Oxidation-Reduction
Reactions
CHE1101, Chapter 4
Learn, 1
Oxidation Reduction Reactions
• In chemical reactions, electrons are
transferred from one atom to another.
• Some atoms in a molecule gain electrons
while other atoms lose electrons
– Referred to as redox reactions
CHE1101, Chapter 4
Learn, 2
Oxidation Numbers
In order to keep
track of what loses
electrons and what
gains them, we
assign oxidation
numbers.
CHE1101, Chapter 4
Learn, 3
Oxidation
• A species is oxidized when it loses electrons.
– Here, zinc loses two electrons to go from neutral
zinc metal to the Zn2+ ion.
CHE1101, Chapter 4
Learn, 4
Reduction
• A species is reduced when it gains electrons.
– Here, each of the H+ gains an electron, and they
combine to form H2.
CHE1101, Chapter 4
Learn, 5
Oxidation and Reduction
• Oxidation and Reduction always occur
together
• The total number of electrons lost by
substances is always the same as the total
number gained others
– Oxidation – loss of electrons
– Reduction – gain of electrons
LEO goes GER
CHE1101, Chapter 4
Learn, 6
Oxidation and Reduction
• The oxidizing agent is the substance that is
reduced.
– H+ oxidizes Zn by taking electrons from it.
• The reducing agent is the substance that is
oxidized.
– Zn reduces H+ by giving it electrons.
CHE1101, Chapter 4
Learn, 7
Hierarchy of Rules for Assigning
Assigning
Oxidation Numbers
Oxidation
Numbers
1. Oxidation numbers must add up to charge on
molecule, formula unit or ion
2. Atoms of free elements have oxidation numbers
of zero
3. Metals in Groups 1A, 2A, and Al have +1, +2,
and +3 oxidation numbers, respectively
4. H and F in compounds have +1 and –1
oxidation numbers, respectively
5. Oxygen has –2 oxidation number
6. Group 7A elements have –1 oxidation number
CHE1101, Chapter 4
Learn, 8
Hierarchy of Rules for Assigning
Assigning
Oxidation Numbers
Oxidation
Numbers
7. Group 6A elements have –2 oxidation number
8. Group 5A elements have –3 oxidation number
9. When there is a conflict between two of these
rules or ambiguity in assigning an oxidation
number, apply rule which comes first and ignore
conflicting rule
Oxidation State
– Used interchangeably with oxidation number
– Indicates charge on monatomic ions
– Iron(III) means +3 oxidation state of Fe or Fe3+
CHE1101, Chapter 4
Learn, 9
Oxidation–Reduction Reactions
Involves 2 processes:
Oxidation = Loss of electrons
Na  Na+ + e–
Oxidation Half-Reaction
Reduction = Gain of electrons
Cl2 + 2e–  2Cl–
Net reaction:
Reduction Half-Reaction
2Na + Cl2  2Na+ + 2Cl–
– Oxidation and reduction always occur together
– Can't have one without the other
CHE1101, Chapter 4
Learn, 10
Redox Reactions
An
Example
Net: 2Mg + O2  2MgO
Oxidation:
Mg  Mg2+ + 2e –
– Loses electrons = oxidized
– Reducing agent
Reduction:
O2 + 4e–  2O2–
– Gains electrons = reduced
– Oxidizing agent
CHE1101, Chapter 4
Learn, 11
Problem
Which species functions as the oxidizing agent
in the following oxidation-reduction reaction?
Zn(s) + Pt2+(aq)  Pt(s) + Zn2+(aq)
A. Pt(s)
B. Zn2+(aq)
C. Pt2+(aq)
D. Zn(s)
E. None of these, as this is not a redox
reaction.
CHE1101, Chapter 4
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Your Turn!
Which species gets oxidized in the following
reaction?
2Ag+(aq) + Zn(s)  Zn2+(s) + 2Ag(s)
A. Ag(s)
B. Ag+(aq)
C. Zn2+(aq)
D. Zn(s)
E. None of these, as this is not a redox reaction
CHE1101, Chapter 4
Learn, 13
Ex. 1 Assigning Oxidation Number
1. Li2O
Li (2 atoms) × (+1) = +2 (Rule 3)
O (1 atom) × (–2) = –2 (Rule 5)
sum = 0 (Rule 1)
+2 –2 = 0 so the charges are balanced to zero
2. CO2
C (1 atom) × (x) = x
O (2 atoms) × (–2) = –4 (Rule 5)
sum = 0 (Rule 1)
x – 4 = 0 or x = +4
C is in +4 oxidation state
CHE1101, Chapter 4
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Learning Check
Assign oxidation numbers to all atoms:
Example 1: ClO4–
O (4 atoms) × (–2) = –8
Cl (1 atom) × (–1) = –1
(molecular ion) sum ≠ –1
(violates Rule 1)
Rule 5 for oxygen comes before Rule 6 for halogens
O (4 atoms) × (–2) = –8
Cl (1 atom) × (x) = x
sum = –1 (Rule 1)
–8 + x = –1 or x = 8 – 1
So x = +7; Cl is oxidation state +7
CHE1101, Chapter 4
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Learning Check
Assign oxidation states to all atoms:
• MgCr2O7
Mg =+2; O = –2; and Cr = x (unknown)
[+2] + [2x] + [7 × (–2)] = 0
2x – 12 = 0
x = +3
Cr is oxidation number of +3
• KMnO4
K =+1; O = – 2; so Mn = x
[+1] + [x] + [4 × (–2)] = 0
x–7=0
x = +7
Mn is oxidation number of +7
CHE1101, Chapter 4
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Your Turn!
What is the oxidation number of each atom
in H3PO4?
A. H = –1; P = +5; O = –2
B. H = 0; P = +3; O = –2
+1 +5 -2
C. H = +1; P = +7; O = –2
H3PO4
D. H = +1; P = +1; O = –1
E. H = +1; P = +5; O = –2
CHE1101, Chapter 4
Learn, 17
Your Turn!
What is the oxidation number of each atom in
sodium chlorate?
A. Na = +1; Cl = +5; O = –2
B. Na = +1; Cl = +7; O = –4
C. Na = +1; Cl = +7; O = –2
D. Na = +1; Cl = +3; O = –1
E. Na = -1; Cl = +5; O = –2
CHE1101, Chapter 4
Learn, 18
+1 +5
NaClO3
-2
Your Turn!
What are the oxidation numbers of potassium
and iodine in potassium triiodide, KI3?
A. K = +1; I = 0
So
I
is
-1
total
3
+1
B. K = +1; I = -1
Therefore, -1/3
C. K = -1; I = -3
for each I atom
KI3
D. K = +1; I = –1/3
E. K = +1; I = –2/3
Oxidation numbers can be fractions because they
represent an average number of ‘excess’ electrons
(or lack thereof) on the atoms – in this case,
one extra electron on the three iodide atoms
CHE1101, Chapter 4
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Your Turn!
Assign oxidation numbers to all atoms in the
following reaction and use them to determine
which species gets reduced.
+2 -2 -1/3
+2.5 -2
-1
2S2O32- + I3–  S4O62- + 3I–
A. I3–
B. I–
C. S4O62D. S2O32CHE1101, Chapter 4
Learn, 20
I is reduced from -1/3 to -1