+(aq) + MnO4−(aq) + 5 e− + 8 H+(aq) → 5 Fe3+(aq) + 5 e− + Mn2+(aq) + 4 H2O(l) 5 Fe2+(aq) + MnO4−(aq) + 5 e− + 8 H+(aq) → 5 Fe3+(aq) + 5 e− + Mn2+(aq) + 4 H2O(l) 5 Fe2+(aq) + MnO4−(aq) + 8 H+(aq) → 5 Fe3+(aq) + Mn2+(aq) + 4 H2O(l)

Balancing Redox Reactions – Half–Cells 1. Write oxidation and reduction half–cells a) Assign oxidation numbers to determine what was oxidized and what was reduced. 2. Balance half–cells by mass. a) First balance elements other than H and O. b) Add H2O where O is needed. c) Add H+ where H is needed. 3. Balance half–cells by charge. a) Balance charge by adding electrons 4. Balance electrons between half–cells. 5. Add half–cells 6. Check by counting atoms and total charge.

Fe2+(aq) + MnO4−(aq) → Fe3+(aq) + Mn2+(aq) +2 +7 +3 +2 ox.: 5x ( 5 Fe2+(aq) → 5 Fe3+(aq) + 5 e− ) red.: MnO4−(aq) + 5 e− + 8 H+(aq) → Mn2+(aq) + 4 H2O(l) 5 Fe2+(aq) + MnO4−(aq) + 5 e− + 8 H+(aq) → 5 Fe3+(aq) + 5 e− + Mn2+(aq) + 4 H2O(l) 5 Fe2+(aq) + MnO4−(aq) + 8 H+(aq) → 5 Fe3+(aq) + Mn2+(aq) + 4 H2O(l)

Balancing Redox Reactions – Half–Cells 1. Write oxidation and reduction half–cells a) Assign oxidation numbers to determine what was oxidized and what was reduced. 2. Balance half–cells by mass. a) First balance elements other than H and O. b) Add H2O where O is needed. c) Add H+ where H is needed. 3. Balance half–cells by charge. a) Balance charge by adding electrons 4. Balance electrons between half–cells. 5. Add half–cells 6. Check by counting atoms and total charge. If the reaction is done in a base, neutralize H+ with OH−.

Fe2+(aq) + MnO4−(aq) → Fe3+(aq) + Mn2+(aq) +2 +7 +3 +2 ox.: 5x ( 5 Fe2+(aq) → 5 Fe3+(aq) + 5 e− ) red.: MnO4−(aq) + 5 e− + 8 H+(aq) → Mn2+(aq) + 4 H2O(l) 5 Fe2+(aq) + MnO4−(aq) + 5 e− + 8 H+(aq) → 5 Fe3+(aq) + 5 e− + Mn2+(aq) + 4 H2O(l) 5 Fe2+(aq) + MnO4−(aq) + 8 H+(aq) → 5 Fe3+(aq) + Mn2+(aq) + 4 H2O(l) + 8 OH– + 8 OH– 5 Fe2+(aq) + MnO4−(aq) + 8 H2O(l) → 5 Fe3+(aq) + Mn2+(aq) + 4 H2O(l) + 8 OH– 5 Fe2+(aq) + MnO4−(aq) + 4 H2O(l) → 5 Fe3+(aq) + Mn2+(aq) + 8 OH–