1. Oxidation-Reduction Reactions

2. LEO

3. LEO says
GER!

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

5. Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2
Oxidation Number
The charge the atom would have in a molecule (or an
ionic compound) if electrons were completely transferred.
Free elements (uncombined state) have an oxidation number of ______.
Na, Be, K, Pb, H2, O2, P4 = 0
In monatomic ions, the oxidation number is equal to the __________________.
Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2
The oxidation number of oxygen is usually _________. In H2O2 and O22- it is __________.

6. Oxidation Number
The oxidation number of hydrogen is ___ except when it is bonded to metals in binary compounds. In these cases, its oxidation number is ___. (LiAlH4)
Group IA metals are ___, IIA metals are ___ and fluorine is always ___.
6. The sum of the oxidation numbers of all the atoms in a molecule or ion is equal to ________________________.

7. CALCULATING OXIDATION STATE - 2
OXIDATION STATES
CALCULATING OXIDATION STATE - 2
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
TiCl4
BrF5

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 Removal (loss) of electrons ‘OIL’
species will get less negative or more positive
REDUCTION Gain of electrons ‘RIG’
species will become more negative or less positive
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. 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
H2O2 —> H2O
Cr2O72- —> Cr3+
Cr2O72- —> CrO42-
SO42- —> SO2

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+ 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
H2O2 —> H2O R to -2
Cr2O72- —> Cr3+ R to +3
Cr2O72- —> CrO42- N to +6
SO42- —> SO2 R to +4

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

13. 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

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

15. 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

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

17. 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.

18. 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+

19. 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

20. 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.

21. 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¯

22. ANSWERS BALANCING REDOX EQUATIONS 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
ANSWERS

23. Electrochemical Cells
_______
__________
_______
__________
spontaneous
redox reaction

24. Electrochemical Cells
The difference in electrical potential between the anode and cathode is called:
____________________
Cell Diagram
Zn (s) + Cu2+ (aq) Cu (s) + Zn2+ (aq)
[Cu2+] = 1 M & [Zn2+] = 1 M
Zn (s) | Zn2+ (1 M) || Cu2+ (1 M) | Cu (s)
anode
cathode

25. Standard Electrode Potentials
Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s)
Anode (oxidation):
Zn (s) Zn2+ (1 M) + 2e-
Cathode (reduction):
2e- + 2H+ (1 M) H2 (1 atm)

26. Standard Electrode Potentials
____________ ____________ ____________ (E0) is the voltage associated with a reduction reaction at an electrode when all solutes are 1 M and all gases are at 1 atm.
Any time you see º, think “standard state conditions”
Reduction Reaction
2e- + 2H+ (1 M) H2 (1 atm)
E0 = 0 V
Standard hydrogen electrode (SHE)

27. Standard Electrode Potentials
E0 = 0.76 V
cell
Standard emf (E0 )
cell
E0 = Ecathode - Eanode
cell
E0 = Ereduct’n - Eoxid
cell
Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s)
E0 = EH /H - EZn /Zn
cell + 2+ 2
0.76 V = 0 - EZn /Zn 2+
EZn /Zn = V 2+
Zn2+ (1 M) + 2e Zn E0 = ____________

28. Standard Electrode Potentials
E0 = 0.34 V
cell
E0 = Ecathode - Eanode
cell
Ecell = ECu /Cu – EH /H 2+ + 2
0.34 = ECu /Cu - 0 2+
ECu /Cu = 0.34 V 2+
Pt (s) | H2 (1 atm) | H+ (1 M) || Cu2+ (1 M) | Cu (s)
Anode (oxidation):
H2 (1 atm) H+ (1 M) + 2e-
Cathode (reduction):
2e- + Cu2+ (1 M) Cu (s)

29. E0 is for the reaction as written
The more positive E0 the greater the tendency for the substance to be reduced
The more negative E0 the greater the tendency for the substance to be oxidized
Under standard-state conditions, any species on the left of a given half-reaction will react spontaneously with a species that appears on the right of any half-reaction located below it in the table (the diagonal rule)

30. The half-cell reactions are reversible
The sign of E0 changes when the reaction is reversed
Changing the stoichiometric coefficients of a half-cell reaction does not change the value of E0

31. Can Sn reduce Zn2+ under standard-state conditions?
How do we find the answer?
Look up the Eº values in Table.
Zn2+(aq) + 2e- —> Zn(s) (Is this oxidation or reduction?)
Which reactions in the table will reduce Zn2+(aq)?

32. Cd is the stronger oxidizer
What is the standard emf of an electrochemical cell made of a Cd electrode in a 1.0 M Cd(NO3)2 solution and a Cr electrode in a 1.0 M Cr(NO3)3 solution?
Cd2+ (aq) + 2e Cd (s) E0 = V
Cd is the stronger oxidizer
Cd will oxidize Cr
Cr3+ (aq) + 3e Cr (s) E0 = V
Anode (oxidation):
Cr (s) Cr3+ (1 M) + 3e-
x 2
Cathode (reduction):
2e- + Cd2+ (1 M) Cd (s)
x 3
E0 = Ecathode - Eanode
cell
E0 = – (-0.74)
cell
E0 = _________
cell

33. Spontaneity of Redox Reactions
DG = -nFEcell
n = number of moles of electrons in reaction
F = 96,500 J
V • mol
DG0 = -nFEcell
= 96,500 C/mol
DG0 = -RT ln K
= -nFEcell
Ecell = RT nF
ln K
(8.314 J/K•mol)(298 K)
n (96,500 J/V•mol)
ln K = = V n
ln K
Ecell = V n
log K
Ecell

34. Spontaneity of Redox Reactions
If you know one, you can calculate the other…
If you know K, you can calculate Eº and Gº
If you know Eº, you can calculate Gº

35. Spontaneity of Redox Reactions
Relationships among G º, K, and Eºcell

36. What is the equilibrium constant for the following reaction at 250C
What is the equilibrium constant for the following reaction at 250C? Fe2+ (aq) + 2Ag (s) Fe (s) + 2Ag+ (aq) = V n
ln K
Ecell
Oxidation:
2Ag Ag+ + 2e-
n = ___
Reduction:
2e- + Fe Fe
E0 = EFe /Fe – EAg /Ag 2+ +
E0 = – (0.80)
E0 = ___________ V
x n E0
cell
exp
K = V
x 2
-1.24 V
= exp
K = ________________

37. Calculate DG0 for the following reaction at 250C.
2Al3+(aq) + 3Mg(s) Al(s) + 3Mg+2(aq)
Oxidation:
2Mg Mg2+ + 6e-
n = ?
Reduction:
6e- + 3Al Al
E0 = Ecathode - Eanode
cell
DG0 = -nFEcell

38. Calculate DG0 for the following reaction at 250C.
2Al3+(aq) + 3Mg(s) Al(s) + 3Mg+2(aq)
Oxidation:
2Mg Mg2+ + 6e-
E0 = __ V
Reduction:
6e- + 3Al Al
E0 = __ V
n = __
E0 = Ecathode - Eanode
cell
= ___ V
DG0 = -nFEcell
= ___ X (96,500 J/V mol) X ___ V
DG0 = _______ kJ/mol

39. The Effect of Concentration on Cell Emf
DG = DG0 + RT ln Q
DG = -nFE
DG0 = -nFE
-nFE = -nFE0 + RT ln Q
E = E0 -
ln Q RT nF
_____________ equation
At 298K - V n
ln Q E
E = - V n
log Q E
E =

40. The Nernst equation enables us to calculate E as a function of [reactants] and [products] in a redox reaction.

41. Fe2+ (aq) + Cd (s) Fe (s) + Cd2+ (aq)
Will the following reaction occur spontaneously at 250C if [Fe2+] = 0.60 M and [Cd2+] = M?
Fe2+ (aq) + Cd (s) Fe (s) + Cd2+ (aq)
Oxidation:
Cd Cd2+ + 2e-
n = ___
Reduction:
2e- + Fe Fe
E0 = EFe /Fe – ECd /Cd 2+
E0 = – (-0.40) - V n
ln Q E
E =
E0 = V - V 2 ln
-0.04 V
E =
0.010
0.60
E = ____________
E ___ 0
________________

42. Batteries Dry cell Leclanché cell Anode: Zn (s) Zn2+ (aq) + 2e-
Cathode:
2NH4 (aq) + 2MnO2 (s) + 2e Mn2O3 (s) + 2NH3 (aq) + H2O (l) +
Zn (s) + 2NH4 (aq) + 2MnO2 (s) Zn2+ (aq) + 2NH3 (aq) + H2O (l) + Mn2O3 (s)

43. Batteries Mercury Battery Anode:
Zn(Hg) + 2OH- (aq) ZnO (s) + H2O (l) + 2e-
Cathode:
HgO (s) + H2O (l) + 2e Hg (l) + 2OH- (aq)
Zn(Hg) + HgO (s) ZnO (s) + Hg (l)

44. Batteries Lead storage battery Anode:
Pb (s) + SO2- (aq) PbSO4 (s) + 2e- 4
Cathode:
PbO2 (s) + 4H+ (aq) + SO2- (aq) + 2e PbSO4 (s) + 2H2O (l) 4
Pb (s) + PbO2 (s) + 4H+ (aq) + 2SO2- (aq) PbSO4 (s) + 2H2O (l) 4

45. Solid State Lithium Battery
Batteries
Solid State Lithium Battery

46. Batteries
A ______ ______ is an electrochemical cell that requires a continuous supply of reactants to keep functioning
Anode:
2H2 (g) + 4OH- (aq) H2O (l) + 4e-
Cathode:
O2 (g) + 2H2O (l) + 4e OH- (aq)
2H2 (g) + O2 (g) H2O (l)

47. Corrosion

48. Cathodic Protection of an Iron Storage Tank

49. _______________ is the process in which electrical energy is used to cause a nonspontaneous chemical reaction to occur.

50. Electrolysis of Water

51. Electrolysis and Mass Changes
charge (C) = current (A) x time (s)
1 mole e- = 96,500 C
So what is the charge on a single electron?