1. Chapter 4 Oxidation Reduction Reactions
Part B

2. Oxidation-Reduction Reactions
Oxidation-reduction (redox) reactions involve the transfer of electrons between reactant species.
2Na(s)+Cl2(g) ⟶ 2NaCl(s)
The reaction can be represented by two half-reactions
2Na(s) ⟶ 2Na+(s)+2e−
oxidation = loss of electrons
reducing agent (Na) = species that is oxidized
Cl2(g)+2e− ⟶ 2Cl−(s)
reduction = gain of electrons
oxidizing agent(Cl2) = species that is reduced

3. Oxidation number
Oxidation number (or oxidation state) of an element in a compound is the charge its atoms would possess if the compound was ionic.
Guidelines to assign oxidation numbers
The oxidation number of an atom in an elemental substance is zero.
The oxidation number of a monatomic ion is equal to the ion’s charge.
Oxidation numbers for common nonmetals are usually assigned as follows:
• Hydrogen: +1 when combined with nonmetals, −1 when combined with metals

4. Oxidation number
• Oxygen: −2 in most compounds; • Halogens: −1 The sum of oxidation numbers for all atoms in a molecule or polyatomic ion equals the charge on the molecule or ion. New definition of Redox reaction Fe2+(aq)+Ag+(aq) ⟶ Fe3+ (aq)+Ag(s) oxidation = increase in oxidation number Fe2+(aq) ⟶ Fe3+ (aq) + e reduction = decrease in oxidation number Ag+(aq) + e ⟶ Ag(s)

5. Single replacement reaction
Single-replacement (displacement) reactions are redox reactions in which an ion in solution is displaced (or replaced) via the oxidation of a metallic element . Examples Zn(s)+2HCl(aq) ⟶ ZnCl2(aq)+H2(g) Zn(s)+2H+ (aq) ⟶ Zn2+ +H2(g) Cu(s)+2AgNO3(aq) ⟶ Cu(NO3)2(aq)+2Ag(s) Cu(s)+2Ag+ (aq) ⟶ Cu2+ (aq)+2Ag(s)

6. Activity Series of Metals6

7. Balancing Redox Reactions via the Half-Reaction Method in Acidic Solution
1. Write the two half-reactions representing the redox process.
2. Balance all elements except oxygen and hydrogen.
3. Balance oxygen atoms by adding H2O molecules.
4. Balance hydrogen atoms by adding H+ ions.
5. Balance charge by adding electrons.
6. If necessary, multiply each half-reaction’s coefficients by the smallest possible integers to yield equal numbers of electrons in each.
7. Add the balanced half-reactions together and simplify by removing species that appear on both sides of the equation.
8. Finally, check to see that both the number of atoms and the total charges are balanced

8. Balancing Redox Reactions in Acidic Solution
Balance in Acidic Solution
Cr2O72– + Fe2+  Cr Fe3+
1. Break into half-reactions
Cr2O72–  Cr3+
Fe2+  Fe3+
2. Balance atoms other than H and O
Cr2O72–  2Cr3+
Put in 2 coefficient to balance Cr
Fe already balanced

9. Balancing Redox Reactions in Acidic Solution
Balance in Acidic Solution
Cr2O72– + Fe2+  Cr Fe3+
1. Break into half-reactions
Cr2O72–  Cr3+
Fe2+  Fe3+
2. Balance atoms other than H and O
Cr2O72–  2Cr3+
Put in 2 coefficient to balance Cr
Fe already balanced

10. Balancing Redox Reactions in Acidic Solution
3. Balance O by adding H2O to the side that needs O
Cr2O72–  2Cr3+
Left side has seven O atoms
Right side has none
Add seven H2O to right side
Fe2+  Fe3+
No O to balance
+ 7H2O

11. Balancing Redox Reactions in Acidic Solution
4. Balance H by adding H+ to side that needs H
Cr2O72–  2Cr H2O
Right side has fourteen H atoms
Left side has none
Add fourteen H+ to left side
Fe2+  Fe3+
No H to balance
14H+ +

12. Balancing Redox Reactions in Acidic Solution
5. Balance net charge by adding electrons.
14H+ + Cr2O72–  2Cr3+ + 7H2O
6 electrons must be added to reactant side
Fe2+  Fe3+
1 electron must be added to product side
Now both half-reactions balanced for mass and charge
6e– +
Net Charge =
14(+1) +(–2) = 12
Net Charge =
2(+3)+7(0) = 6
+ e–

13. Balancing Redox Reactions in Acidic Solution
6. Make electron gain equal electron loss; then add half-reactions
6e– + 14H+ + Cr2O72–  2Cr3+ + 7H2O
Fe2+  Fe3+ + e–
7. Cancel anything that's the same on both sides
6[ ]
6Fe3+ + 2Cr H2O + 6e–
6e– + 6Fe H+ + Cr2O72– 
6Fe H+ + Cr2O72– 
6Fe3+ + 2Cr3+ + 7H2O

14. Balancing Redox Reactions in Basic Solution
1. Write the two half-reactions representing the redox process.
2. Balance all elements except oxygen and hydrogen.
3. Balance oxygen atoms by adding H2O molecules to the side that needs oxygen or by adding H2O molecules to the side that has excess oxygen and two OH- to the opposite side of the arrow: (O) + H2O = 2 OH-
4. Balance hydrogen atoms by adding OH- to the side that has excess hydrogen and H2O to the opposite side of the arrow. (H) + OH- = H2O
5. Balance charge by adding electrons.
6. If necessary, multiply each half-reaction’s coefficients by the smallest possible integers to yield equal numbers of electrons in each.
7. Add the balanced half-reactions together and simplify by removing species that appear on both sides of the equation.
8. Finally, check to see that both the number of atoms and the total charges are balanced

15. Balancing Redox Reactions in Basic Solution
Cl−(aq)+MnO4−(aq) ⟶ ClO3−(aq)+MnO2(s)
1. Break into half-reactions
MnO4− ⟶ MnO2
Cl−⟶ ClO3−
2. Balance O atoms by adding H2O
MnO4− ⟶ MnO2+ 2 H2O
Cl− + 3 H2O ⟶ ClO3−
3. Balance H atoms by adding OH- and H2O
MnO4−+ 4 H2O ⟶ MnO2 + 2 H2O + 4OH-
MnO4−+ 2 H2O ⟶ MnO2 + 4OH-

16. Balancing Redox Reactions in Basic Solution
3. Balance H atoms
Cl−+ 3 H2O + 6OH- ⟶ ClO3− + 6H2O
Cl−+ 6OH- ⟶ ClO3− + 3 H2O
4. Balance net charge by adding electrons.
MnO4−+ 2 H2O + 3e ⟶ MnO2 + 4OH-
Cl−+ 6OH- ⟶ ClO3− + 3 H2O + 6e
5. Make electron gain equal electron loss; then add half-reactions and simplify
2MnO4−+ H2O +Cl−⟶ ClO3− + 2MnO2 + 2OH-

17. Balancing Redox Reactions via the Half-Reaction Method in Basic Solution
Another way (presented in OpenStax on page 908) to generate the balanced overall equation in basic solution is to start with the balanced equation in acidic solution, then “convert” it to the equation for basic solution by performing additional steps. a. Add OH− ions to both sides of the equation in numbers equal to the number of H+ ions. b. On the side of the equation containing both H+ and OH− ions, combine these ions to yield water molecules. c. Simplify the equation by removing any redundant water molecules It is necessary to exercise caution when doing this, as many reactants behave differently under basic conditions

18. a. Add H+ to the side that needs hydrogen (left for example)
Balancing Redox Reactions via the Half-Reaction Method in Basic Solution
Actually this method is the same as the one balancing hydrogen by adding OH- to the side that has excess hydrogen and water to the opposite side.
a. Add H+ to the side that needs hydrogen (left for example)
b. Add OH− ions to both sides of the equation .
H+ + OH− → OH−
c. Replace H+ and OH− ions by water molecules.
H2O → OH−
That is balancing hydrogen by adding H2O to the side that needs hydrogen(left) and OH- to the side that has excess hydrogen.
H2O → OH− + (H in excess)