Download Oxidation Number

Chemistry Specialist: Alicia Hart
Chem 2A Workshop
Oxidation State (Oxidation Number)
Simply put, it is the degree to which an element is oxidized in a chemical compound. We assign a
number to the element according to how many e- s the element appears to lose, gain or contribute
to a bond in a chemical compound. This is said to be a hypothetical charge because the oxidation
state of the atom is not always consistent with its actual charge. Below are the rules that govern
the assignment of oxidation numbers.
Assigning Oxidation Numbers
1. An atom in its elementary (uncombined) state is assigned an oxidation number of zero
(e.g. O2).
2. The oxidation number of a monatomic ion corresponds to its charge (e.g. Fe3+ Ox. # is
+3).
3. The algebraic sum of the elemental oxidation states in a neutral molecule is zero (e.g.
CO2, C is +4 and O is -2; since there are two O’s [2(-2) = -4], the Ox. #’s sum to zero).
In contrast, the algebraic sum of the elemental oxidation states in a polyatomic ion is = to
the net charge on the molecule (e.g. CO32-, since O is -2, C must be +4 because [3(-2) +
x= -2]).
When in combination with other atoms:
4. Typically, group 1 elements (i.e. alkali metals) are assigned an oxidation state of +1 (Ag
is also assigned +1) and group 2 elements (i.e. alkaline earth metals) are assigned an
oxidation state of +2 (Zn, Cd also assigned +2)
5. Halogens are assigned an oxidation number of -1. EXCEPTION: Oxidation number
varies when combined with oxygen (F always assigned -1).
6. Hydrogen is assigned an oxidation number of +1. EXCEPTION: In metal hydrides,
hydrogen is assigned an oxidation number of -1 (e.g. NaH).
7. Oxygen is assigned an oxidation number of -2. EXCEPTION: Oxygen is assigned an
oxidation state of -1 when forming peroxides (e.g. H2O2).
* Note: the more electronegative atom is assigned the (-) oxidation state.
OIL RIG
Oxidation Is Loss Reduction Is Gain (in e- s)
Reduction: occurs when the oxidation number of an atom becomes more negative.
Oxidation: occurs when the oxidation number of an atom becomes more positive.
Reducing agent: this is the chemical species that causes the reduction. In other words, the
reducing agent is the compound that contains the element being oxidized.
Oxidizing agent: this is the chemical species that causes the oxidation. In other words, the
oxidizing agent is the compound that contains the element being reduced.
Balancing Redox Reactions
Acidic Solution:
1.
2.
3.
4.
5.
6.
Break up the chemical reaction into two half-reactions (red-ox)
Balance all atoms other than H or O
Balance O’s by adding H2O to the side with less O
Balance H’s by adding H+ to the side with less H
Balance charge by adding e- s to appropriate side
Combine the half-reactions canceling any like terms (1 H+ and 1 –OH = 1 H2O) and
balance the overall charge by multiplying each half-reaction by the appropriate integer.
7. Check your work and then you’re done!
Example:
SO2(aq) + Cr2O72-(aq)  SO42-(aq) + Cr3+(aq)
* remember: reaction carried out in acidic conditions
1. SO2  SO42Cr2O72-  Cr3+
S from +4 to +6 (oxidation)
Cr from +6 to +3 (reduction)
2. Cr2O72-  2Cr3+
3. SO2 + 2 H2O  SO42Cr2O72-  2 Cr3+ + 7 H2O
4. SO2 + 2 H2O  SO42- + 4 H+
Cr2O72- + 14 H+  2 Cr3+ + 7 H2O
5. SO2 + 2 H2O  SO42- + 4 H+ + 2 eCr2O72- + 14 H+ + 6 e-  2 Cr3+ + 7 H2O
6. 3 (SO2 + 2 H2O
 SO42- + 4 H+ + 2 e-)
2+
Cr2O7 + 14 H + 6 e  2 Cr3+ + 7 H2O ____
3 SO2 + Cr2O72- + 2 H+  3 SO42- + 2 Cr3+ + H2O
Basic Solution:
Nearly identical procedure, but we need to add a step between steps 4 and 5. We shall
label this step 4B: after balancing the H+, add the same amount of –OH to both sides of
each of the half-reactions.
Example:
MnO4-(aq) + H2O2(aq)  MnO2(s) + O2(g)
* remember: reaction is carried out in basic conditions
1.
MnO4-  MnO2
H 2O2  O2
2.
Not necessary; all atoms other than H and O are balanced.
3.
MnO4-  MnO2 + 2 H2O
Mn from +7 to +4 (reduction)
O from -1 to 0 (oxidation)
4A. MnO4- + 4 H+  MnO2 + 2 H2O
H 2O2  O2 + 2 H+
4B. MnO4- + 4 H+ + 4 -OH  MnO2 + 2 H2O + 4 -OH
H2O2 + 2 -OH  O2 + 2 H+ + 2 -OH
5.
MnO4- + 4 H+ + 4 -OH + 3 e-  MnO2 + 2 H2O + 4 -OH
H2O2 + 2 -OH  O2 + 2 H+ + 2 -OH + 2 e-
6.
2 (MnO4- + 4 H+ + 4 -OH + 3 e-  MnO2 + 2 H2O + 4 -OH)
3 (H2O2 + 2 -OH  O2 + 2 H+ + 2 -OH + 2 e-)
2 MnO4- + 3 H2O2  2 MnO2 + 3 O2 + 2 -OH + 2 H2O
Jim and Rolf’s alternative method:
Balancing Redox Reactions
A six step method:
HElOHEN
H
El
O
Half rxn's:
Elements:
Oxygen:
break up into balance the Elements balance O
Half (1/2) rxn's except O and H
with H2O
Example: Balance the following reaction in acid:
1) H: 1/2 rxn's
Cr2O7-2
Þ
H 2O 2
Þ
Cr+3
O2
red (+6 Þ +3)
ox (-1 Þ 0)
2) El: elements
Cr2O7-2
Þ
H 2O 2
Þ
2 Cr+3
O2
red
ox
3) O: oxygen
Cr2O7-2
Þ
H 2O 2
Þ
2 Cr+3 + 7 H2O red
O2
ox
H
Hydrogen:
balance H
with H+
E
N
Electrons:
Number of electrons:
add the e- to multiply so the Number
balance charge of e- is the same in both
Cr2O7-2 + H2O2 Þ Cr+3 + O2
4) Cr2O7-2 + 14 H+ Þ 2 Cr+3 + 7 H2O red
H 2O 2
Þ O2 + 2 H+ ox
5) E: electrons Be sure to use the net (total) charge on each sides. Multiply coefficients by charges to get
the net.
Cr2O7-2 + 14 H+ + 6e- Þ 2 Cr+3 + 7 H2O red
H 2O 2
Þ 2 e- + O2 + 2 H+ ox
6) N: number
Cr2O7-2 + 14 H+ + 6e- Þ 2 Cr+3 + 7 H2O red
3 H 2O 2
Þ 6 e- + 3 O2 + 6 H+ ox
7) Now add them up.
Cr2O7-2 + 8 H+ + 3 H2O2 Þ 2 Cr+3 + 7 H2O + 3 O2
8) Always check that in your answer all electrons cancel and atoms and charges balance! Also check
to see if you can divide by a common divisor (see if you have the lowest possible coefficients) and if you
can collect terms.
Balancing in Base
The procedure is the same as with acid except that after you add the H+ to balance the hydrogens, add the
same number of OH- to both sides and form water on one side.
Example: Balance the following 1/2 rxn in base:
BrO3-
Þ
BrO4-
1) H: (The 1/2 rxn is given in this case)
BrO3Þ BrO42) El: elements
BrO3Þ BrO4Bromines are already balanced.
3) O: oxygen
BrO3- + H2O Þ BrO44) H: hydrogen (but because its really in base, after adding H+, then add OH- to both sides.)
BrO3- + H2O + 2 OH- Þ BrO4- + 2H+ + 2 OHIf you put H+ and OH- together, they will form water.
BrO3- + H2O + 2 OH- Þ BrO4- + (2H+ + 2 OH- Þ 2 H2O)
Cancel the waters and continue.
BrO3- + 2 OH- Þ BrO4- + H2O
5) E: electrons
BrO3- + 2 OH- Þ BrO4- + H2O + 2 e-