Download Assigning Oxidation Numbers and Definitions

Oxidation and Reduction Reactions
(called “redox”)
OBJECTIVES
Define the terms oxidation
and reduction.
Don’t Write This..
Early chemists saw “oxidation” reactions
only as the combination of a material with
oxygen to produce an oxide.
But, not all oxidation processes that use
oxygen involve burning. (Rust, H2O2)
A process called “reduction” is the
opposite of oxidation, and originally meant
the loss of oxygen from a compound.
Do Write This!
Oxidation and reduction reactions
always occur simultaneously – some
element oxidizes while the other reduces.
The definitions have become more
broad over time.
Oxidized is when a substance loses
electrons.
Reduced Is when the substance gains
electrons. (Sounds backwards right?)
Oxidation and Reduction (Redox)
Redox reactions mean that electrons
are transferred between reactants
Mg
+
S→
Mg2+
+
S2-
(MgS)
•The magnesium atom (which has zero charge)
changes to a magnesium ion by losing 2
electrons, and is oxidized to Mg2+.
(The charge increased!)
•The sulfur atom (which has no charge) is
changed to a sulfide ion by gaining 2 electrons,
and is reduced to S2- (The charge was reduced!)
0
1
0
1
2 Na  Cl 2  2 Na Cl
Half Reactions: We’ll talk about this later
Each sodium atom loses one electron:
1
0
Na  Na  e

OXIDIZIED!
Each chlorine atom gains one electron:
0

1
Cl  e  Cl
REDUCED!
How can you remember the terms?
LEO the lion says GER :
Lose Electrons = Oxidation
0
1
Na  Na  e

Gain Electrons = Reduction
0

1
Cl  e  Cl
Here’s where it get confusing…
- Magnesium is oxidized. But it causes the
Sulfur to be reduced. Therefore
Magnesium is the reducing agent.
- Sulfur is being reduced. But it causes the
Magnesium to be oxidized. Therefore
Sulfur is the oxidizing agent.
Mg is the
reducing
agent
Mg is oxidized: loses e-, becomes a Mg2+ ion
Mg(s) + S(s) → Mg2+(aq) + S2- (aq)
S is the oxidizing agent
S is reduced: gains e- = S2- ion
Practice Problems
Identify the substance being reduced and
oxidized: Mg + Cl  Mg+1(aq) Cl2-1(aq)
Magnesium = oxidized
Chlorine = reduced
Identify the reduction agent and
oxidizing agent: Zn +2Ag1+ Zn2+ +2Ag
Zinc = oxidized = Reducing agent
Silver = reduced = Oxidizing agent
Oxidation Numbers
OBJECTIVES
Determine the oxidation
number of each atom that
make up a molecule or
chemical compound.
Assigning Oxidation Numbers
• An “oxidation number” is a positive or
negative number assigned to an atom to
indicate its gain or loss of an electron.
• Most elements will be the same as they
are on the periodic table.
• However, some elements will always
have certain oxidation numbers. Some
elements can have different oxidation
numbers. (Transition Metals, C, N, S, P)
Rules for Assigning Oxidation Numbers
1) The oxidation number of any element by
itself (not in a compound) is zero.
0
1
0
1
2 Na  Cl 2  2 Na Cl
2) The oxidation number of a monatomic
ion is the same as its charge.


Ag  e  Ag
3) The sum of the charges of all the atoms
in the compound or molecule must
equal 0.
H2O
2(+1) + (-2) = 0
H
O
Ca(O H ) 2
(+2) + 2(-2) + 2(+1) = 0
Ca
O
H
4) Oxygen’s oxidation number is usually -2.
EXCEPT in a peroxide molecules where
it is -1.
5) Hydrogen is usually a +1 charge.
EXCEPT in metal hydrides where it is -1.
NORMAL
EXCEPTION
EXCEPTION
H2O
H 2 O2
Na H
1
2
1
1
1
1
6) The sum of the charges in a polyatomic
ion is equal to its charge.
?
N O3

?
S O4
2
X + 3(-2) = -1
N
O
X + 4(-2) = -2
S
O
thus N = +5
thus S = +6
7) Not on your sheet! Halogens are
normally -1, except when they are next
to an oxygen in a polyatomic ion!
HClO 2
(+1) + X + (-4) = 0
H
Cl
O
Thus Cl = +3
oxidation #s
oxidation #s
oxidation #s
+6 -2
2SO4
+5 -2
3PO4
+1
-2
Na2C4O8
oxidation #s
+2
-1
2Zn(OH)4
-1
+2? -2
oxidation #s +1
Na2O2
+2
+2?-1
+1
oxidation #s
CaH2
EXCEPTION!
EXCEPTION!
Not All Reactions are Redox Reactions
- Reactions in which there has been no
change in oxidation number are NOT
redox reactions.
Examples:
1 5 2
1
1
1
1
1 5 2
Ag N O3 (aq)  Na Cl (aq)  Ag Cl ( s)  Na N O3 (aq)
1 2 1
1
6 2
1
6 2
1
2
2 Na O H (aq)  H 2 S O 4 (aq)   Na 2 S O 4 (aq)  H 2 O(l )
ASSIGN AN OXIDATION NUMBER / STATE TO EACH ATOM IN :
Cl2
Cl(0)
CO32-
Ca2+
Ca(+2)
SO32-
O(-2)  S(+4)
Al3+
Al(+3)
ClO-
O(-2)  Cl(+1)
H2O
H(+1)  O(-2)
IO4-
O(-2)  I(+7)
O(-2)  C(+4)
CO2
O(-2)  C(+4)
ClF
F(-1)  Cl(+1)
MnO4-
NO3-
O(-2)  N(+5)
Na2S4O6
CuCl
Cl(-1)  Cu(+1)
CuBr2
Br(-1)  Cu(+2)
N(0)
C2O42-
O(-2)  C(+3)
N2
BrF5
F(-1)  Br(+5)
SF6
F(-1)  S(+6)
S2-
S(-2)
VCl2
Cl(-1)  V(+2)
CH4
Mn2O3
H(+1)  C(-4)
O(-2)  Mn(+7)
Na(+1) & O(-2)  S(+2.5)
O(-2)  Mn(+3)
CO
O(-2)  C(+2)
BrF
F(-1)  Br(+1)
Na2S
Na(+1)  S(-2)
NO2-
O(-2)  N(+3)
BrO3-
O(-2)  Br(+5)
NH4+
H(+1)  N(-3)
H2SO4
O(-2) & H(+1)  S(+6)
SO42-
O(-2)  S(+6)
I-
S2O32-
O(-2)  S(+2)
NH3
H(+1)  N(-3)
CCl4
Cl(-1)  C(+4)
Cr2O72-
O(-2)  Cr(+6)
I(-1)