1. Electro-chemistry CHAPTER 15
15.2 Oxidation-Reduction (Redox) Reactions

2. 4Fe(s) + 3O2(g) → 2Fe2O3(s) Fe Fe3+ Rusting is an oxidation process
loses 3 electrons

3. 4Fe(s) + 3O2(g) → 2Fe2O3(s) Fe Fe3+ Rusting is an oxidation process
Early chemists talked about “reducing” compounds into pure metals long before they understood the chemistry of what was happening Fe
Fe3+
loses 3 electrons
Pure metal
gains 3 electrons

4. Redox reactions Zn(s) + CuSO4 → ZnSO4(aq) + Cu(s) CuSO4(aq) Zn(s)
Cu(s) deposit

5. Redox reactions Zn(s) + CuSO4 → ZnSO4(aq) + Cu(s)
We can’t actually observe the transfer of electrons, but we know that:
Cu2+ was reduced back to Cu
Zn was oxidized by losing electrons and becoming Zn2+
Oxidation: Loss of electrons. The oxidized element becomes more positive.
Reduction: Gain of electrons. The reduced element becomes more negative.

6. How can we determine if a reaction is a redox reaction or not?
If it is, how can we determine which elements are reduced, and which elements are oxidized?
Oxidation: Loss of electrons. The oxidized element becomes more positive.
Reduction: Gain of electrons. The reduced element becomes more negative.

7. Track the transfer of electrons
How can we determine if a reaction is a redox reaction or not?
If it is, how can we determine which elements are reduced, and which elements are oxidized?
Track the transfer of electrons
Oxidation: Loss of electrons. The oxidized element becomes more positive.
Reduction: Gain of electrons. The reduced element becomes more negative.

8. Cl2 + H2 → 2HCl
The oxidation number of Cl is –1
because Cl “pulls” an electron from H
The oxidation number of H is +1
because HCl is neutral overall
oxidation number: the number of electrons that an element has lost or gained in forming a chemical bond with another element.

9. Cl2 + H2 → 2HCl Cl is reduced because it gains an electron
H is oxidized because it lost an electron
The oxidation number of Cl is –1
because Cl “pulls” an electron from H
The oxidation number of H is +1
because HCl is neutral overall
oxidation number: the number of electrons that an element has lost or gained in forming a chemical bond with another element.

10. The oxidation number is different from the real charge of the atom
oxidation number: the number of electrons that an element has lost or gained in forming a chemical bond with another element.

11. Oxidation number rules
Oxidation number of an atom in a pure element is 0
Ex: in Cl2, nCl = 0
n = 0
Rule 2
Sum of oxidation numbers in a neutral molecule is 0
Ex: nCO2 = nC + 2(nO) = 0
Rule 3
Sum of oxidation numbers in an ion is equal to the charge of the ion. Ex: n for SO42– = –2
n = charge of ion
Rule 4
Metals have positive oxidation numbers according to their group. Ex: Group 1A metals (Na, K…) have n = +1
Group 2A metals (Mg, Ca…) have n = +2
n = +(group number)
Rule 5
For non-metals:
- Fluorine (F)
- Hydrogen (H), except in hydrides like LiH or NaH
- Oxygen (O)
- Group 7A (Cl, Br…), group 6A (S, Se…), and group 5A (N, P…), respectively
n = –1
n = +1
n = –2
n = –1, –2, and –3

12. Oxidation number rules
Oxidation number of an atom in a pure element is 0
Ex: in Cl2, nCl = 0
n = 0
Rule 2
Sum of oxidation numbers in a neutral molecule is 0
Ex: nCO2 = nC + 2(nO) = 0
Rule 3
Sum of oxidation numbers in an ion is equal to the charge of the ion. Ex: n for SO42– = –2
n = charge of ion
Rule 4
Metals have positive oxidation numbers according to their group. Ex: Group 1A metals (Na, K…) have n = +1
Group 2A metals (Mg, Ca…) have n = +2
n = +(group number)
Rule 5
For non-metals:
- Fluorine (F)
- Hydrogen (H), except in hydrides like LiH or NaH
- Oxygen (O)
- Group 7A (Cl, Br…), group 6A (S, Se…), and group 5A (N, P…), respectively
n = –1
n = +1
n = –2
n = –1, –2, and –3

13. Oxidation number rules
Oxidation number of an atom in a pure element is 0
Ex: in Cl2, nCl = 0
n = 0
Rule 2
Sum of oxidation numbers in a neutral molecule is 0
Ex: nCO2 = nC + 2(nO) = 0
Rule 3
Sum of oxidation numbers in an ion is equal to the charge of the ion. Ex: n for SO42– is –2
n = charge of ion
Rule 4
Metals have positive oxidation numbers according to their group. Ex: Group 1A metals (Na, K…) have n = +1
Group 2A metals (Mg, Ca…) have n = +2
n = +(group number)
Rule 5
For non-metals:
- Fluorine (F)
- Hydrogen (H), except in hydrides like LiH or NaH
- Oxygen (O)
- Group 7A (Cl, Br…), group 6A (S, Se…), and group 5A (N, P…), respectively
n = –1
n = +1
n = –2
n = –1, –2, and –3

14. Oxidation number rules
Oxidation number of an atom in a pure element is 0
Ex: in Cl2, nCl = 0
n = 0
Rule 2
Sum of oxidation numbers in a neutral molecule is 0
Ex: nCO2 = nC + 2(nO) = 0
Rule 3
Sum of oxidation numbers in an ion is equal to the charge of the ion. Ex: n for SO42– is –2
n = charge of ion
Rule 4
Metals have positive oxidation numbers according to their group. Ex: Group 1A metals (Na, K…) have n = +1
Group 2A metals (Mg, Ca…) have n = +2
n = +(group number)
Rule 5
For non-metals:
- Fluorine (F)
- Hydrogen (H), except in hydrides like LiH or NaH
- Oxygen (O)
- Group 7A (Cl, Br…), group 6A (S, Se…), and group 5A (N, P…), respectively
n = –1
n = +1
n = –2
n = –1, –2, and –3

15. Oxidation number rules
Oxidation number of an atom in a pure element is 0
Ex: in Cl2, nCl = 0
n = 0
Rule 2
Sum of oxidation numbers in a neutral molecule is 0
Ex: nCO2 = nC + 2(nO) = 0
Rule 3
Sum of oxidation numbers in an ion is equal to the charge of the ion. Ex: n for SO42– is –2
n = charge of ion
Rule 4
Metals have positive oxidation numbers according to their group. Ex: Group 1A metals (Na, K…) have n = +1
Group 2A metals (Mg, Ca…) have n = +2
n = +(group number)
Rule 5
For non-metals:
- Fluorine (F)
- Hydrogen (H), except in hydrides like LiH or NaH
- Oxygen (O)
- Group 7A (Cl, Br…), group 6A (S, Se…), and group 5A (N, P…), respectively
n = –1
n = +1
n = –2
n = –1, –2, and –3

16. Oxidation number rules
Oxidation number of an atom in a pure element is 0
Ex: in Cl2, nCl = 0
n = 0
Rule 2
Sum of oxidation numbers in a neutral molecule is 0
Ex: nCO2 = nC + 2(nO) = 0
Rule 3
Sum of oxidation numbers in an ion is equal to the charge of the ion. Ex: n for SO42– = –2
n = charge of ion
Rule 4
Metals have positive oxidation numbers according to their group. Ex: Group 1A metals (Na, K…) have n = +1
Group 2A metals (Mg, Ca…) have n = +2
n = +(group number)
Rule 5
For non-metals:
- Fluorine (F)
- Hydrogen (H), except in hydrides like LiH or NaH
- Oxygen (O)
- Group 7A (Cl, Br…), group 6A (S, Se…), and group 5A (N, P…), respectively
n = –1
n = +1
n = –2
n = –1, –2, and –3
The oxidation number rules apply in the order they appear on the list

17. Oxidation number rules
Find the oxidation number of each element in potassium peroxide (K2O2).
= 2(nK) + 2(nO)
= 2(+1) + 2(–2)
= +2 – 4
= –2
Rule 4: nK = +1
Rule 5: nO = –2
But according to Rule 2:
the sum of oxidation numbers in a neutral molecule is zero
What steps do we follow to make sure the rule works?

18. Oxidation number rules
Find the oxidation number of each element in potassium peroxide (K2O2).
1) Rule 2: Sum of oxidation numbers in a neutral molecule is 0
2(nK) + 2(nO) = 0
2) Rule 4: Metals have positive oxidation numbers according to their group.
K belongs to Group 1A, so nK = +1
3) Solve: 2(nK) + 2(nO) = 0
2(+1) + 2(nO) = 0
nO = –2 / 2 = –1
Answer: nK = +1 and nO = –1

19. Find the oxidation number of each element in carbon dioxide (CO2).

20. Find the oxidation number of each element in carbon dioxide (CO2).
1) Rule 2: Sum of oxidation numbers in a neutral molecule is 0
nC + 2(nO) = 0

21. Find the oxidation number of each element in carbon dioxide (CO2).
1) Rule 2: Sum of oxidation numbers in a neutral molecule is 0
nC + 2(nO) = 0
2) Rule 5: nO = –2

22. Find the oxidation number of each element in carbon dioxide (CO2).
1) Rule 2: Sum of oxidation numbers in a neutral molecule is 0
nC + 2(nO) = 0
2) Rule 5: nO = –2
3) Solve: (nC) + 2(nO) = 0
(nC) + 2(–2) = 0
nC = +4

23. Find the oxidation number of each element in carbon dioxide (CO2).
1) Rule 2: Sum of oxidation numbers in a neutral molecule is 0
nC + 2(nO) = 0
2) Rule 5: nO = –2
3) Solve: (nC) + 2(nO) = 0
(nC) + 2(–2) = 0
nC = +4
Answer: nC = +4 and nO = –2

24. Notice that the oxidation number for an element can be different from one compound to another
In K2O2, nO = –1
In CO2, nO = –2

25. Find the oxidation number of each element in the nitrite ion (NO2–).
1) Rule 3: Sum of oxidation numbers in an ion is equal to the charge of the ion
nN + 2(nO) = –1
2) Rule 5: nO = –2
3) Solve: nN + 2(nO) = –1
nN + 2(–2) = –1
nN = +4 – 1 = +3
Answer: nN = +3 and nO = –2

26. Find the oxidation number of each element in the compound C3H4.
1) Rule 2: Sum of oxidation numbers in a neutral molecule is 0
3(nC) + 4(nH) = 0
2) Rule 5: nH = +1
3) Solve: 3(nC) + 4(nH) = 0
3(nC) + 4(+1) = 0
nC = –4/3
Answer: nC = –4/3 and nH = +1
Oxidation numbers can be fractions!

27. We can use oxidation numbers to answer these questions
How can we determine if a reaction is a redox reaction or not?
2) If it is, how can we determine which elements are reduced, and which elements are oxidized?
We can use oxidation numbers to answer these questions

28. We can use oxidation numbers to answer these questions
How can we determine if a reaction is a redox reaction or not?
If there is a change in oxidation numbers from reactants to products, it is a redox reaction
2) If it is, how can we determine which elements are reduced, and which elements are oxidized?
We can use oxidation numbers to answer these questions

29. We can use oxidation numbers to answer these questions
How can we determine if a reaction is a redox reaction or not?
If there is a change in oxidation numbers from reactants to products, it is a redox reaction
2) If it is, how can we determine which elements are reduced, and which elements are oxidized?
If the oxidation number increases, that element is reduced
If the oxidation number decreases, that element is oxidized
We can use oxidation numbers to answer these questions

30. Find the element that is oxidized and the element that is reduced in the reaction of iron with oxygen resulting in rust (iron oxide, Fe2O3):
4Fe(s) + 3O2(g) → 2Fe2O3(s)

31. Find the element that is oxidized and the element that is reduced in the reaction of iron with oxygen resulting in rust (iron oxide, Fe2O3):
4Fe(s) + 3O2(g) → 2Fe2O3(s)
Asked: The element that is oxidized, and the element that is reduced
Given: The balanced equation, and the oxidation number rules
Relationships: The oxidation number rules
Solve: 1) Find the oxidation numbers for each element in the reactants
2) Find the oxidation numbers for each element in the product
3) Find for which element the oxidation number increased or decreased

32. 4Fe(s) + 3O2(g) → 2Fe2O3(s)
Asked: The element that is oxidized, and the element that is reduced
Given: The balanced equation, and the oxidation number rules
Relationships: The oxidation number rules
Solve: 1) Find the oxidation numbers for each element in the reactants
2) Find the oxidation numbers for each element in the product
3) Find for which element the oxidation number increased or decreased
Rule 1: The oxidation number of an atom in a pure element is zero
nFe = 0 and nO = 0

33. 4Fe(s) + 3O2(g) → 2Fe2O3(s)
Asked: The element that is oxidized, and the element that is reduced
Given: The balanced equation, and the oxidation number rules
Relationships: The oxidation number rules
Solve: 1) Find the oxidation numbers for each element in the reactants
2) Find the oxidation numbers for each element in the product
3) Find for which element the oxidation number increased or decreased
nFe = 0
nO = 0
Rule 2: Sum of oxidation numbers in a neutral molecule is zero
2(nFe) + 3(nO) = 0
Rule 5: nO = –2
Solve: 2(nFe) + 3(–2) = 0 so nFe = +6 / 2 = +3

34. 4Fe(s) + 3O2(g) → 2Fe2O3(s) 3O2(g) + 4Fe(s) → 2Fe2O3(s)
Asked: The element that is oxidized, and the element that is reduced
Given: The balanced equation, and the oxidation number rules
Relationships: The oxidation number rules
Solve: 1) Find the oxidation numbers for each element in the reactants
2) Find the oxidation numbers for each element in the product
3) Find for which element the oxidation number increased or decreased
3O2(g) + 4Fe(s) → 2Fe2O3(s)
oxidation numbers –2
oxidation
reduction

35. O is the oxidizing agent
oxidation
reduction
3O2(g) + 4Fe(s) → 2Fe2O3(s) –2
oxidation numbers
Fe is oxidized by O
O is the oxidizing agent

36. Fe is the reducing agent
oxidation
reduction
3O2(g) + 4Fe(s) → 2Fe2O3(s) –2
oxidation numbers
Fe is oxidized by O
O is the oxidizing agent
O is reduced by Fe
Fe is the reducing agent

37. 3O2(g) + 4Fe(s) → 2Fe2O3(s)
oxidation numbers –2
oxidation
reduction
Where there is a reduction reaction, there is also an oxidation reaction, and vice versa.
This is because electrons that are lost by one element must be gained by another element

38. Find the reducing and the oxidizing agents in:
2Mg(s) + O2(g) → 2MgO(s)

39. Find the reducing and the oxidizing agents in:
2Mg(s) + O2(g) → 2MgO(s)
Relationships: The balanced chemical reaction and the oxidation number rules
Solve:
O2(g) + 2Mg(s) → 2MgO(s)
oxidation numbers –2
oxidation
reduction

40. Find the reducing and the oxidizing agents in:
2Mg(s) + O2(g) → 2MgO(s)
Relationships: The balanced chemical reaction and the oxidation number rules
Solve:
O2(g) + 2Mg(s) → 2MgO(s)
oxidation numbers –2
oxidation
reduction
Answer: Mg is oxidized by O O is the oxidizing agent
O is reduced by Mg Mg is the reducing agent

41. Rules to assign oxidation numbers:
- the rules must be followed in the order they are presented
- the oxidation number of an element can be different from one compound to another
- the oxidation number of an element can be a fraction

42. Rules to assign oxidation numbers:
- the rules must be followed in the order they are presented
- the oxidation number of an element can be different from one compound to another
- the oxidation number of an element can be a fraction
Assigning oxidation numbers helps to:
- determine if a chemical reaction is a redox reaction or not
- determine which elements are reduced, and which elements are oxidized
- determine which elements or compounds are oxidizing agents, and which are reducing agents