1. Oxidation-Reduction Reactions
LESSON OBJECTIVES
You will learn:
1) The purpose for oxidation numbers
2) The rules for oxidation numbers
3) How to find oxidation numbers, example questions
4) How to use oxidation numbers to determine if a reaction is oxidation or reduction

2. MOLECULES: SO VERY HARD TO FOLOW THE e-
Compare and ionic reaction to a molecular one:
Ionic: Na(s) + Cl2(g)  2 NaCl(s)
Following the electrons we see Na (s) has picked up an e- and Cl has gained it.
Molecular: H2(s) + O2(g)  2H2O(g)
Ok, so were did the electrons go? You cannot tell.
So we need a way to FOLLOW the e-
The way we “follow” electrons is to use oxidation numbers, or rules invented to help us to “see” the electrons when they are invisible!

3. Oxidation Numbers for Bookkeeping
Hydrogen has "lost" 1 electron to oxygen. Oxygen has "gained" 2 electrons
Partial charges given to atoms in a molecule in this way are called oxidation numbers.
We can use oxidation numbers to keep track of where electrons are in a molecule
Oxygen has an oxidation number = -2 because the oxygen atom has "gained" 2 e-
Each H still has an oxidation number = Oxygen, however, now has an oxidation # of -1 because each oxygen gains 1 electron from each hydrogen (it equally shares the other so it does not “go” to anyone in particular) 

4. Oxidation Number Rules
The charge the atom would have in a molecule if electrons were completely transferred.
Free elements (uncombined state); oxidation number = 0.
Na, Be, K, Pb, H2, O2, P4 = 0
In monatomic ions, the oxidation # is equal to the charge.
Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2
The oxidation # of oxygen is usually - 2. In H2O2 , oxygen = -1.
The oxidation number of hydrogen is +1 except when it is bonded to metals (as in metal hydrides) in binary compounds. In these cases, its oxidation number is : Ex: (LiH4)

5. Oxidation Numbers The sum of the oxidation numbers of all the atoms in
a molecule or ion is equal to the charge.
Ex: Cr2O72- all the ox #’s must add to equal -2
thus 2 Cr + 7(-2) = -2
2 Cr = -2
2 Cr = -12
Cr = +6
If all else fails play the electronegativity's game, whichever has the higher electronegativity, wins its charge.
Ex: PCl3 notice we have no rule for either P or Cl
Thus we look up their electronegativity's, P= 2.1 and Cl =3.0
Cl “wins” or it has a higher electronegaivirty thus we give it
whatever charge it would have if it were ionic thus:
Cl = -1 and P has to be -3 in PCl3

6. CALCULATING OXIDATION STATE
OXIDATION STATES
CALCULATING OXIDATION STATE
Q. What is the oxidation state of each element in the following compounds/ions ?
CH4
PCl3
NCl3
CS2
ICl5
BrF3
PCl4+
H3PO4
NH4Cl
H2SO4
MgCO3
SOCl2

7. CALCULATING OXIDATION STATE
OXIDATION STATES
CALCULATING OXIDATION STATE
Q. What is the oxidation state of each element in the following compounds/ions ?
CH4 C = - 4 H = +1
PCl3 P = +3 Cl = -1
NCl3 N = +3 Cl = -1
CS2 C = +4 S = -2
ICl5 I = +5 Cl = -1
BrF3 Br = +3 F = -1
PCl4+ P = +5 Cl = -1
H3PO4 P = +5 H = +1 O = -2
NH4Cl N = -3 H = +1 Cl = -1
H2SO4 S = +6 H = +1 O = -2
MgCO3 Mg = +2 C = +4 O = -2
SOCl2 S = +4 Cl = -1 O = -2

8. THE ROLE OF OXIDATION STATE IN NAMING SPECIES
OXIDATION STATES
THE ROLE OF OXIDATION STATE IN NAMING SPECIES
To avoid ambiguity, the oxidation state is often included in the name of a species
manganese(IV) oxide shows that Mn is in the +4 oxidation state in MnO2
sulphur(VI) oxide for SO3 S is in the +6 oxidation state
dichromate(VI) for Cr2O72- Cr is in the +6 oxidation state
phosphorus(V) chloride for PCl5 P is in the +5 oxidation state
phosphorus(III) chloride for PCl3 P is in the +3 oxidation state
Name the following... PbO2
Lead (IV) Oxide
SnCl2
Tin (II) Chloride
SbCl3
Antimony (III) Chloride
TiCl4
Titanium (IV) Chloride

9. REDOX REACTIONS OXIDATION AND REDUCTION IN TERMS OF ELECTRONS
Oxidation and reduction are not only defined as changes in O and H
REDOX When reduction and oxidation take place
OXIDATION LEO
REDUCTION GER
REDUCTION in O.S. Species has been REDUCED
e.g. Cl is reduced to Cl¯ (0 to -1)
INCREASE in O.S. Species has been OXIDISED
e.g. Na is oxidised to Na+ (0 to +1)

10. GER! LEO says Loss of Electrons = Oxidation
Gain of Electrons = Reduction

11. OXIDATION AND REDUCTION IN TERMS OF ELECTRONS
REDOX REACTIONS
OXIDATION AND REDUCTION IN TERMS OF ELECTRONS
REDUCTION in O.S. INCREASE in O.S.
Species has been REDUCED Species has been OXIDISED
Q State if the changes involve oxidation (O) or reduction (R) or neither (N)
Fe2+ —> Fe3+
I2 —> I¯
F2 —> F2O
C2O42- —> CO2
H2O —> O2
H2O —> H2O
Cr2O72- —> Cr3+
Cr2O72- —> CrO42-
SO42- —> SO2

12. OXIDATION AND REDUCTION IN TERMS OF ELECTRONS
REDOX REACTIONS
OXIDATION AND REDUCTION IN TERMS OF ELECTRONS
REDUCTION in O.S. INCREASE in O.S.
Species has been REDUCED Species has been OXIDISED
O = Oxidation , R = Reduction
Fe2+ —> Fe3+ O +2 to +3
I2 —> I¯ R to -1
F2 —> F2O R to -1
C2O42- —> CO2 O +3 to +4
H2O —> O2 O to 0
H2O —> H2O N to -1 for H and -2 to -2 for O
Cr2O72- —> Cr3+ R to +3
Cr2O72- —> CrO42- N to +6
SO42- —> SO2 R to +4

13. Exercise
For each of the following reactions find the element oxidized and the element reduced
Cl KBr  KCl Br2

14. So this is a reduction reaction
Exercise
For each of the following reactions find the element oxidized and the element reduced
Cl KBr  KCl Br2  
Br increases from –1 to oxidized
Cl decreases from 0 to – Reduced
So this is a reduction reaction
K remains unchanged at +1

15. Exercise
For each of the following reactions find the element oxidized and the element reduced
Cu HNO3  Cu(NO3) NO2 + H2O

16. Exercise
For each of the following reactions find the element oxidized and the element reduced
Cu HNO3  Cu(NO3) NO2 + H2O –
Cu increases from 0 to It is oxidized
Only part of the N in nitric acid changes from +5 to +4. It is reduced
The nitrogen that ends up in copper nitrate remains unchanged

17. Exercise
For each of the following reactions find the element oxidized and the element reduced
HNO I  HIO NO2

18. Exercise
For each of the following reactions find the element oxidized and the element reduced
HNO I  HIO NO2
N is reduced from +5 to +4. It is reduced
I is increased from 0 to +5 It is oxidized
The hydrogen and oxygen remain unchanged.

19. Oxidation half-reaction (lose e-)
Electrochemical processes are oxidation-reduction reactions in which:
the energy released by a spontaneous reaction is converted to electricity or
electrical energy is used to cause a nonspontaneous reaction to occur 2+ 2-
2Mg (s) + O2 (g) MgO (s)
2Mg Mg2+ + 4e-
Oxidation half-reaction (lose e-)
O2 + 4e O2-
Reduction half-reaction (gain e-)

20. Balancing Redox Equations LESSON OBJECTIVES
You will learn:
How to balance redox equations with acids

21. Balancing Redox Equations
Balance the oxidation of Fe2+ to Fe3+ by Cr2O72- in acid solution
Write the unbalanced equation for the reaction in ionic form.
Fe2+ + Cr2O Fe3+ + Cr3+
Separate the equation into two half-reactions.
Fe Fe3+ +2 +3
Oxidation:
Cr2O Cr3+ +6 +3
Reduction:
Balance the atoms other than O and H in each half-reaction.
Cr2O Cr3+

22. Balancing Redox Equations
For reactions in acid, add H2O to balance O atoms and H+ to balance H atoms.
Cr2O Cr3+ + 7H2O
14H+ + Cr2O Cr3+ + 7H2O
Add electrons to one side of each half-reaction to balance the charges on the half-reaction.
Fe Fe3+ + 1e-
6e- + 14H+ + Cr2O Cr3+ + 7H2O
If necessary, equalize the number of electrons in the two half-reactions by multiplying the half-reactions by appropriate coefficients.
6Fe Fe3+ + 6e-
6e- + 14H+ + Cr2O Cr3+ + 7H2O

23. Balancing Redox Equations
Add the two half-reactions together and balance the final equation by inspection. The number of electrons on both sides must cancel.
Oxidation:
6Fe Fe3+ + 6e-
Reduction:
6e- + 14H+ + Cr2O Cr3+ + 7H2O
14H+ + Cr2O Fe Fe3+ + 2Cr3+ + 7H2O
Verify that the number of atoms and the charges are balanced.
14x1 – 2 + 6x2 = 24 = 6x3 + 2x3
For reactions in basic solutions, add OH- to both sides of the equation for every H+ that appears in the final equation.

24. COMBINING HALF EQUATIONS
A combination of two ionic half equations, one involving oxidation and the other reduction, produces a REDOX equation. The equations are balanced as follows...
Step 1 Write out the two half equations
Step 2 Multiply the equations so that the number of electrons in each is the same
Step 3 Add the two equations and cancel out the electrons on either side
Step 4 If necessary, cancel any other species which appear on both sides
Q. Construct balanced redox equations for the reactions between...
Mg and H+  H2 and Mg2+
Cr2O72- and Fe2+  Fe Cr3+
H2O2 and MnO4¯  Mn O2
C2O42- and MnO4¯.  CO Mn2+
S2O32- and I2.  S4O I¯
Cr2O72- and I¯

25. ANSWERS Mg ——> Mg2+ + 2e¯ (x1) H+ + e¯ ——> ½ H2 (x2)
Mg + 2H+ ——> Mg H2
Cr2O H+ + 6e¯ ——> 2Cr H2O (x1)
Fe2+ ——> Fe3+ + e¯ (x6)
Cr2O H+ + 6Fe2+ ——> 2Cr Fe H2O
MnO4¯ e¯ H+ ——> Mn H2O (x2)
H2O2 ——> O H e¯ (x5)
2MnO4¯ H2O H+ ——> 2Mn O H2O
MnO4¯ e¯ H+ ——> Mn H2O (x2)
C2O42- ——> 2CO e¯ (x5)
2MnO4¯ + 5C2O H+ ——> 2Mn CO2 + 8H2O
2S2O ——> S4O e¯ (x1)
½ I e¯ ——> I¯ (x2)
2S2O I2 ——> S4O I¯
Cr2O H+ + 6e¯ ——> 2Cr H2O (x1)
½ I e¯ ——> I¯ (x6)
Cr2O H I2 ——> 2Cr I ¯ H2O

26. Balancing redox reactions by the half-reaction method
PROBLEM:
Permanganate ion is a strong oxidizing agent and its deep purple color makes it useful as an indicator in redox titrations. It reacts in basic solution with the oxalate anion to form carbonate anion and solid manganese dioxide. Balance the skeleton ionic reaction that occurs between KMnO4 and Na2C2O4 in basic solution.
MnO4-(aq) + C2O42-(aq) MnO2(s) + CO32-(aq)
PLAN:
Proceed in acidic solution and then neutralize with base.
SOLUTION:
Red Ox
MnO MnO2
C2O CO32- +7 +4 +3 +4
MnO MnO2
C2O CO32-
4H+ +
MnO MnO2
+ 2H2O
C2O CO32-
+ 2H2O
+ 4H+
+ 3e-
+ 2e-

27. (continued) x2 x3
C2O H2O CO H+ + 2e-
4H+ + MnO4- +3e MnO2+ 2H2O
8H+ + 2MnO4- + 6e MnO2+ 4H2O
3C2O H2O CO H+ + 6e-
8H+ + 2MnO4- + 6e MnO2 + 4H2O
3C2O H2O CO H+ + 6e-
2MnO2-(aq) + 3C2O42-(aq) + 2H2O(l) MnO2(s) + 6CO32-(aq) + 4H+(aq)
+ 4OH-
+ 4OH-
2MnO2-(aq) + 3C2O42-(aq) + 4OH-(aq) MnO2(s) + 6CO32-(aq) + 2H2O(l)

28. Disproportionation
Disproportionation is the simultaneous oxidation and reduction of the same species. 3 Cl2 + 6 OH− → 5 Cl− + ClO3− + 3 H2O The chlorine gas reactant is in oxidation state 0. In the products, the chlorine in the Cl− ion has an oxidation number of −1, having been reduced. Whereas the oxidation number of the chlorine in the ClO3− ion is +5, indicating that it has been oxidized.