Two Types of Chemical Rxns
1. Exchange of Ions – no change in
charge/oxidation numbers
Oxidation/Reduction
Chapter 20
Two Types of Chemical Rxns
– Precipitation Rxns
– Acid/Base Rxns
NaOH + HCl
Two Types of Chemical Rxns
2. Exchange of Electrons – changes in oxidation
numbers/charges
Pb(NO3)2(aq) + KI(aq)
– Dissolving Rxns
CaCl2(s) 
Review of Oxidation Numbers
Oxidation numbers – the charge on an ion or an
assigned charge on an atom.
Al Cl2 P4
Mg2+ Cl-
Fe(s) + CuSO4(aq)  FeSO4(aq) + Cu(s)
Remove spectator ions
Fe + Cu2+  Fe2+ + Cu
Protons
Electrons
Review of Oxidation Numbers
Calculate the oxidation numbers for:
HClO
S8
Mn2O3
KMnO4
HSO4-
Cr3+
Fe2(SO4)3
SO32NO3-
1
Oxidation
1. Classical Definition –addition of oxygen
2. Modern Definition – an increase in oxidation
number
Oxidation
Fe + O2 
(limited oxygen)
Fe + O2 
(excess oxygen)
Fe + O2  Fe2O3
CO + O2  CO2
CH3CH2OH  CH3CHO  CH3COOH
Oxidation
C + O2 
C + O2 
(limited oxygen)
(excess oxygen)
Oxidizing/Reducing Agents
Reduction
1. Classical –addition of hydrogen
2. Modern –decrease (reduction) in oxidation
number
N2 + 3H2  2NH3
(Haber process)
R-C=C-R
+ H2 
| |
H H
(unsaturated fat)
(saturated fat)
Oxidizing/Reducing Agents
Oxidation and Reduction always occur together.
Oxidizing Agents
– Get reduced
– Gain electrons
0 Got oxidized, reducing agent
Cu
+
O2

0
+2
CuO
-2
Got reduced, oxidizing agent
Reducing Agents
– Get oxidized
– Lose electrons
2
Identify the Oxidizing/Reducing Agents in the
following (Calculate the ox. numbers also).
Cu + S8  Cu2S
H2 + O2  H2O
Cu + AgNO3  Cu(NO3)2 + Ag
H2O + Al + MnO4-  Al(OH)4- + MnO2
Balancing Redox Reactions
Half-Reaction Method
• Break eqn into oxidation half and reduction
half
• Easy Examples:
Balancing Redox Reactions
What’s really happening:
Cu2+ + Zn  Cu + Zn2+
– Al + Fe2+  Fe + Al3+
– Cu + Zn2+  Cu2+ + Zn
– Mg + Na+  Mg2+ + Na
Balancing Redox Reactions
Steps for more complicated examples
1. Balance all atoms except H and O
2. Balance charge with electrons
3. Balance O with water
4. Balance H with H+
===================================
5. (Add OH- to make water in basic solutions)
3
Balancing Redox Reactions
Example 1:
MnO4- + C2O42-  Mn2+ + CO2
1. Separate into half reactions
MnO4-  Mn2+
C2O42-  CO2
Balancing Redox Reactions
Balancing Redox Reactions
MnO4-  Mn2+
+7
+2
C2O42-  CO2
C2O42-  2CO2
+3
+4 (1 e- per carbon)
5e- + MnO4-  Mn2+
5e- + MnO4-  Mn2+ + 4H2O
C2O42-  2CO2 + 2e-
8H+ + 5e- + MnO4-  Mn2+ + 4H2O
Balancing Redox Reactions
8H+
+
5e-
C2O42-  2CO2 + 2e+ MnO4-  Mn2+ + 4H2O
(X 5)
(X 2)
5C2O42-  10CO2 + 10e16H+ + 10e- + 2MnO4-  2Mn2+ + 8H2O
16H++5C2O42-+2MnO4-  2Mn2+ + 10CO2 + 8H2O
Balancing Redox Reactions (Acidic
Solutions)
Cr2O72- + Cl-  Cr3+ + Cl2
14 H+ + Cr2O72- + 6Cl-  2Cr3+ + 3Cl2 + 7H2O
Cu +
14H+ +
Cu + NO3-  Cu2+ + NO2
+ 2NO3-  Cu2+ + 2NO2 + 2H2O
4H+
Mn2+ + BiO3-  Bi3+ + MnO42Mn2+ + 5BiO3-  5Bi3+ + 2MnO4- + 7H2O
4
Balancing Redox Reactions (Basic
Solutions)
Just add enough OH- to each side to make all of the
H+ to water.
NO2- + Al  NH3 + Al(OH)45H2O + OH- + NO2- + 2Al  NH3 + 2Al(OH)4Cr(OH)3 + ClO-  CrO42- + Cl2
2Cr(OH)3 + 6ClO-  2CrO42- + 3Cl2 + 2OH- + 2H2O
Balance in Both Acidic and Basic Solutions
F- + MnO4-  MnO2 + F2
HNO2 + H2O2  O2 + NO
H is +1
Voltaic (Galvanic) Cells
F- + MnO4-  MnO2 + F2
+
8H + 6F + 2MnO4-  2MnO2 + 3F2 + 4H2O
4H2O + 6F- + 2MnO4-  2MnO2 + 3F2 + 8OHHNO2 + H2O2  O2 + NO
2HNO2 + H2O2  O2 + 2NO + 2H2O
Voltaic (Galvanic) Cells
History
• Galvani (died 1798)– uses static electricity to
move the muscles of dead frogs
• Volta (1800) – Created the first battery
• Voltaic(Galvanic) Cells – redox reactions that
produce a voltage
– Spontaneous reactions ( G<0)
– Voltage of the cell (Eocell) is positive
– Batteries
• Electrolytic cells – redox reactions that must
have a current run through them.
– G>0 and Eocell is negative.
– Often used to plate metals
Voltaic (Galvanic) Cells
Voltaic cell
1. Anode – Oxidation site
2. Cathode – Reduction site (RC cola)
3. Salt bridge – completes the circuit
5
Voltaic (Galvanic) Cells
Voltaic (Galvanic) Cells
• Cell Notation
Zn | Zn2+(aq) ||Cu2+(aq) | Cu
• Anode Zn
 Zn2+ + 2e2+
• Cathode Cu + 2e  Cu
• Cell
Cu2+ + Zn  Cu + Zn2+
Hydrogen Electrode
1. Standard Electrode
2. Voltage(potential) = 0 Volts
2H+(aq) + 2e-  H2(g)
H2(g)  2H+(aq) + 2e-
0 volts
0 volts
3. Often used in electrodes (like pH)
Standard Reduction Potentials
• Always written as a reduction
• If cell is positive, produces a voltage
• Rules
– Flipping an equation changes the sign of E
– Multiplying an equation does not change the
magnitude of E
Calculating Cell Potential
A cell is composed of copper metal and Cu2+(aq)
on one side, and zinc metal and Zn2+(aq) on
the other. Calculate the cell potential.
Zn2+ + 2e-  Zn
-0.76 V
Cu2+ + 2e-  Cu
+0.34 V
flip the zinc equation
Zn
 Zn2+ + 2e+0.76 V
2+
Cu + 2e  Cu
+0.34 V
Zn + Cu2+  Zn2+ + Cu
+1.10 V
6
Example 1
Example 2
What is the cell emf of a cell made using Cu and
Cu2+ in one side and Al and Al3+ in the other?
Write the complete cell reaction.
Calculate the standard emf for the following
reaction. Hint: break into half-reactions.
2Al(s) + 3I2(s)  2Al3+(aq) + 6I-(aq)
ANS: 2.2 V
Example 3
Example 4
A voltaic cell is based on the following half
reactions.
In+(aq)  In3+(aq) + 2eBr2(l) + 2e-  2Br-(aq)
Calculate the standard emf for the following
reaction.
Cr2O72- + 14H+ + 6I-  2Cr3+ + 3I2 + 7H2O
+1.06 V
If the overall cell voltage is 1.46 V, what is the
reduction potential for In3+?
Two half reactions in a voltaic cell are:
Zn2+(aq) + 2e-  Zn(s)
Li+(aq) + e-  Li(s)
a) Calculate the cell emf.
b) Which is the anode? Which is the cathode?
c) Which electrode is consumed?
d) Which electrode is positive?
e) Sketch the cell, indicating electron flow.
Given the following half-reactions:
Pb2+ + 2e- Pb
Ni2+ + 2e-  Ni
a. Calculate the cell potential (Eo).
b.
c.
d.
e.
f.
Label the cathode and anode.
Identify the oxidizing and reducing agents.
Which electrode is consumed?
Which electrode is plated?
Sketch the cell, indicating the direction of
electron flow.
7
Strengths of Oxidizing and
Reducing Agents
• larger the reduction potential, stronger the
oxidizing agent
Stronger
– Wants to be reduced, can oxidize something else.
• lower the reduction potential, stronger the
reducing agent
– Would rather be oxidized
Oxidizing
Agents
Example 1
Cr2O72-
 2F 2Cl-
+2.87 V
+1.36 V
.
.
.
.
.
Li+ + e-
 Li
-3.05V
Example 2
Which of the following is the strongest oxidzing
agent? Which is the strongest reducing
agent?
NO3-
F2 + 2eCl2 + 2e-
Ag+
Which of the following is the strongest reducing
agent? Which is the strongest oxidizing
agent?
I2(s)
Example 3
Can copper metal (Cu(s)) act as an oxidizing
agent?
Fe(s)
Mn(s)
Spontaneity
Voltaic Cells
•Positive emf
•Spontaneous
•Can produce
electric current
•Batteries
Electrolytic Cells
•Negative emf
•Not spontaneous
•Must “pump”
electricity in
•Electrolysis
8
Example 1
Are the following cells spontaneous as written?
a) Cu + 2H+  Cu2+ + H2
b) Cl2 + 2I-  2Cl- + I2
c) I2 + 5Cu2+ + 6H2O  2IO3- + 5Cu +
12H+
d) Hg2+ + 2I-  Hg + I2
Ex 1
Calculate the cell potential and free energy
change for the following reaction:
4Ag + O2 + 4H+  4Ag+ + 2H2O
ANS: +0.43 V, -170 kJ/mol
EMF and K
Go = -RTlnK
-nFEo = -RTlnK
lnK = nFEo
RT
log K = nEo
0.0592
( Go = -nFEo)
(assume 298 K)
EMF and Go
G = -nFE
n = number of electrons transferred
E = Cell emf
F = 96,500 J/V-mol (Faraday’s Constant)
Positive Voltage gives a negative G (spont)
Ex 2
Calculate G and the EMF for the following
reaction. Also, calculate the K.
3Ni2+ + 2Cr(OH)3 + 10OH-  3Ni + 2CrO42- + 8H2O
ANS: +87 kJ/mol, -0.15 V, 6 X 10-16
Example 1
Calculate G, cell voltage and the equilibrium
constant for the following cell:
O2 + 4H+ + 4Fe2+  4Fe3+ + 2H2O
ANS: -177 kJ/mol, 0.459 V, 1 X 1031
9
Example 2
Concentration Cells: Nernst Equation
If the equilibrium constant for a particular
reaction is 1.2 X 10-10, calculate the cell
potential. Assume n = 2.
G = Go + RT lnQ
-nFE = -nFEo + RT lnQ
E = Eo -
RT lnQ
(assume 298 K)
nF
E = Eo 0.0592 log Q
n
Can adjust the voltage of any cell by changing
concentrations
Using the Nernst Eqn
Suppose in the following cell, the concentration
of Cu2+ is 5.0 M and the concentration of Zn2+
is 0.050 M. Calculate the cell voltage.
Zn(s) + Cu2+(aq)Zn2+(aq) + Cu(s)
+1.10 V
E = Eo -
0.0592 V log Q
n
E = 1.10V - 0.0592 V log [Cu][Zn2+]
2
[Cu2+][Zn]
E = 1.10V - 0.0592 V log [Zn2+]
2
[Cu2+]
E = 1.10V - 0.0592 V log [0.050]
2
[5.0]
E = +1.16 V
Example 1
Example 2
Calculate the emf at 298 K generated by the
following cell (Eo= 0.79 V) where: [Cr2O72-]=
2.0 M, [H+ ]=1.0 M, [I-]=1.0 M and [Cr3+ ]= 1.0
X 10-5M.
Calculate the emf at 298 K generated by the
following cell (Eo= 2.20 V) where: [Al3+]= 0.004
M and [I- ]=0.010 M.
2-
Cr2O7
+
14H+
+
6I-

2Al(s) + 3I2(s)  2Al3+(aq) + 6I-(aq)
2Cr3+
+ 3I2 + 7H2O
ANS: +2.36 V
ANS: 0.89 V
10
Example 3
If the voltage of a Zn-H+ cell is 0.45 V at 298 K
when [Zn2+]=1.0 M and PH2=1.0 atm, what is
the concentration of H+? Note that atm can
be used just like molarity.
Zn(s) + 2H+(aq)  Zn2+(aq) + H2(g)
Example 4
What pH is required if we want a voltage of
0.542 V and [Zn2+]=0.10 M and PH2=1.0 atm?
Zn(s) + 2H+(aq)  Zn2+(aq) + H2(g)
ANS: 5.84 X 10-5M, pH = 4.22
ANS: 5.2 X 10-6M
Batteries
Alkaline Batteries
• Basic
• Zinc can acts as the anode
Lead Acid Battery
• 12 Volt DC
• Discharges when starting the
car, recharges as you drive
(generator). Running
reaction backward.
PbO2(s) + Pb(s) +2HSO4
-(aq)
+
2H+(aq)
2MnO2(s)+2H2O(l)+2e-2MnO(OH)(s) + 2OH-(aq)
Zn(s) + 2OH-(aq) Zn(OH)2(s) + 2e2PbSO4(s) +2H2O(l)
Rechargable uses Ni-Cd
Corrosion
• Iron rusts in acidic solns (not above pH=9)
• Water needs to be present
• Salts accelerate the process
O2 + 4H+ + 4e-  2H2O
Fe  Fe2+ + 2e(The Fe2+ eventually goes to Fe3+, Fe2O3)
11
Preventing Corrosion
• Paint
• Sometimes oxide layer(Al2O3)
• Galvanizing (coating Fe with Zn)
Fe2+ + 2e-  Fe E = -0.44 V
Zn2+ + 2e-  Zn E = -0.76 V
Zinc is more easily oxidized
(Zn  Zn2+ + 2eE = +0.76 V)
Example 1
An iron gutter is nailed using aluminum nails.
Will the nail or the iron gutter corrode first?
• Cathodic
protection
(sacrificial anode)
• Magnesium used
in water pipes
• Magnesium rods
used in hot water
heaters
Fe2+ + 2e-  Fe
Al3+ + 3e-  Al
Al will corrode first (Al  Al3+ + 3e- ,E = +1.66 V)
Ex 2
Electrolysis and Electroplating
Which of the following metals could provide
cathodic protection to iron:
Al
E = -0.44 V
E = -1.66 V
Cu
Ni
• Plating of silver on silverware
Zn
12
Electrolytic cells
• Must run electricity through them
• Running a voltaic cell backwards
• Used to produce sodium metal
Na+(aq) + e-  Na (s)
Cl2(g) + 2e-  2Cl-(aq)
-2.71 V
+1.36 V
• As a voltaic cell
2Na(s)
 2Na+(aq) + 2e- +2.71 V
Cl2(g) + 2e  2Cl-(aq)
+1.36 V
2Na(s) + Cl2 (s)2NaCl(aq)
+4.07 V
• As an electrolytic cell
2Na+(aq) + 2e-  2Na (s)
-2.71 V
2Cl-(aq)  Cl2(g) + 2e-1.36 V
2NaCl(aq)  2Na(s) + Cl2 (s) -4.07 V
Quantitative Electrolysis
• Electric current = Amperes
• 1 ampere = 1Coloumb
1 second
• 1 F = 96,500 C/mol
I = Q
t
– One mole of electrons has a charge of 96,500 C
– One electron has a charge of 1.602 X 10-19 C
Example 1
What mass of aluminum can be produced in
1.00 hour by a current of 10.0 A?
Al3+ + 3e-  Al
Q= I t
Q = (10.0 A)(3600 s) = 36,000 C
Example 2
Moles of e- = (36,000C)(1 mol e-) = 0.373 mol e(96,500 C)
Mg2+ + 2e-  Mg
Al3+ + 3e-  Al
0.373 mol
Al3+ + 3e- 
0.373 mol
What mass of magnesium can be produced in 4000
s by a current of 60.0 A?
Al
0.124 mol Al  3.36 g Al
ANS: 30.2 g Mg
13
Example 3
What current is required to plate 6.10 grams of
gold in 30.0 min?
How long would it take to plate 50.0 g of
magnesium from magnesium chloride if the
current is 100.0 A?
Au3+ + 3e-  Au
(6.10 grams Au)(1mol Au) = 0.0310 mol Au
(196.97g Au)
Q=It
I = Q/t (Actually I = dQ/dt)
I = 8966 C = 5.0 amps
1800 s
Au3+ + 3e- 
Au
0.0310 mol Au
Au3+ + 3e- 
0.0929 mol e
Au
0.0310 mol Au
(0.0929 mol e)(96,500 C) = 8966 C
(1 mol e)
Given the following:
Ag+(aq) + e-  Ag(s)
+0.799V
Fe3+(aq) + e-  Fe2+(aq)
+0.771 V
a. Write the reaction that occurs.
b. Calculate the standard cell potential.
c. Calculate Grxn for the reaction from the cell
potential.
d. Calculate for the reaction.
e. Predict the sign of Srxn.
f. Sketch the cell, labeling anode, cathode, and
the direction of electron flow.
Do SO3 and SO32- have the same molecular
shape? How about SO2?
14
16. a) Not redox
b) I oxidized (-1 to +5) , Cl reduced (+1 to -1)
c) S oxidized (+4 to +6), N reduced (+5 to +2)
d) Br oxidized (-1 to 0), S reduced (+6 to +4)
20 a. Mo3+ + 3e-  Mo
b. H2O + H2SO3  SO42- + 2e- + 4H+
c. 4H+ + 3e- + NO3-  NO + 2H2O
d. 4H+ + 4e- + O2  2H2O
e. 4OH- + Mn2+  MnO2 + 2e- + 2H2O
f. 5OH- + Cr(OH)3  CrO42- + 3e- + 4H2O
g. 2H2O + 4e- + O2  4OH-
26. a) Al oxidzed, Ni2+ reduced
b) Al  Al3+ + 3e- Ni2+ + 2e-  Ni
c) Al anode, Ni cathode
d) Al negative, Ni positive
e) Electrons flow towards the Ni electrode
f) Cations migrate towards Ni electrode
22. a. 3NO2- + Cr2O72- +8H+  3NO3- + 2Cr3+ +
4H2O
b. 2HNO3 + 2S +H2O  2H2SO3 + N2O
c. 2Cr2O72- + 3CH3OH + 16H+  4Cr3+ 3HCO2H +
11H2O
d. 2MnO4- + 10Cl- + 16H+  2Mn2+ + 5Cl2 + 8H2O
e. NO2- + 2Al + 2H2O  NH4+ + 2AlO2f. H2O2 + 2ClO2 + 2OH-  O2 + 2ClO2- + 2H2O
34 a) Cd is anode, Pd is cathod
b) Ered = 0.63 V
36 a) 2.87 V b) 3.21 V
c) -1.211 V d) 0.636V
38 a) 1.35 V
b) 0.29 V
41 a) Mg
b) Ca
c) H2 d) H2C2O4
2+
42 a) Cl2
b) Cd
c) BrO3- d) O3
3+
44. a) Ce (weak reductant)
b) Ca (strong reductant)
c) ClO3- (strong oxidant)
d) N2O5 ( oxidant)
46 a) H2O2 strongest oxidizing agent
b) Zn strongest reducing agent
50.a) 3.6 X 108
b) 1041
c) 10103
52. 0.292 V
54 a) 4 X 1015 b) 2 X1065 c) 7.3 X1049
62a) 2.35 V
b) 2.48 V c) 2.27 V
64. a) 0.771 V b) 1.266 V
88. a) 173 g
b) 378 min
90.E = 1.10 V Wmax = -212 kJ/mol Cu
W = -1.67 X 105 J
15
1a) 14H+ + Cr2O72- + 3Fe  2Cr3+ + 3Fe2+ + 7H2O
b) 2Br- + F2  2F- + Br2
c) 4OH- + 2Cr(OH)3 + ClO3-  2CrO42- + Cl- + 5H2O
2b) 0.463 V
c) -89.4 kJ/mol d) 4.4 X 1015
e) 0.442 V
3) F2 is str. oxidizing agent, Li, str. reducing agent
4) b) 78 minutes c) 1.19 g d) 0.695 g
In a measuring cup:
• 5 mL of oil
• 5 mL of ethanol
• 5 mL of 50% NaOH solution (approximately 30
drops).
Place in beaker
• Heat the mixture, stirring with popsicle stick.
• Remove from heat. After ~5 minutes, add 10 mL of
saturated salt solution.
• Collect some of the solid and test the pH of your
soap. Compare the pH to that of commercial bar
soap and liquid detergent solution. See if it
lathers.
16