1. Oxidation-Reduction Reactions

2. 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

3. Oxidation Numbers
In order to keep track of what loses electrons and what gains them, we assign oxidation numbers.

4. Oxidation A species is oxidized when it loses electrons.
Here, zinc loses two electrons to go from neutral zinc metal to the Zn2+ ion.

5. Reduction A species is reduced when it gains electrons.
Here, each of the H+ gains an electron, and they combine to form H2.

6. 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

7. 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.

8. Hierarchy of Rules for Assigning Oxidation Numbers
Oxidation numbers must add up to charge on molecule, formula unit or ion
Atoms of free elements have oxidation numbers of zero
Metals in Groups 1A, 2A, and Al have +1, +2, and +3 oxidation numbers, respectively
H and F in compounds have +1 and –1 oxidation numbers, respectively
Oxygen has –2 oxidation number
Group 7A elements have –1 oxidation number

9. Hierarchy of Rules for Assigning Oxidation Numbers
Group 6A elements have –2 oxidation number
Group 5A elements have –3 oxidation number
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+

10. Oxidation–Reduction Reactions
Involves 2 processes:
Oxidation = Loss of electrons
Na  Na+ + e– Oxidation Half-Reaction
Reduction = Gain of electrons
Cl2 + 2e–  2Cl– Reduction Half-Reaction
Net reaction:
2Na + Cl2  2Na+ + 2Cl–
Oxidation and reduction always occur together
Can't have one without the other

11. An Example Redox Reactions Net: 2Mg + O2  2MgO Mg  Mg2+ + 2e –
Oxidation:
Mg  Mg2+ + 2e –
Loses electrons = oxidized
Reducing agent
Reduction:
O2 + 4e–  2O2–
Gains electrons = reduced
Oxidizing agent

12. Problem
Which species functions as the oxidizing agent in the following oxidation-reduction reaction?
Zn(s) Pt2+(aq)  Pt(s) + Zn2+(aq)
Pt(s)
Zn2+(aq)
C. Pt2+(aq)
D. Zn(s)
E. None of these, as this is not a redox reaction.

13. Your Turn! Which species gets oxidized in the following reaction?
2Ag+(aq) + Zn(s)  Zn2+(s) + 2Ag(s)
Ag(s)
Ag+(aq)
C. Zn2+(aq)
D. Zn(s)
E. None of these, as this is not a redox reaction

14. Ex. 1 Assigning Oxidation Number
Li2O
Li (2 atoms) × (+1) = (Rule 3)
O (1 atom) × (–2) = –2 (Rule 5)
sum = (Rule 1)
+2 –2 = 0 so the charges are balanced to zero
CO2
C (1 atom) × (x) = x
O (2 atoms) × (–2) = –4 (Rule 5)
sum = (Rule 1)
x – 4 = 0 or x = +4
C is in +4 oxidation state

15. 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
Cl (1 atom) × (x) = x
sum = –1 (Rule 1)
–8 + x = –1 or x = 8 – 1
So x = +7; Cl is oxidation state +7

16. Assign oxidation states to all atoms: MgCr2O7
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

17. 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 C. H = +1; P = +7; O = –2 D. H = +1; P = +1; O = –1 E. H = +1; P = +5; O = –2
H3PO4 +5

18. 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
NaClO3 +5

19. Your Turn!
What are the oxidation numbers of potassium and iodine in potassium triiodide, KI3?
A. K = +1; I = 0
B. K = +1; I = -1
C. K = -1; I = -3
D. K = +1; I = –1/3
E. K = +1; I = –2/3
So I3 is -1 total +1
Therefore, -1/3
for each I atom
KI3
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

20. Your Turn!
Assign oxidation numbers to all atoms in the following reaction and use them to determine which species gets reduced. /
2S2O32- + I3–  S4O I–
A. I3–
B. I–
C. S4O62-
D. S2O32-
I is reduced from -1/3 to -1