1. Oxidation-Reduction Reactions

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 = +4 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
Q. Name the following... PbO2
SnCl2
SbCl3 NEXT SLIDE FOR ANSWERS
TiCl4
BrF5

9. 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
Q. Name the following... PbO2 Lead (IV) Oxide
SnCl2 Tin (II) Chloride
SbCl3 Antimony (III) Chloride
TiCl4 Titanium (IV) Chloride

10. 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)

11. LEO

12. LEO says
GER!

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

14. 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

15. 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

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

17. 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

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

19. 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

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

21. 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.

22. 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+

23. 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

24. 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.

25. 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+
Cr2O72- and Fe2+
H2O2 and MnO4¯
C2O42- and MnO4¯
S2O32- and I2
Cr2O72- and I¯

26. 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

27. Disproportionation
Disproportionation is the simultaneous oxidation and reduction of the same species. Cu+  Cu2+ + e oxidation Cu+ + e  Cu reduction Mn3+ + 2H2O  MnO2 + 4H+ + e oxidation Mn3+ + e  Mn2+ reduction