Chapter 4
Reduction-Oxidation Reactions
Redox Reactions
Sodium chloride
Sodium chloride
Na  Na+ + e
Cl2 + 2e  2Cl
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
4
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–
Your Turn!
Which species functions as the oxidizing agent
in the following oxidation-reduction reaction?
Zn(s) + Pt2+(aq)  Pt(s) + Zn2+(aq)
A. Pt(s)
Zn2+(aq)
C. Pt2+(aq)
D. Zn(s)
B.
E.
None of these, as this is not a redox reaction.
6
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
7
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
8
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.
9
Rules for Oxidation States (Numbers)
1. The sum of Oxidation numbers equals to the
charge on molecule, formula unit or ion.
2. The oxidation state of elements is zero.
3. Oxidation state for monoatomic ions are the
same as their charge.
4. In its compounds fluorine is always –1.
5. Hydrogen is assigned the oxidation state +1.
6. Oxygen is assigned an oxidation state of -2 in
its covalent compounds (except as a peroxide).
7. If two rules conflict, apply higher rule.
Oxidation States

Assign the oxidation states to each element in
the following.

CO2
NO3H2SO4
Fe2O3
Fe3O4
Cr2O72O2F2
H2O2
LiH
BaO2









Oxidation-Reduction
Transfer electrons, so the oxidation states change.
Ox
0
1
0
1
Na  Cl 2  Na Cl
Oxidation
Reduction
Red
Increase in
decrease in
Ox
Oxid. number
0
Oxid. number
2 2
0
Mg  O 2  Mg O
Red
Ox
0
4 2
0
S  O 2  S O2
Red
Ox
4 1
4 2
0
1
2
C H 4  O2  C O2  H 2 O
C (CH4) oxidized →
CH4 reducing agent
Red
Red
O2 reduced →
O2 oxidizing agent
Red
Ox
PbS has been oxidized, PbS is the reducing agent.
O2 has been reduced, O2 is the oxidizing agent.
Red
Ox
PbO has been reduced, PbO is the oxidizing agent.
CO has been oxidized, CO is the reducing agent.
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
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 halfreactions
7. Cancel anything that is the same on both sides
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
Fe2+  Fe3+
– Fe already balanced
3. Balance O by adding H2O to the side that needs O.
Cr2O72  2Cr3+ + 7 H2O
• Right side has 7 O atoms
• Left side has none
• Add 7 H2O to left side
Fe2+  Fe3+
• No O to balance
4. Balance H by adding H+ to side that needs H
14H+ + Cr2O72  2Cr3+ + 7H2O
• Left side has 14 H atoms
• Right side has none
• Add 14 H+ to right side
Fe2+  Fe3+
• No H to balance
5. Balance net charge by adding electrons.
6e + 14H+ + Cr2O72  2Cr3+ + 7H2O
Net Charge =
14(+1) (–2) = 12
Net Charge =
2(+3)+7(0) = 6
– 6 electrons must be added to reactant side
Fe2+  Fe3+ + e
– 1 electron must be added to product side

Now both half-reactions balanced for mass &
charge
6. Make e– gain equal e– loss; then add half-reactions
6e + 14H+ + Cr2O72–  2Cr3+ + 7H2O
6[ Fe2+  Fe3+ + e ]
6e + 6Fe2+

+ 14H+ + Cr2O72
6Fe3+ + 2Cr3+
+ 7H2O + 6e
7. Cancel anything that's the same on both sides
6Fe2+ + 14H+ 
+ Cr2O72
6Fe3+ + 2Cr3+
+ 7H2O
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
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
Basic Solution

Ag + CN- +O2 Ag(CN)2-

Cr(OH)3 + OCl- + OH- CrO42- + Cl- + H2O
 CrI3 + Cl2  CrO4 + IO4 + Cl