1. :-) Emoticons Are Born :-( (1982)
Reading for Thursday: Chapter 5 section 3
HOMEWORK – DUE Thursday 9/21/17
BW 4b: CH 4 #'s (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

2. 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

3. 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)

4. 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

5. 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

6. 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

7. 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

8. 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

9. 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−

10. 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.

11. 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.: x ( 5 Fe2+(aq) → 5 Fe3+(aq) e− )
red.: MnO4−(aq) e− H+(aq) → Mn2+(aq) H2O(l)

12. 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.

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

14. 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.

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

16. 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.

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

18. 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

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

20. 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.

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

22. 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

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

24. 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.

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

26. 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−.

27. Fe2+(aq) + MnO4−(aq) → Fe3+(aq) + Mn2+(aq)+2 +7 +3 +2
ox.: x ( 5 Fe2+(aq) → 5 Fe3+(aq) e− )
red.: MnO4−(aq) e− H+(aq) → Mn2+(aq) 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–