CHAPTER 17
ELECTROCHEMISTRY
Oxidation and Reduction (Redox)
Electrons are transferred
Spontaneous redox rxns can transfer
energy
Electrons (electricity)
Heat
Non-spontaneous redox rxns can be
made to happen with electricity
Oxidation and Reduction
An old memory device for oxidation
and reduction goes like this…
LEO says GER
Lose Electrons = Oxidation
Gain Electrons = Reduction
Oxidation and Reduction
Another memory device (my favorite!)
for oxidation and reduction goes like
this…
“OIL RIG”
Oxidation Is Loss
Reduction Is Gain
Steps for acidic solution
1.
Write two separate equations for oxidation
and reduction
–
–
include any compound containing the element
involved
does not need to include everything
2
3
2
MnO (aq)  Fe (aq)  Fe (aq)  Mn (aq)
4
2
reduction : MnO (aq)  Mn (aq)
2
3
oxidation : Fe (aq)  Fe (aq)
4
Steps for acidic solution
2. For each half-reaction:
– balance all elements but H and O
– balance O using H2O
– balance H using H+
– balance charge using electrons (e-)


4
2
5e  8H (aq)  MnO (aq)  Mn (aq)  4H 2O(l)
-
2
3
Fe (aq)  Fe (aq)  1e

Steps for acidic solution
• If needed, multiply one or both half
reactions by an integer so that number
of electrons is equal in both half
reactions
2
3

5(Fe (aq)  Fe (aq)  1e )
2
3
5Fe (aq)  5Fe (aq)  5e

Steps for acidic solution
4. Add the half-reactions together and
simplify
5. Check to be sure all is balanced.

2
5e  8H (aq)  MnO (aq)  5Fe (aq) 
-
4
2
3
Mn (aq)  4H 2O  5Fe (aq)  5e

2
8H (aq)  MnO (aq)  5Fe (aq) 
4
2
3
Mn (aq)  4H 2O  5Fe (aq)
-
Figure 17.1 A Method to Separate the Oxidizing
and Reducing Agents of a Redox Reaction
Electrochemistry
Electrochemistry Terminology #1
Oxidation – A process in which an
element attains a more positive
oxidation state
Na(s)  Na+ + eReduction – A process in which an
element attains a more negative
oxidation state
Cl2 + 2e-  2Cl-
Electrochemistry Terminology #3
 Oxidizing agent
The substance that is reduced is the
oxidizing agent
 Reducing agent
The substance that is oxidized is the
reducing agent
Electrochemistry Terminology #4
 Anode
The electrode where
oxidation occurs
 Cathode
The electrode where
reduction occurs
Memory device:
Reduction
at the
Cathode
Figure 17.3 a-b The Electrode Compartment in
Which Oxidation occurs is called the Anode; the
Electrode Compartment in which Reduction Occurs is
Called the Cathode
Cell Potential (Ecell) = electromotive force (emf)
Figure 17.4 Digital Voltmeters Draw
only a Negligible Current and are
Convenient to Use
17–15
Figure 17.5 a-b Reaction in a
Galvanic Cell
Copyright © Houghton
Mifflin Company. All
17–16
Figure 17.6 A Galvanic Cell
involving the Half-Reactions
17–17
Table of
Reduction
Potentials
Measured
against
the
Standard
Hydrogen
Electrode
Figure 17.7 Schematic of
Galvanic Cell Involving the HalfReactions
Copyright © Houghton
Mifflin Company. All
17–19
Figure 17.8 Schematic of
Galvanic Cell Based on HalfReactions
17–20
Measuring
Standard
Electrode
Potential
Potentials are measured against a hydrogen ion
reduction reaction, which is arbitrarily
assigned a potential of zero volts.
Galvanic (Electrochemical) Cells
Spontaneous redox
processes have:
A positive
cell potential, E0
A negative free
energy change, (-G)
Zn - Cu
Galvanic
Cell
From a table
of reduction
potentials:
Zn2+ + 2e-  Zn
Cu2+ + 2e-  Cu
E = -0.76V
E = +0.34V
Zn - Cu
Galvanic
Cell
The less positive,
or more negative
reduction potential
becomes the
oxidation…
Zn  Zn2+ + 2eCu2+ + 2e-  Cu
E = +0.76V
E = +0.34V
Zn + Cu2+  Zn2+ + Cu
E0 = + 1.10 V
Line
Notation
An abbreviated
representation of
an electrochemical
cell
Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)
Anode
Anode
|
material
solution
||
Cathode
solution
|
Cathode
material
Calculating G0 for a Cell
G0 = -nFE0
n = moles of electrons in balanced redox equation
F = Faraday constant = 96,485 coulombs/mol e-
Zn + Cu2+  Zn2+ + Cu
E0 = + 1.10 V
coulombs
Joules
G   (2 mol e )(96 485
)(1.10
)

mol e
Coulomb
0

G   212267 Joules   212 kJ
0
???
Concentration
Cell
Both sides have
the same
components but
at different
concentrations.
Step 1: Determine which side undergoes
oxidation, and which side undergoes reduction.
???
Anode
Concentration
Cell
Cathode
Both sides have
the same
components but
at different
concentrations.
The 1.0 M Zn2+ must decrease in concentration, and
the 0.10 M Zn2+ must increase in concentration
Zn2+ (1.0M) + 2e-  Zn
(reduction)
Zn  Zn2+ (0.10M) + 2eZn2+ (1.0M)  Zn2+ (0.10M)
(oxidation)
???
Concentration Cell
Anode
Cathode
Concentration
Cell
Both sides have
the same
components but
at different
concentrations.
Step 2: Calculate cell potential using the Nernst
Equation (assuming 25 C).
Zn2+ (1.0M)  Zn2+ (0.10M)
0.0591
EE 
log(Q)
n
0
Nernst Calculations
Zn2+ (1.0M)  Zn2+ (0.10M)
0.0591
EE 
log(Q)
n
0
E  0.0 Volts
0
n2
(0.10)
Q
(1.0)
0.0591
0.10
E  0.0 
log(
)  0.030 Volts
2
1.0
Electrolytic
Processes
Electrolytic
processes are
NOT spontaneous.
They have:
A negative
cell potential, (-E0)
A positive free
energy change, (+G)
Electrolysis of
Water
In acidic solution
Anode rxn:
2 H 2O  O2  4 H   4e 


4
H
O

4
e

2
H

4
OH
Cathode rxn:
2
2
2H 2O  2H 2  O2
-1.23 V
-0.83 V
-2.06 V
Electroplating of
Silver
Anode reaction:
Ag  Ag+ + eCathode reaction:
Ag+ + e-  Ag
Electroplating requirements:
1. Solution of the plating metal
2. Anode made of the plating metal
3. Cathode with the object to be plated
4. Source of current
Solving an Electroplating Problem
Q: How many seconds will it take to plate out
5.0 grams of silver from a solution of AgNO3
using a 20.0 Ampere current?
Ag+ + e-  Ag
5.0 g
1 mol Ag 1 mol e-
96 485 C 1 s
20.0 C
1
mol
e
107.87 g 1 mol Ag
= 2.2 x 102 s
Figure 17.9 Concentration Cell
17–35
Figure 17.10 Concentration Cell
Copyright © Houghton
Mifflin Company. All
17–36
Figure 17.11 Schematic
Diagram of the Cell Described
in Sample Exercise 17.7
Copyright © Houghton
Mifflin Company. All
17–37