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ELECTROCHEMISTRY
Chapter 18
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Electron Transfer Reactions
• Electron transfer reactions are oxidationreduction or redox reactions.
• Results in the generation of an electric
current (electricity) or be caused by
imposing an electric current.
• Therefore, this field of chemistry is often
called ELECTROCHEMISTRY.
Terminology for Redox
Reactions
• OXIDATION—loss of electron(s) by a species;
increase in oxidation number; increase in
oxygen.
• REDUCTION—gain of electron(s); decrease in
oxidation number; decrease in oxygen;
increase in hydrogen.
• OXIDIZING AGENT—electron acceptor;
species is reduced.
• REDUCING AGENT—electron donor; species
is oxidized.
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You can’t have one… without
the other!
• Reduction (gaining electrons) can’t happen without an
oxidation to provide the electrons.
• You can’t have 2 oxidations or 2 reductions in the same
equation. Reduction has to occur at the cost of
oxidation
LEO the lion says GER!
o l x
s e i
e c d
t a
r t
o i
n o
s n
GER!
a l e
i e d
n c u
t c
r t
o i
n o
s n
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Another way to remember
• OIL RIG
e
x s o
i
s
d s
a
t
i
o
n
s a
d
i
u
n
c
t
i
o
n
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Oxidation Reduction Reactions
(Redox)
0
0

1

1
2
Na

Cl

2
Na
Cl
2
Each sodium atom loses one electron:
0

1

Na

Na

e
Each chlorine atom gains one electron:
0


1
Cl

e
Cl
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LEO says GER :
Lose Electrons = Oxidation
0

1

Na

Na

e Sodium is oxidized
Gain Electrons = Reduction
0


1
Cl

e
Cl
Chlorine is reduced
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Not All Reactions are Redox Reactions
Reactions in which there has been no
change in oxidation number are not redox
rxns.
Examples:

1

5

2

1

1

1

11


5

2
A
g
N
O
(
a
q
)(

N
a
C
l
a
q
)

A
g
C
l
(
s
)

N
a
N
O
(
a
q
)
3
3

1

2

1
1

6

2

1

6

2
1

2
2
N
a
O
H
(
a
q
)

H
S
O
(
a
q
)


N
a
S
O
(
a
q
)
H
O
(
l
)
2
2
4
2
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Rules for Assigning Oxidation Numbers
Rules 1 & 2
1. The oxidation number of any uncombined
element is zero
2. The oxidation number of a monatomic ion
equals its charge
0
0

1

1
2
Na

Cl

2
Na
Cl
2
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Rules for Assigning Oxidation Numbers
Rules 3 & 4
3. The oxidation number of oxygen in
compounds is -2
4. The oxidation number of hydrogen in
compounds is +1
1
2
H2 O
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Rules for Assigning Oxidation
Number Rule 5
5. The sum of the oxidation numbers in
the formula of a compound is 0
1
2
H2 O
2(+1) + (-2) = 0
H
O

2 
2
1
Ca
(O
H
)2
(+2) + 2(-2) + 2(+1) = 0
Ca
O
H
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Rules for Assigning Oxidation Numbers
Rule 6
6. The sum of the oxidation numbers in the
formula of a polyatomic ion is equal to
its charge
? 2
? 2
NO
SO

3
X + 3(-2) = -1
N
O
 X = +5
2
4
X + 4(-2) = -2
S
O
 X = +6
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PRACTICE THIS AT HOME
See Set 1
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OXIDATION-REDUCTION
REACTIONS
Direct Redox Reaction
Oxidizing and
reducing agents in
direct contact.
Cu(s) + 2 Ag+(aq) --->
Cu2+(aq) + 2 Ag(s)
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OXIDATION-REDUCTION
REACTIONS
Indirect Redox Reaction
A battery functions by transferring electrons
through an external wire from the reducing
agent to the oxidizing agent.
Why Study Electrochemistry?
• Batteries
• Corrosion
• Industrial
production of
chemicals such as
Cl2, NaOH, F2 and
Al
• Biological redox
reactions
The heme group
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Electrochemical Cells
• An apparatus that allows
a redox reaction to occur
by transferring electrons
through an external
connector.
• Product favored reaction
---> voltaic or galvanic cell
----> electric current
• Reactant favored reaction
---> electrolytic cell --->
electric current used to
cause chemical change.
Batteries are voltaic
cells
Basic Concepts
of Electrochemical Cells
Anode
Cathode
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CHEMICAL CHANGE --->
ELECTRIC CURRENT
With time, Cu plates out
onto Zn metal strip, and
Zn strip “disappears.”
•Zn is oxidized and is the reducing agent
Zn(s) ---> Zn2+(aq) + 2e•Cu2+ is reduced and is the oxidizing agent
Cu2+(aq) + 2e- ---> Cu(s)
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CHEMICAL CHANGE --->
ELECTRIC CURRENT
•To obtain a useful
current, we separate the
oxidizing and reducing
agents so that electron
transfer occurs thru an
external wire.
This is accomplished in a GALVANIC or
VOLTAIC cell.
A group of such cells is called a battery.
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Discover:electrochemistry movie
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Zn --> Zn2+ + 2e-
Cu2+ + 2e- --> Cu
Oxidation
Anode
Negative
Reduction
Cathode
Positive
<--Anions
Cations-->
RED CAT
•Electrons travel thru external wire.
•Salt bridge allows anions and cations to move
between electrode compartments.
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Terms Used for Voltaic Cells
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CELL POTENTIAL, E
• For Zn/Cu cell, potential is +1.10 V at 25 ˚C
and when [Zn2+] and [Cu2+] = 1.0 M.
• This is the STANDARD CELL
POTENTIAL, Eo
• —a quantitative measure of the tendency of
reactants to proceed to products when all
are in their standard states at 25 ˚C.
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Calculating Cell Voltage
• Balanced half-reactions can be added
together to get overall, balanced
equation.
Zn(s) ---> Zn2+(aq) + 2eCu2+(aq) + 2e- ---> Cu(s)
-------------------------------------------Cu2+(aq) + Zn(s) ---> Zn2+(aq) + Cu(s)
If we know Eo for each half-reaction, we
could get Eo for net reaction.
TABLE OF STANDARD
REDUCTION POTENTIALS
oxidizing
ability of ion
Eo (V)
Cu2+ + 2e-
Cu
+0.34
2 H+ + 2e-
H2
0.00
Zn2+ + 2e-
Zn
-0.76
To determine an oxidation
from a reduction table, just
take the opposite sign of the
reduction!
reducing ability
of element
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Zn/Cu Electrochemical Cell
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+
Anode,
negative,
source of
electrons
Cathode,
positive,
sink for
electrons
Zn(s) ---> Zn2+(aq) + 2eEo = +0.76 V
Cu2+(aq) + 2e- ---> Cu(s)
Eo = +0.34 V
--------------------------------------------------------------Cu2+(aq) + Zn(s) ---> Zn2+(aq) + Cu(s)
Eo = +1.10 V
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Eo
for a Voltaic Cell
Cd --> Cd2+ + 2eor
Cd2+ + 2e- --> Cd
Fe --> Fe2+ + 2eor
Fe2+ + 2e- --> Fe
All ingredients are present. Which way does
reaction proceed?
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Eo for a Voltaic Cell
From the table, you see
• Fe is a better reducing
agent than Cd
• Cd2+ is a better
oxidizing agent than
Fe2+
Overall reaction
Fe + Cd2+ ---> Cd + Fe2+
Eo = E˚cathode + E˚anode
= (-0.40 V) + - (-0.44 V)
= +0.04 V
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More About
Calculating Cell Voltage
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Assume I- ion can reduce water.
2 H2O + 2e- ---> H2 + 2 OHCathode
2 I- ---> I2 + 2eAnode
------------------------------------------------2 I- + 2 H2O --> I2 + 2 OH- + H2
Assuming reaction occurs as written,
E˚ = E˚cat+ E˚an= (-0.828 V) + (- 0.535 V) = -1.363 V
Minus E˚ means rxn. occurs in opposite direction
(the connection is backwards or you are
recharging the battery)
Charging a Battery
When you charge a battery, you are
forcing the electrons backwards (from
the + to the -). To do this, you will
need a higher voltage backwards than
forwards. This is why the ammeter in
your car often goes slightly higher
while your battery is charging, and
then returns to normal.
In your car, the battery charger is
called an alternator. If you have a
dead battery, it could be the
battery needs to be replaced OR
the alternator is not charging the
battery properly.
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Dry Cell Battery
Anode (-)
Zn ---> Zn2+ + 2eCathode (+)
2 NH4+ + 2e- --->
2 NH3 + H2
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Alkaline Battery
Nearly same reactions as
in common dry cell, but
under basic conditions.
Anode (-): Zn + 2 OH- ---> ZnO + H2O + 2eCathode (+): 2 MnO2 + H2O + 2e- --->
Mn2O3 + 2 OH-
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Mercury Battery
Anode:
Zn is reducing agent under basic conditions
Cathode:
HgO + H2O + 2e- ---> Hg + 2 OH-
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Lead Storage Battery
Anode (-) Eo = +0.36 V
Pb + HSO4- ---> PbSO4 + H+ + 2eCathode (+) Eo = +1.68 V
PbO2 + HSO4- + 3 H+ + 2e---> PbSO4 + 2 H2O
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Ni-Cad Battery
Anode (-)
Cd + 2 OH- ---> Cd(OH)2 + 2eCathode (+)
NiO(OH) + H2O + e- ---> Ni(OH)2 + OH-
H2 as a Fuel
Cars can use electricity generated by H2/O2
fuel cells.
H2 carried in tanks or generated from
hydrocarbons
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Line Notation For Voltaic Cells
• Voltaic cells can be described by a line notation based
on the following conventions.
• A single vertical line indicates a change in state or
phase. Within a half-cell, the reactants are listed before
the products.
• A double vertical line is used to indicate the junction
between the half-cells.
• The line notation for the anode (oxidation) is written
before the line notation for the cathode (reduction).
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• The line notation for a standard-state Daniell
cell is written as follows.
• Zn|Zn2+||Cu2+|Cu
• anode(oxidation)
cathode(reduction)
• Electrons flow from the anode to the cathode
in a voltaic cell. (They flow from the electrode
at which they are given off to the electrode at
which they are consumed.)
• Reading from left to right, this line notation
therefore corresponds to the direction in
which electrons move
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Write the line notation for the
following cell
Answer
The cell is based on the following half-reactions.
Oxidation: Zn  Zn2+ + 2 e• Reduction :2 H+ + 2 e-  H2
• Since we list the reactants before the products in a
half-cell, we write the following line notation for the
anode.anode: Zn/Zn2+
• The line notation for the cathode has to indicate that
H+ ions are reduced to H2 gas on a platinum metal
surface. This can be done as follows.cathode:
H+|H2|Pt
• Since line notation is read in the direction in which
electrons flow, we place the notation for the anode
before the cathode, as follows.
• Zn/Zn2+||H+/H |Pt
2
• Anode cathode
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Balancing Equations
for Redox Reactions
Some redox reactions have equations that must be balanced by
special techniques.
MnO4- + 5 Fe2+ + 8 H+
---> Mn2+ + 5 Fe3+ + 4 H2O
Mn = +7
Fe = +2
Mn = +2 Fe = +3
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Balancing Equations
Consider the
reduction of Ag+
ions with copper
metal.
Cu + Ag+
--give--> Cu2+ + Ag
Balancing Equations
Step 1:
Divide the reaction into half-reactions, one
for oxidation and the other for reduction.
Ox
Cu ---> Cu2+
Red
Ag+ ---> Ag
Step 2:
Balance each element for mass. Already
done in this case.
Step 3:
Balance each half-reaction for charge by
adding electrons.
Ox
Cu ---> Cu2+ + 2eRed
Ag+ + e- ---> Ag
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Balancing Equations
Step 4:
Multiply each half-reaction by a factor so
that the reducing agent supplies as many electrons
as the oxidizing agent requires.
Reducing agent
Cu ---> Cu2+ + 2eOxidizing agent
2 Ag+ + 2 e- ---> 2 Ag
Step 5:
Add half-reactions to give the overall
equation.
Cu + 2 Ag+ ---> Cu2+ + 2Ag
The equation is now balanced for both
charge and mass.
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Balancing Equations
Balance the following in acid solution—
VO2+ + Zn ---> VO2+ + Zn2+
Step 1:
Write the half-reactions
Ox
Zn ---> Zn2+
Red
VO2+ ---> VO2+
Step 2:
Balance each half-reaction for
mass.
Ox
Zn ---> Zn2+
Red
2 H+ + VO2+ ---> VO2+ + H2O
Add H2O on O-deficient side and add H+
on other side for H-balance.
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Balancing Equations
Step 3:
Balance half-reactions for charge.
Ox
Zn ---> Zn2+ + 2eRed
e- + 2 H+ + VO2+ ---> VO2+ + H2O
Step 4:
Multiply by an appropriate factor.
Ox
Zn ---> Zn2+ + 2eRed
2e- + 4 H+ + 2 VO2+
---> 2 VO2+ + 2 H2O
Step 5:
Add balanced half-reactions
Zn + 4 H+ + 2 VO2+
---> Zn2+ + 2 VO2+ + 2 H2O
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Tips on Balancing Equations
• Never add O2, O atoms, or
O2- to balance oxygen.
• Never add H2 or H atoms to
balance hydrogen.
• Be sure to write the correct
charges on all the ions.
• Check your work at the end
to make sure mass and
charge are balanced.
• PRACTICE!
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Practice
• 2. Balance the following reactions. Label the
oxidation and reduction reactions. What
element is being oxidized? Reduced? What is
the reducing agent? Oxidizing agent?
Co3+ (aq)+Ni (s)  Co2+(aq) +Ni2+(aq)
(acidic environment)
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Balance in Acid
•
•
•
•
•
Fe2O3+ C  Fe + CO2
What is Oxidized?
What is reduced?
What is the oxidizing agent, Reducing Agent?
Balance in Acid
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Balancing in base
• Solve as if the reaction is in ACID.
• After getting the balanced half reactionfor
acid.
• Where there are H+ neutralize them with the
same number of OH-, this creates water
• Make sure you add the same number of OHto each side.
• Cancelling waters keeping the OH• Final check of mass and charge.
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MnO4- + CN- --> MnO2 + CNO- (in base)
• Half reactions:
• MnO4- --> MnO2
• CN- --> CNOBalance each half reaction as if in Acid and add
to get net reaction
2H+ +3CN- +2MN04-  3CNO-+2MnO2+ 1H2O
Now replace the H+ with the same number of
OH- on each side.
Cancel the water and re add showing the OHCheck Mass and charge.
Balancing Redox equations using the Oxidation number method
(Basic solution is demonstrated)
Al(s) + OH-(aq)H2O
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Al(OH)-4(aq) + H2(g)
1. What are the reduction and oxidation pairs?
a) Al(s) and Al(OH)-4(aq) (oxidized)
b) ? and H2(g) (reduced)
Hint: 1. The reaction is taking place in a basic, aqueous media.
2. Look for a reduction potential for H2(g) in a table.
2. Calculate the Oxidation numbers
-3e-
0
2H2O
+
Al(s) +
OH-(aq)
and transfer to the redox partner
-3
H2O
2 Al(OH)-4(aq) + 3 H2(g)
2 x -1e-
2 x +1
2x0
3. Mass Balance
4. Charge Balance
6 H2O + 2 Al(s) +
H2(g)
OH-(aq)
H2O
2 Al(OH)-4(aq) + 3
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Change in Oxidation Procedure:
a. Write out as much of the unbalanced reaction as
possible
b. Assign oxidation numbers
c. Draw brackets to connect the atoms that are oxidized
and the atoms that are reduced. Write the net increases
and decreases in electrons.
d. Find the factors that create the least common multiple
and use these to assign balanced stoichiometry for the
reaction.
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http://www.mpcfaculty.net/mark_
bishop/redox_balance_oxidation
.htm