1. Reduction-Oxidation Reactions Redox Reactions
Chapter 4
Reduction-Oxidation Reactions
Redox Reactions

2. Sodium chloride

3. Sodium chloride
Na  Na+ + e
Cl2 + 2e  2Cl

4. 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 & reduction always occur together
Can't have one without the other

5. Oxidation Reduction Reaction
Oxidizing Agent
- Substance that accepts e's
Accepts e's from another substance
Substance that is reduced
Cl2 + 2e  2Cl–
Reducing Agent
- Substance that donates e's
Releases e's to another substance
Substance that is oxidized
Na  Na+ + e–

6. Your Turn!
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)
Pt2+(aq)
Zn(s)
None of these, as this is not a redox reaction.

7. Redox Reactions Very common
Batteries—car, flashlight, cell phone, computer
Metabolism of food
Combustion
Chlorine Bleach
Dilute NaOCl solution
Cleans through redox
reaction
Oxidizing agent
Destroys stains by oxidizing them

8. Redox Reaction Ex. Fireworks displays Net: 2Mg + O2  2MgO Oxidation:
Mg  Mg2+ + 2e
Loses electrons = Oxidized
Reducing agent
Reduction:
O2 + 4e  2O2
Gains electrons = Reduced
Oxidizing agent

9. Redox Reaction? S + O2  SO2
Combustion: Oxidation in old sense, reaction with oxygen
But n no ions in SO2
How can we decide which loses and which gains!!
Oxidation number (state)!
If compound were ionic, what would the charges have been.

10. Rules for Oxidation States (Numbers)
The sum of Oxidation numbers equals to the charge on molecule, formula unit or ion.
The oxidation state of elements is zero.
Oxidation state for monoatomic ions are the same as their charge.
In its compounds fluorine is always –1.
Hydrogen is assigned the oxidation state +1.
Oxygen is assigned an oxidation state of -2 in its covalent compounds (except as a peroxide).
If two rules conflict, apply higher rule.

11. Oxidation States
Assign the oxidation states to each element in the following.
CO2
NO3-
H2SO4
Fe2O3
Fe3O4
Cr2O72-
O2F2
H2O2
LiH
BaO2

12. Oxidation-Reduction
Transfer electrons, so the oxidation states change. Ox
Red
Oxidation
Increase in
Oxid. number
Reduction
decrease in
Oxid. number Ox
Red

13. Ox
Red Ox
Red
Red
C (CH4) oxidized →
CH4 reducing agent
O2 reduced →
O2 oxidizing agent

14. PbS has been oxidized, PbS is the reducing agent.
O2 has been reduced, O2 is the oxidizing agent.

15. PbO has been reduced, PbO is the oxidizing agent.
CO has been oxidized, CO is the reducing agent.

16. Identify the 1) Oxidizing agent 2) Reducing agent
3) Substance oxidized
4) Substance reduced
in the following reactions
Fe (s) + O2(g) ® Fe2O3(s)
Fe2O3(s)+ 3 CO(g) ® 2 Fe(l) + 3 CO2(g)
SO3- + H+ + MnO4- ® SO4- + H2O + Mn+2

17. Balancing Redox Reactions Ion-Electron Method – Acidic Solution
1. Divide equation into 2 half-reactions
2. Balance atoms other than H & O
3. Balance O by adding H2O to side that needs O
4. Balance H by adding H+ to side that needs H
5. Balance net charge by adding e–
6. Make e– gain equal e– loss; then add half- reactions
7. Cancel anything that is the same on both sides

18. Balance in Acidic Solution Cr2O72– + Fe2+  Cr3+ + Fe3+
1. Break into half-reactions
Cr2O72  Cr3+
Fe2+  Fe3+
2. Balance atoms other than H & O
Cr2O72  2Cr3+
Put in 2 coefficient to balance Cr
Fe already balanced
Start Mon 1110

19. 3. Balance O by adding H2O to the side that needs O. Cr2O72  2Cr3+
Right side has 7 O atoms
Left side has none
Add 7 H2O to left side
Fe2+  Fe3+
No O to balance
+ 7 H2O

20. Cr2O72  2Cr3+ + 7H2O 14H+ + Fe2+  Fe3+
4. Balance H by adding H+ to side that needs H
Cr2O72  2Cr H2O
Left side has 14 H atoms
Right side has none
Add 14 H+ to right side
Fe2+  Fe3+
No H to balance
14H+ +

21. 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 & charge
6e +
Net Charge =
14(+1) (–2) = 12
Net Charge =
2(+3)+7(0) = 6
+ e

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

23. Practice
The following reactions occur in acidic aqueous solution. Balance them:
MnO4- + Fe2+ ® Mn2+ + Fe3+
Cu + NO3- ® Cu2+ + NO(g)
Pb + PbO2 + SO42- ® PbSO4
Mn2+ + NaBiO3 ® Bi3+ + MnO4-
Cr2O72- + C2H5OH ® Cr3+ + CO2

24. Ion-Electron method in Basic Solution
The simplest way to balance an equation in basic solution
Use steps 1-7 above, then
8. Add the same number of OH– to both sides of the equation as there are H+.
9. Combine H+ & OH– to form H2O
10. Cancel any H2O that you can from both sides

25. Basic Solution Ag + CN- +O2 ® Ag(CN)2-
Cr(OH)3 + OCl- + OH- ® CrO42- + Cl- + H2O
CrI3 + Cl2 ® CrO4- + IO4- + Cl-