1. Oxidation-Reduction Reactions “Redox”
LEO SAYS GER

2. Oxidation and Reduction (Redox)
Electrons are transferred
Spontaneous redox rxns can transfer energy
Electrons (electricity)
Heat
Non-spontaneous redox rxns can be made to happen with electricity

3. Oxidation Reduction Reactions (Redox)
Each sodium atom loses one electron:
Each chlorine atom gains one electron:

4. LEO says GER :
Lose Electrons = Oxidation
Sodium is oxidized
Gain Electrons = Reduction
Chlorine is reduced

5. Rules for Assigning Oxidation Numbers Rules 1 & 2
The oxidation number of any uncombined element is zero
2. The oxidation number of a monatomic ion
equals its charge

6. Rules for Assigning Oxidation Numbers Rules 3 & 4
3. The oxidation number of oxygen in
compounds is -2
4. The oxidation number of hydrogen in compounds is +1

7. Rules for Assigning Oxidation Number Rule 5
5. The sum of the oxidation numbers in the formula of a compound is 0
2(+1) + (-2) = 0
H O
(+2) + 2(-2) + 2(+1) = 0
Ca O H

8. Rules for Assigning Oxidation Numbers Rule 6
6. The sum of the oxidation numbers in the
formula of a polyatomic ion is equal to
its charge
X + 4(-2) = -2
S O
X + 3(-2) = -1
N O
 X = +5
 X = +6

9. The Oxidation Number Rules - SIMPLIFIED
1. The sum of the oxidation numbers in
ANYTHING is equal to its charge
2. Hydrogen in compounds is +1
3. Oxygen in compounds is -2

10. Not All Reactions are Redox Reactions
Reactions in which there has been no change in oxidation number are not redox rxns.
Examples:

11. Reducing Agents and Oxidizing Agents
This slide refers to vocabulary that has been excluded from AP Chemistry by the College Board, and will not be tested.
Reducing Agents and Oxidizing Agents
The substance reduced is the oxidizing agent
The substance oxidized is the reducing agent
Sodium is oxidized – it is the reducing agent
Chlorine is reduced – it is the oxidizing agent

12. Trends in Oxidation and Reduction
Active metals:
Lose electrons easily
Are easily oxidized
Active nonmetals:
Gain electrons easily
Are easily reduced

13. Redox Reaction Prediction #1
Reduced in reaction
Formed in reaction
MnO4- (acid solution)
MnO4- (basic solution)
MnO2 (acid solution)
Cr2O72- (acid)
CrO42-
HNO3, concentrated
HNO3, dilute
H2SO4, hot conc
Metallic Ions
Free Halogens
HClO4
Na2O2
H2O2
Mn(II)
MnO2
Cr(III)
NO2 NO
SO2
Metallous Ions
Halide ions
Cl-
OH- O2

14. Redox Reaction Prediction #2
Oxidized in reaction
Formed in reaction
Halide Ions
Free Metals
Metalous Ions
Nitrite Ions
Sulfite Ions
Free Halogens (dil, basic sol)
Free Halogens (conc, basic sol)
C2O42-
Halogens
Metal Ions
Metallic ions
Nitrate Ions
SO42-
Hypohalite ions
Halate ions
CO2

15. For any equation to be balanced:
1. The number of atoms of each type on the left side of the arrow must equal the number of atoms of each type to the right of the arrow.
2. The total charges of all the ions on the left side of the arrow must equal the total charges of all the ions to the right of the arrow.

16. In addition, for redox reactions:
3. The electrons lost (during oxidation) must equal the electrons gained (during reduction).

17. Balancing Oxidation-Reduction Reactions:
1. Assign oxidation numbers to every atom in the reaction.
Cr2O72- + C2O42-  Cr3+ + CO2

18. Balancing Oxidation-Reduction Reactions:
1. Assign oxidation numbers to every atom in the reaction.
Cr2O72- + C2O42-  Cr3+ + CO2
Oxygen, in a compound or ion, is -2

19. Balancing Oxidation-Reduction Reactions:
1. Assign oxidation numbers to every atom in the reaction.
Cr2O72- + C2O42-  Cr3+ + CO2
Use the combined ‘charges’ of the oxygens in each ion or compound to determine the oxidation number of Cr or C.

20. Balancing Oxidation-Reduction Reactions:
1. Assign oxidation numbers to every atom in the reaction.
Cr2O72- + C2O42-  Cr3+ + CO2
Use the combined ‘charges’ of the oxygens in each ion or compound to determine the oxidation number of Cr or C.

21. Balancing Oxidation-Reduction Reactions:
1. Assign oxidation numbers to every atom in the reaction.
Cr2O72- + C2O42-  Cr3+ + CO2
The sum of the oxidation numbers of the other element must add up to the charge on the ion or molecule.

22. Balancing Oxidation-Reduction Reactions:
1. Assign oxidation numbers to every atom in the reaction.
Cr2O72- + C2O42-  Cr3+ + CO2
The sum of the oxidation numbers of the other element must add up to the charge on the ion or molecule.

23. Balancing Oxidation-Reduction Reactions:
1. Assign oxidation numbers to every atom in the reaction.
Cr2O72- + C2O42-  Cr3+ + CO2
Divide the sum of the charges by the number of atoms to get the oxidation number for chromium and carbon in the reactants.

24. Balancing Oxidation-Reduction Reactions:
1. Assign oxidation numbers to every atom in the reaction.
Cr2O72- + C2O42-  Cr3+ + CO2
+12/2= /2=
Divide the sum of the charges by the number of atoms to get the oxidation number for chromium and carbon in the reactants.

25. Balancing Oxidation-Reduction Reactions:
2. Write ‘bare bones’ half reactions.
Include only the atom, ion or element that changes oxidation number.
Cr e-  Cr+3
C+3  C e-
Remember that each half reaction must also be balanced for charge. The total charges on the left must equal the total charges on the right.

26. Balancing Oxidation-Reduction Reactions:
3. Take into account any subscripts in the formulas of reactants and products, and multiply the half reactions accordingly.
Cr2O72- + C2O42-  Cr3+ + CO2
2[Cr e-  Cr+3]
2[C+3  C e-]

27. Balancing Oxidation-Reduction Reactions:
3. Take into account any subscripts in the formulas of reactants and products, and multiply the half reactions accordingly.
Cr2O72- + C2O42-  Cr3+ + CO2
2Cr e-  2 Cr+3
2C+3  2 C e-

28. Balancing Oxidation-Reduction Reactions:
4. Multiply each half reaction by the appropriate factor so that the number of electrons lost = number of electrons gained.
1[2Cr e-  2 Cr+3]
3[2C+3  2 C e-]

29. Balancing Oxidation-Reduction Reactions:
4. Multiply each half reaction by the appropriate factor so that the number of electrons lost = number of electrons gained.
2Cr e-  2 Cr+3
6C+3  6 C e-

30. Balancing Oxidation-Reduction Reactions:
4. Add the two half reactions together.
2Cr e-  2 Cr+3
6C+3  6 C e-
2Cr C+3  2 Cr C+4

31. Balancing Oxidation-Reduction Reactions:
4. Add the two half reactions together.
2Cr e-  2 Cr+3
6C+3  6 C e-
2Cr C+3  2 Cr C+4
At this point, the electrons lost = the electrons gained during the reaction.

32. Balancing Oxidation-Reduction Reactions:
5. You now have the number of each atom that undergoes oxidation or reduction in the balanced equation. Take any subscripts into account when inserting coefficients.
2Cr C+3  2 Cr C+4
Cr2O C2O42- 2Cr3+ + 6CO2

33. Balancing Oxidation-Reduction Reactions:
6. Balance the reaction for charge, using OH- (if in base) or H+ (if in acid).
The equation below takes place in acid:
Cr2O C2O42- 2Cr3+ + 6CO2
Charges: = -8 (left)  +6 (right)

34. Balancing Oxidation-Reduction Reactions:
Cr2O C2O42- 2Cr3+ + 6CO2
Charges: = -8 (left)  +6 (right)
Since the reaction takes place in acid, you need to add 14 H+ to the left side so that the charges become equal.

35. Balancing Oxidation-Reduction Reactions:
14 H+ + Cr2O C2O42- 2Cr3+ + 6CO2
Charges on left = +6 = Charges on right

36. Balancing Oxidation-Reduction Reactions:
7. Balance for H and O by adding water to the appropriate side of the reaction.
14 H+ + Cr2O C2O42- 2Cr3+ + 6CO2 + 7 H2O

37. Balancing Oxidation-Reduction Reactions:
8. Check the balance for all atoms in the reaction.
14 H+ + Cr2O C2O42- 2Cr3+ + 6CO2 + 7 H2O
Left: 14 H Right: 14 H
2 Cr Cr
19 O O
6 C C

38. Redox Stoichiometry
Calculations involving concentrations and redox reactions are quite common. Many ores containing metals are analyzed using redox titrations. Since many compounds change color as they are oxidized or reduced, one of the reactants may serve as the indicator in the titration.

39. Redox Stoichiometry
The concentration of iron(II) can be determined by titration with bromate ion, in acid. The products are iron(III) ion and the bromide ion.
What is the concentration of iron(II) ion if mL of 0.105M potassium bromate is required to completely react with mL of the iron solution.

40. Redox Stoichiometry
The concentration of iron(II) can be determined by titration with bromate ion, in acid. The products are iron(III) ion and the bromide ion.
1. Write the balanced chemical reaction.
Fe2+(aq) + BrO31-(aq)  Fe3+(aq) + Br1-(aq)