3. Oxidation-Reduction Reactions
Electrochemistry:
Oxidation-Reduction Reactions
gaining to 2e-
Zn(s) + Cu+2 (aq)  Zn2+(aq) + Cu(s)
loss of 2e-
Copper is reduced because it goes down
in charge
Zinc is oxidized - it goes up in charge

4. Messing with electrons
Electrochemistry
Messing with electrons
What is the charge on each atom +1 +7 -8 +3 -2
= 1 +3 -3 +1 +7 -2 +3 -2 +1 +3 -1
KMnO Cr(OH) Fe(OH)2+1

5. Zn(s) + 2H+(aq)  Zn2+(aq) + H2(aq)
Electrochemistry: Oxidation-Reduction Reactions
By writing the oxidation number of each element under the
reaction equation, we can easily see the oxidation state
changes that occur
Zn(s) + 2H+(aq)  Zn2+(aq) + H2(aq)
In any oxidation-reduction reaction (redox),
both oxidation and reduction must occur.

6. Balancing Oxidation-Reduction Reactions
Electrochemistry:
Balancing Oxidation-Reduction Reactions
Oxidation Number Method
-3e’s
X 4 = -12
Al(l) + MnO2  Al2O3 + Mn 4 3 2 3
+4e’s
X 3 = +12

7. Balancing Oxidation-Reduction Reactions
Electrochemistry:
Balancing Oxidation-Reduction Reactions
Sample problem: +2 +4
-2e’s
X 5 = -10
I2O5(s) + CO(g)  I2(s) + CO2(g) 5 5
+5e’s
X 2 = 10 +5

8. MnO4-(aq) + C2O42-(aq)  Mn2+(aq) + CO2(g) (acid)
Half-Reaction Method :
MnO4-(aq) + C2O42-(aq)  Mn2+(aq) + CO2(g) (acid)

9. Half-Reaction Method (in acid):
MnO4-(aq) + C2O42-(aq)  Mn2+(aq) + CO2(g)
Step 1: Divide into two half-reactions,
MnO4-  Mn2+
C2O42-  CO2
Step 2: Balance main element
MnO4-  Mn2+
C2O42-  2CO2

10. MnO4-  Mn2+ + 4H2O C2O42-  2CO2 MnO4-  Mn2+ + 4H2O 8H+ +
Step 3: Balance the O atoms by adding H2O : :
MnO4-  Mn2+
+ 4H2O
C2O42-  2CO2
Step 4: Balance the H atoms by adding H+
8H+ +
MnO4-  Mn2+ + 4H2O
C2O42-  2CO2
Step 5: Balance charge with electrons +7 +2
5e- +
8H++ MnO4-  Mn2+ + 4H2O -2
C2O42-  2CO2
+ 2e-

11. 5e- + 8H++ MnO4-  Mn2+ + 4H2O C2O42-  2CO2 + 2e-
Step 6: Make the electrons balance :
5e- + 8H++ MnO4-  Mn2+ + 4H2O
C2O42-  2CO2 + 2e-
Step 7: Cancel and add
10e H++ 2MnO4-  2Mn2+ + 8H2O +
5C2O42-  10CO2 +10e-
16H+ + 2MnO4- + 5C2O42-  2Mn CO2 + 8H2O

12. CN- + MnO4-  CNO- + MnO2 -1 +1 H2O + CN-  CNO- + 2H+ + 2e- 3e-+
Balancing in base
CN- + MnO4-  CNO- + MnO2 -1 +1 3 3 6 3 6
H2O +
CN-  CNO-
+ 2H+
+ 2e- 6 2 8 2 4
3e-+
4H+ +
MnO4-  MnO2
+ 2H2O +3
2H+ + 3CN- + 2MnO4-  3CNO- + 2MnO2 + H2O
+ 2OH-
2OH-
2H2O 1

13. Cr(OH)3 + ClO  CrO4-2 + Cl2 Have at it Balancing in base
4Cr(OH)3 + 6ClO + 8OH- 4CrO Cl2 + 10H2O

14. metals ordered from the highest to lowest
Note that the activity series is simply the oxidation half-reactions of the
metals ordered from the highest to lowest

15. Using the hydrogen electrode as a reference
2H+ (1 M) + 2e-  H2 (1 atm, 25 °C) E°red = 0 V

16. Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
Cell EMF or Voltage
From the table:
Zn+2 + 2e-  Zn red= -.76V
Flip
Cu+2 + 2e-  Cu red= .34V
Zn  Zn+2 + 2e- ox = .76V
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
E°cell = 1.10 V
E ° ox + E ° red = E°cell

17. Voltaic or Galvanic Cells-
oxidation +
reduction
I love M dawg
Voltaic or Galvanic Cells

18. Sample Problem. Draw a voltaic cell
using the following equation and label
all parts include line notation.
Cr2O72-(aq) + I-(aq)  Cr3+(aq) + I2(s)

20. Spontaneity and Extent of Redox Reactions
A positive emf indicates a spontaneous process, and
a negative emf indicates a nonspontaneous one.
Sample Problem: Using the standard electrode potentials,
determine whether the following reaction are spontaneous
A. Cu(s) + 2H+(aq)  Cu2+(aq) + H2(g)
Cu(s)  Cu2+(aq) + 2e Eox = -.34V
2H+(aq) + 2e-  H2(g) Ered = 0.0V
Ecell = -.34V
B. Cl2(g) + 2I-(aq)  2Cl-(aq) + I2(s)
Cl2(g) + 2e-  2Cl-(aq) Ered = 1.36V
2I-(aq)  2e- + I2(s) Eox = -.54V
Ecell = .82V

21. EMF and Free Energy Change
Any redox reaction involves free energy change
(G) which also may be used as a measure of
spontaneity or work (max or min.)
G° = -nE °
n = # of moles of e-s transferred
 = C the charge of one mole of e-s or 96,500 J/V-mol e-
Note that because n and  are both positive values, a positive value in E leads to a negative value of G.

22. Use the standard electrode potentials to
calculate the standard free energy
change, G°, for the following reaction
2Br-(aq) + F2(g) Br2(l) + 2F-(aq)
2Br-(aq)  Br2(l) + 2e E°ox = -1.06
F2(g) + 2e-  2F-(aq) E°red = 2.87
2Br-(aq) + F2(g) Br2(l) + 2F-(aq) E°cell = 1.81 V
G = -nE
G = -(2 mol e- )(96,500 J/volt-mol e-)( 1.81 V)
G° = x 105 J = -349 kJ

23. EMF and Equilibrium Constant
“Remember in Chapter 19, we related G° to the equilibrium constant, K?”
G° = -RT lnK
G ° = -nE °
-nE° = -RT lnK
Simplify
E ° = logK n

24. Calculate the K for the following reaction
O2 (g) + 4H+ (aq) + 4Fe2+ (aq)  4Fe3+ (aq) + 2H2O (l)
O2 (g) + 4H+ (aq) + 4e-  2H2O (l) E°red = 1.23 V
4Fe2+ (aq)  4Fe3+ (aq) + 4e E°ox = V
O2 (g) + 4H+ (aq) + 4Fe2+ (aq)  4Fe3+ (aq) + 2H2O (l)
E°cell = 0.46 V
n E °
log K =
4(0.46V) = =
31.1
0.0591V
0.0591V
K = x 1031

25.  = .0591 log k n .86V = .0591 log k 10 Log K = 145.5 K = 10145.5
Calculate the equilibrium constant for the reaction
IO3- (aq) + 5Cu (s) + 12H+ (aq) I2 (s) + 5Cu2+ (aq) + 6H2O (l)
2IO3- (aq) + 12H+ (aq) + 10e-  I2 (s) + 6H2O (l) red= 1.20V
5Cu (s)  5Cu2+ (aq) + 10e ox= -.34V
 = log k n
cell= .86V
.86V = log k 10
Log K = 145.5
K =

26. Lead Storage Battery
anode: Pb (s) + SO42- (aq)  PbSO4 (s) + 2e E °= V
cathode:PbO2(s)+ SO42-(aq)+ H+(aq)+ 2e-  PbSO4(s)+ 2H2O(l) E °= V
Pb(s)+ PbO2(s)+ 4H+ +2SO4(aq) 2PbSO4(s)+ 2H2O(l) E °= V
Note that one advantage of the lead storage battery is that it can be recharged because the PbSO4 produced during discharge adheres to the electrodes

27. Dry Cell alkaline anode: Zn (s)  Zn 2+ (aq) + 2e-
cathode:2NH4+(aq)+2MnO2(s)+ 2e  Mn2O3(s) + 2NH3(aq) + H2O (l)

28. Fuel Cells anode: 2H2(g)+ 4OH-(aq) 4H2O(l) + 4e-
cathode: 4e- + O2(g) + H2O(l)  4OH-(aq)
2H2(g) + O2(g)  2H2O(l)

29. Electrolytic Cells: Electrolysis
Notice 
voltage source acts
like an electron
pump
(- red)
(+ ox.)
Eº = ( -)
These electrodes
are inert
Electrolysis is driven by an outside energy source

30. 2H2O + 2e-  H2 + 2OH- E°red = -0.83 Na+ + e-  Na(s) E°red = -2.71
Electrolysis of Aqueous Solutions
Sodium cannot be prepared by electrolysis of aqueous solutions of NaCl, because water
is more easily reduced than Na+:
Possible Cathode reactions
(Note: Not in table on AP exam)
2H2O + 2e-  H2 + 2OH- E°red = -0.83
Na+ + e-  Na(s) E°red = -2.71
The possible Anode reactions are:
2Cl-  Cl E°ox = -1.36
2H2O  4H+ + O2 + 4e E°ox = -1.23
Therefore
2Cl-(aq)  Cl2(g) E°ox =
2H2O + 2e-  H2 + 2OH- E°red =
E°cell >

31. Electrolysis of With Active Electrodes
Possibilities
Ni(s)  Ni2+(aq) + 2e E°ox = 0.28
2H2O  4H+ + O2 + 4e- E°ox = -1.23
Ni(s)  Ni2+(aq) + 2e-
anode:
cathode:
Ni2+(aq) + 2e-  Ni(s)
Electroplating creates a silver lining

32. = 3.36 g of Al Quantitative Aspects of Electrolysis
Calculate the mass of aluminum produced in 1.00 hr. by the electrolysis of molten AlCl3 if the current is 10.0 A. (C = amperes x seconds)
(10.0A)
(1.00 hr)
(3600 sec)
(1C)
(1)
(1molAl)
(27.0 g Al)
(1 hr)
(1 A-s)
(96,500C)
(3)
(1 mol Al)
= 3.36 g of Al

33. 30.2g Mg Quantitative Aspects of Electrolysis
The half-reaction for the formation of magnesium metal upon electrolysis of molten MgCl2 is
Mg2+ + 2e-  Mg. Calculate the mass of magnesium formed upon passage of 60.0 A for a period of 4000 s
(60.0A)(4000s)(1C)(1) (mole Mg)(24.3g)
(A•s)(96,500C) (2) (mole Mg)
30.2g Mg

34. Electrical Work since G= wmax and G = - nE then wmax= - nE
The unit employed by electric
utilities is kilowatt-hour
(kWh =3.6 x 106J)

35. Corrosion reactions are redox reactions in which a metal is attacked by some substance in its environment and converted to an unwanted compound.
All metals except gold and platinum are thermodynamically
capable of undergoing oxidation in air at room temperature

36. The rusting of iron is known to require oxygen; iron does not rust in water unless O2
is present.
E°red = 1.23 V
E°red = 0.44 V
Note that as the pH increases, the reduction of O2 becomes less favorable

37. The corrosion of Iron:Protecting the surface with tin
Tin has lower E°ox so less likely to oxidize until the
surface is broken then it accelerates as it makes a
voltaic cell with iron.
Fe(s)  Fe2+ + 2e- E°ox = 0.44 V
Sn(s)  Sn2+ + 2e- E°ox = 0.14 V

38. Galvanization The corrosion of Iron:Cathodic Protection
Fe(s)  Fe2+(aq) + 2e- E°ox = 0.44 V
Zn(s)  Zn2+(aq) + 2e- E°ox = 0.76V
Zinc is more
positive and
goes first and has
an oxide coat
that seals.
sacrificial
anode
Galvanization

39. The corrosion of Iron:Cathodic Protection