George Mason University
General Chemistry 212
Chapter 21
Electrochemistry
Acknowledgements
Course Text: Chemistry: the Molecular Nature of Matter and
Change, 7th edition, 2011, McGraw-Hill
Martin S. Silberberg & Patricia Amateis
The Chemistry 211/212 General Chemistry courses taught at George
Mason are intended for those students enrolled in a science
/engineering oriented curricula, with particular emphasis on
chemistry, biochemistry, and biology The material on these slides is
taken primarily from the course text but the instructor has modified,
condensed, or otherwise reorganized selected material.
Additional material from other sources may also be included.
Interpretation of course material to clarify concepts and solutions to
problems is the sole responsibility of this instructor.
2/18/2015
1
Electrochemistry


Redox Reactions and Electrochemical Cells

Review of Oxidation Reduction Concepts

Half-Reaction Method for Balancing Redox
Reactions

Electrochemical Cells
Voltaic Cell: Using Spontaneous Reactions to
Generate Electrical Energy

Construction and Operation

Cell Notation

Why Does the Cell Work
2/18/2015
2
Electrochemistry
Cell Potential: Output of a Voltaic Cell
 Standard Cell Potentials
 Strengths of Oxidizing and Reducing Agents
 Free Energy and Electrical Work
 Standard Cell Potential
 Effect of Concentration of Ecell
 Changes in Ecell During Cell Operation
 Concentration Cells
 Electrochemical Processes in Batteries
 Primary (Nonrechargeable Batteries)
 Fuel Cells

2/18/2015
3
Electrochemistry


Corrosion: A case of Environmental
Electrochemistry

Corrosion of Iron

Protecting Against Corrosion
Electrolytic Cells: Using electrical energy to drive
Nonspontaneous Reactions

Construction and Operation

Predicting Electrolysis Products

Stoichiometry of Electrolytes
2/18/2015
4
Electrochemistry

Electrochemistry


The study of the relationship between chemical
change (reactions) and the flow of electrons
(electrical work)
Electrochemical Systems

Electrolytic – Work done by absorbing free
energy from a source (passage of an electrical
current through a solution) to drive a
nonspontaneous reaction

Voltaic/Galvanic – Release of free energy from
a spontaneous reaction to produce electricity
(Batteries)
2/18/2015
5
Electrochemistry

Oxidation-Reduction Concepts Review

Oxidation – Loss of Electrons

Reduction – Gain of Electrons

Oxidizing Agent – Species that causes another species
to be oxidized (lose electrons)
Oxidizing agent is reduced (gains e-)

Reducing Agent – Species that cause another species to
be reduced (gain electrons)
Reducing agent is oxidized (loses e-)


2/18/2015
Oxidation (e- loss) always accompanies
Reduction (e- gain)
Total number of electrons gained by the atoms/ions of
the oxidizing agent always equals the total number of
electrons lost by the reducing agent
6
Electrochemistry
2/18/2015
7
Electrochemistry

Oxidation Number

A number equal to the magnitude of the charge an
atom would have if its shared electrons were held
completely by the atom that attracts them more
strongly

The oxidation number in a binary ionic compound
equals the ionic charge

The oxidation number for each element in a covalent
compound (or polyatomic ion) are assigned according
to the relative attraction of an atom for electrons

See next slide for a summary of the rules for assigning
oxidation numbers
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8
Electrochemistry
Rules for Assigning an oxidation Number (O.N.)
2/18/2015
9
Electrochemistry


Balancing Redox Reactions

Oxidation Number Method

Half-Reaction Method
The balancing process must insure that:
The number of electrons lost by the
reducing agent equals the number of
electrons gained by the oxidizing agent
2/18/2015
10
Electrochemistry

Oxidation Number Method

Assign oxidation numbers to all elements in the
reaction

From changes in oxidation number of given
elements, identify oxidized and reduced species

For each element that undergoes a change of
oxidation number, compute the number of
electrons lost in the oxidation and gained in the
reduction from the oxidation number change (Draw
tie-lines between these atoms)

Multiply one or both these number by appropriate
factors to make the electrons lost equal to the
electrons gained

Use factors as coefficients in reaction equation
2/18/2015
11
Practice Problem
Balance equation with Oxidation Number method:
Cu(s) + HNO3 (aq)  Cu(NO3 )2 + NO2 (g) + H 2O(l)
0
+5
+1 -2
+5
+2
-2
-2
+4
+1 -2
Cu(s) + HNO3 (aq)  Cu(NO3 )2 + NO2 (g) + H 2O(l)
Loses 2e-
Cu(s) + HNO3 (aq)  Cu(NO3 )2 + NO2 (g) + H 2O(l)
Gains 1e-
Balance the Equation
Accounts for the two electrons
needed to balance the 2 electrons
from the Copper oxidation
Cu(s) + 2HNO3 (aq)  Cu(NO3 )2 + 2NO2 (g) + H 2O(l)
Cu(s) + 4HNO3 (aq)  Cu(NO3 )2 + 2NO2 (g) + 2H 2O(l)
2/18/2015
12
Electrochemistry


Half-Reaction Method

Applicable to Acid or Base solutions

Does not usually require Oxidation Numbers (ON)
Procedure

Divide the overall reaction into:

Oxidation Half-Reaction

Reduction Half-Reaction

Balance each half-reaction for atoms & charge

Multiply one or both reactions by some integer to make
electrons gained equal to electrons lost

Recombine to given balanced redox equation
2/18/2015
13
Electrochemistry

Redox Half-Reaction Method – Example


26
Cr2 O7 (aq)

+ I- (aq)  Cr 3 (aq) + I 2 (s)
Divide steps into Half-Reactions


6 2 2Cr2 O7
 Cr 3 (Cr gains e- - reduction)
I -  I 2 (Iodine loses e- - oxidation)
2/18/2015
14
Electrochemistry

Balance Atoms & Charges for Cr2O72- / Cr3+

 Cr

O 
2
6
Cr2 O7
6
2
14H
6e -6
+
 2Cr 3+
2-
7
+

Add 7 Water molecules
to balance Oxygen
 2Cr 3+ + 7H 2O

2Cr26 O7
+ 14H+ +

+
(+12
(+14)
+
 2Cr
Cr26 O7

2-
3+
Add 14 H+ ions on left to
balance 14 H on right
+ 7H 2O
 2Cr 3+ + 7H 2O
-14)
=
+6
Add 6 electrons (e-) on left
to balance reaction charges
(6 electrons gained  this is the reduction reaction

Balance Atoms & Charges for I- / I2
2I -  I 2
2I
2/18/2015
-
 I 2 + 2e
No need to add H2O or H+
-
Add 2 electrons (e-) on right
to balance reaction charges
15
Electrochemistry

Redox Half-Reaction Method – Example (con’t)

Multiply each half-reaction, if necessary, by an integer
to balance electrons lost/gained


2 e- lost in oxidation reaction and 6 e- gained in
reduction
 Multiply oxidation half-reaction by 3
3(2I- )  3I 2 + 3(2e- )
6I-  3I 2 + 6e
Add 2 half-reactions together
6e -
+ 14H+
+
 Cr2O7  2-
 2Cr 3+ + 7H 2O
6I - 
6I 2/18/2015
+ 14H+
+
 Cr2O7  2-

3I 2 + 6e3I 2
+ 2Cr 3+ + 7H 2O
16
Electrochemistry

Half-Reaction Method in a “Basic” solution
Sodium Permanganate & Sodium Oxalate
 Mn
7
O42

NaMnO4
-
(aq) +


3 2 2C2 O 4
Na2C2O4
 Mn
4
O2 2 (s)

4
+ C O3

2-
(aq)
Half-Reactions
C2O42- 
MnO4-  MnO2
CO32-
MnO4-  MnO2 + 2H2O
2 H 2O + C2O42-  2CO32-
4H+ + MnO4-  MnO2 + 2H2O
2 H2O + C2O42-  2CO32- + 4H+
3e- + 4H + + MnO 4-  MnO 2 + 2H 2O
2 H 2O + C2O 4 2-  2CO 3 2- + 4H + + 2e -
(reduction)
(oxidation)
Multiply each reaction by appropriate integer
6e- + 8H+ + 2MnO4-  2MnO2 + 4H 2O
2/18/2015
6H2O + 3 C2O42-  6CO32- + 12H + + 6e17
Electrochemistry

Sodium Permanganate & Sodium Oxalate (con’t)

Add reactions
6e- + 8H+ + 2MnO4-  2MnO2 + 4H 2O
6H2O + 3 C2O42-  6CO32- + 12H + + 6e2MnO4- + 2H2O + 3 C2O42-  2MnO2

+ 6CO32- + 4H+
Add OH- to neutralize H+ , balance H2O, and form “basic” solution
2MnO4- + 2H2O + 3 C2O42- + 4OH-  2MnO2 + 6CO32- + 4H+ + 4OH2MnO4- + 2H2O + 3 C2O42- + 4OH-  2MnO2 + 6CO32- + 4H2O
2MnO4- + 3 C2O42- + 4OH-  2MnO2 + 6CO32- + 2H2O
2/18/2015
18
Electrochemistry

Electrochemical Cells

2/18/2015
Voltaic (Galvanic) Cells

Use spontaneous reaction (G < 0) to generate
electrical energy

Difference in Chemical Potential energy between
higher energy reactants and lower energy products is
converted to electrical energy to power electrical
devices

Thermodynamically - The system does work on the
surroundings
19
Electrochemistry

Electrochemical Cells

2/18/2015
Electrolytic Cells

Uses electrical energy to drive nonspontaneous
reaction (G > 0)

Electrical energy from an external power supply
converts lower energy reactants to higher energy
products

Thermodynamically – The surroundings do work on
the system

Examples – Electroplating and recovering metals
from ores
20
Electrochemistry
2/18/2015
21
Electrochemistry
Electrochemical Cells
 Cell notation is used to describe the structure of a
voltaic (galvanic) cell
 For the Zn/Cu cell, the cell notation is:
Zn(s) Zn2+(aq) Cu2+(aq) Cu(s)
= phase boundary (solid Zn vs. Aqueous Zn2+)
= salt bridge
 Anode reaction (oxidation)
is left of the salt bridge
 Cathode reaction (reduction) is right of the salt bridge
 Half-cell components usually appear in the same order
as in the half-reactions (Zn(s) + 2e-  Zn2+).
 Zinc solid loses 2 e- (oxidized) to produce zinc(II) at the
negative ANODE
 Copper(II) gains 2e- (reduced) to form copper metal at
22
positive CATHODE
2/18/2015

Electrochemistry

Voltaic (Galvanic) Cells

Zinc metal (Zn) in solution of Cu++ ions
Cu 2+ (aq) + 2e-  Cu(s)
[reduction]
Zn(s)  Zn 2+ (aq) + 2e-
[oxidation]
Zn(s) + Cu 2+ (aq)  Zn 2+ + Cu(s)

Construction of a Voltaic Cell

The oxidizing agent (Zn) and reducing agent (Cu2+)
in the same beaker will not generate electrical
energy

Separate the half-reactions by a barrier and connect
them via an external circuit (wire)
Set up salt bridge between chambers to maintain
neutral charge in electrolyte solutions
2/18/2015

23
Electrochemistry


2/18/2015
Oxidation Half-Cell

Anode Compartment – Oxidation of Zinc (An Ox)

Zinc metal in solution of Zn2+ electrolyte (ZnSO4)

Zn is reactant in oxidation half-reaction

Conducts released electrons (e-) out of its half-cell
Reduction Half-Cell

Cathode Compartment – Reduction of Copper (Red
Cat)

Copper bar in solution of Cu2+ electrolyte (CuSO4)

Copper metal is product in reduction half-cell reaction

Conducts electrons into its half-cell
24
Electrochemistry
Zinc-Copper Voltaic Cell
2/18/2015
25
Electrochemistry

Relative Charges on the Anode/Cathode
electrodes
 Electrode
charges are determined by the source
of the electrons and the direction of electron
flow
 Zinc
atoms are oxidized (lose 2 e-) to form Zn2+
at the anode

Anode – negative charge (e- rich)
 Released
electrons flow to right toward
cathode to be accepted by Cu2+ to form Cu(s)

2/18/2015
Cathode – positive charge (e- deficient)
26
Electrochemistry

Purpose of Salt Bridge

Electrons from oxidation of Zn leave neutral ZnSO4
solution producing net positive charge

Incoming electrons to CuSO4 solution would
produce net negative charge in solution as copper
ions are reduced to copper metal

Resulting charge imbalance would stop reaction

Salt bridge provides “liquid wire” allowing ions to
flow through both compartments completing circuit

Salt bridge constructed of an inverted “U-tube”
containing a solution of non-reacting
Na+ & SO42- ions in a gel
2/18/2015
27
Electrochemistry

Active vs Inactive Electrodes

2/18/2015
Active Electrodes

Electrodes in Zn/Cu2+ cell are active

Zinc & Copper bars are components of the
cell reactions

Mass of Zn bar decreases as Zn2+ ions in cell
solution increase

Mass of Copper bar increases as Cu2+ ions
accept electron to form more copper metal
28
Electrochemistry

Active vs Inactive Electrodes

2/18/2015
Inactive Electrodes

In many Redox reactions, one or the other
reactant/product is not capable of serving as
an electrode

Inactive electrodes - Graphite or Platinum

Can conduct electrons into and out of halfcells

Cannot take part in the half-reactions
29
Electrochemistry
Voltaic Cell
with
Inactive Graphite
Electrodes
2/18/2015
30
Practice Problem
A mercury battery, used for hearing aids and electric
watches, delivers a constant voltage (1.35 V) for long
periods. The half reactions are given below. Which half
reaction occurs at the Anode and which occurs at the
Cathode? What is the overall cell reaction?
HgO(s) + H2O(l) + 2e-  Hg(l) + 2 OH-(aq)
Zn(s) + 2 OH-(aq)  Zn(OH)2(s) + 2eAns: Reduction occurs at Cathode (Red Cat)
Hg2+ gains 2 e- (reduced) to form Hg
Oxidation occurs at the Anode (An Ox)
Zn loses 2 e- (oxidized) to form Zn2+
2/18/2015
HgO(s) + Zn(s) + H 2O  Zn(OH)2 + Hg(l)
31
Practice Problem
Write the cell notation for a voltaic cell with the
following cell reaction
Ni(s) + Pb (aq)  Ni (aq) + Pb(s)
2+
2+
Ans:
Pb 2+ (aq) + 2e-  Pb(s)
Ni(s)  Ni 2+ + 2e-
Reduction @ Cathode (Red Cat)
Oxidation @ Anode (An Ox))
Ni(s) I Ni 2+ (aq)
Anode (oxidation) is
represented on left side
of Cell notation
2/18/2015
Pb 2+ (aq) I Pb(s)
Cathode (reduction) is
represented on right side
of cell notation
32
Practice Problem
Write the cell reaction for the following voltaic cell
Pt|H2(g) | H+(aq) ║ Br2(l) | Br-(aq)|Pt
Note: Platinum (Pt) serves as a reaction site at the anode,
but does not participate in the reaction
Ans:
Br2 (l) + 2e-  2Br - (aq)
gain 2 e-
Reduction @ Cathode (Red Cat)
H2 (g)  2H+ (aq) + 2e-
lose 2 e-
Oxidation @ Anode (An Ox)
PtIH2 (g) + Br2 (l)  2H + (aq) + 2Br - (aq)IPt
2/18/2015
33
Electrochemistry

Cell Potential

2/18/2015
The movement of electrons is analogous to
the pumping of water from one point to another

Water moves from a point of high pressure to
a point of lower pressure. Thus, a pressure
difference is required

The work expended in moving the water
through a pipe depends on the volume of
water and the pressure difference
34
Electrochemistry

Cell Potential

Movement of Electrons

An electric charge moves from a point of high
electrical potential (high electrical pressure)
to one of lower electrical potential

The work expended in moving the electrical
charge through a conductor depends on the
potential difference and the amount of charge
Work(w) = Potential difference (E)  Charge
w = Ecell  Charge
2/18/2015
35
Electrochemistry

Cell Potential

Purpose of a voltaic cell is to convert the free
energy of a spontaneous reaction into the
kinetic energy of electrons moving through an
external circuit (electrical energy)

Electrical energy is proportional to the
difference in the electrical potential between
the two cell electrodes
Cell Potential
2/18/2015
36
Electrochemistry

Cell Potential

Positive Cell Potential – Electrons flow spontaneously
from the negative electrode (Anode) to the positive
electrode (Cathode)
Ecell > 0 for a spontaneous process

Negative cell potential is associated with a
“nonspontaneous” cell reaction
Ecell < 0 for a nonspontaneous process

Cell potential for a cell reaction at equilibrium would be
“0”
Ecell = 0 for an equilibrium process

2/18/2015
As with Entropy, there is a clear relationship between
Ecell , K, and G
37
Electrochemistry

Units of Cell Potential

The SI (metric) unit of electrical charge is the:
Coulomb (C)

The SI (metric) unit of current is the:
Ampere (A)
1 coulomb
1 ampere =
second

1A = 1 C / s
The SI (metric) unit of electrical potential is the:
“Volt (V)”

By definition, the energy released by a potential difference of
one volt moving between the anode and cathode of a voltaic
cell releases 1 joule of work per coulomb of charge
1 volt =
2/18/2015
1J
C
1C =
1J
V
38
Electrochemistry

The charge (F) that flows through a cell equals the
number of moles of electrons (n) transferred times the
charge of 1 mol of electrons
charge
Charge = moles of e 
mol e-
Moles e- = n
Charge
Mol e -
96, 485 C
= F = Faraday Constant =
mol e -
Charge
= nF
F =
96, 485 C
mol e-
F = 96, 485 ×
2/18/2015
J
C - Coulomb   , SI unit of charge
V
J
J
4
=
9.6×10
×
V • mol eV • mol e 39
Electrochemistry

The total amount of a substance undergoing a ReDox
reaction in a cell depends on the total charge flowing
through the cell, which is a function of the current
(amperes) and the time (seconds)

The ratio of the total charge and the charge per mole of
electrons determines the number moles of substance
that will react

Thus,
Moles of substance =
Charge (C)
moles e- (n)  Charge / moles e - (F)
Moles of substance =
Current (i) × Time (t)
=
n×F
Moles of substance =
2/18/2015
i ×t
n×F
amperes  seconds
moles e96, 485 Coulombs
×
moles substance
mole e-
n = total moles electrons
40
Electrochemistry
How long in seconds would it take to deposit 38.93 g of Pb
from an aqueous solution of Pb2+ with a current of 1.679 A?
Moles =
Time =
Time =
it
nF
=
Current (Amperes) × Time (seconds)
(moles e- ) × (Faradays (Coulombs / mole e - ))
 Moles   mole e-   Faradays 
Current

1 mol Pb   2 mol e 39.93 g Pb

207.2
g

  mol Pb
  96, 485 C 

- 
  mol e 
1.679 A
Time = 2.215 × 104 sec
2/18/2015
41
Electrochemistry

Standard Cell Potential

Eocell – The potential measured at a specific
temperature (298 K) with no current flowing
and all concentrations in their “Standard States”

1 atm for gases

1 M for solutions

Pure solids for electrodes
Zn(s) + Cu2+ (aq; 1M)  Zn2+ (aq; 1M) + Cu(s)
2/18/2015
Eocell = 1.10V
42
Electrochemistry

Standard Electrode Half-Cell Potentials
o
 E half-cell – Potential associated with a given half-cell
reaction (electrode compartment) when all components
are in “Standard States”


Standard Electrode Potential for a half-cell reaction,
whether anode (oxidation) or cathode (reduction) is
written and presented in Appendix D as a “reduction”
2+
Cu
(aq)
+
2e
 Cu(s) [reduction - cathode]
Ex.
Zn(s)  Zn 2+ (aq) + 2e- [oxidation - anode]
would be written in the table as:
Cu2+ (aq) + 2e-  Cu(s) EoCopper (Eocathode ) [reduction]
Zn2+ (aq) + 2e-  Zn(s)
2/18/2015
Eo Zinc
(Eoanode )
[reduction]
43
Electrochemistry

Standard Electrode Half-Cell Potentials

Electrons flow spontaneously from Anode (negative) to
Cathode (positive)

Cathode must have a more “Positive” Eohalf-cell than the
Anode

For a “positive” Eocell ; i.e., a spontaneous reaction
Eocell = (Eocathode (reduction) - Eoanode (oxidation) ) > 0
Eocell = Eocopper - Eo Zinc

The standard cell potential is the difference between
the standard electrode potential of the “Cathode”
(reduction) half-cell and the standard electrode
potential of the “Anode” (oxidation) half-cell

Standard half-cell potentials are “intensive” properties,
thus their values do NOT have to be adjusted for
stoichiometry (# of moles)
2/18/2015
44
Practice Problem
Write out the overall equation for the cell reaction and
determine the standard cell potential for the following
galvanic cell
[Eo (Ag+/Ag) = 0.80 V; Eo (Ni2+/Ni) = - 0.26 V]
Ni(s) I Ni 2+ (aq)
Ag + (aq) I Ag(s)
2Ag + (aq) + 2e-  2Ag(s)
Ni(s)  Ni 2+ (aq) + 2e-
[reduction (cathode)]
[oxidation (anode)]
Ni(s) + 2Ag + (aq)  Ni 2+ + 2Ag(s)
Write both reactions in "reduction" form
2Ag + (aq) + 2e-  2Ag(s)
Ni 2+ (aq) + 2e-

Ni(s)
[reduction (cathode)]
[reduction (anode)]
Eocell = EoSilver(Red) - Eo Nickel(Ox)
2/18/2015
Eocell = 0.80
- (-0.26) = 1.06 V
45
Electrochemistry

The Standard Hydrogen Electrode

Half-cell potentials are not absolute quantities

The values found in tables are determined relative to a
“Standard”

The Standard Electrode potential is defined as zero
(Eoreference) = 0.00

The “standard reference half-cell” is a standard
“Hydrogen” electrode

Specially prepared Platinum electrode immersed in a
1 M aqueous solution of a strong acid through which H2
gas at 1 atm is bubbled
2H+ (aq; 1 M) + 2e-
2/18/2015
H2 (g); 1 atm)
Eoreference = 0.00V
46
Electrochemistry

Reference Half-Cell and Unknown Half-Cell


The “Standard” electrode can act as either the “Anode”
or the “Cathode”
Oxidation of H (lose e-) at anode half-cell and
2
reduction of unknown at cathode half-cell
Eocell = Eocathode - Eoanode
Eounknown - Eoreference
Eocell = Eounknown - 0.00 V = Eounknown

Reduction of H+ (gain e-) at cathode half-cell and
oxidation of unknown at anode half-cell
Eocell = Eocathode - Eoanode
Eoreference - Eounknown
Eocell = 0.00 V - Eounknown
2/18/2015
= - Eounknown
47
Practice Problem
Determine the standard electrode potential, Eozinc, using a
voltaic cell consisting of the Zn/Zn2+ half-reaction and the
H+/H2 half-reaction.
Eocell = + 0.76 V
Ans: Zinc is being oxidized (loses 2e-) producing electrons at
the negative anode; H+ gains e- at positive cathode
2H+ (aq) + 2e- 
Zn(s) 
H2 (g)
Zn2+ (aq) + 2e-
Eoreference = 0.00 V
Eo zinc = ? V
Zn(s) + 2H+ (aq)  Zn2+ (aq) + H2 (g)
Eocell = Eocathode - Eoanode
Eocell = 0.76V
Eoreference - Eo Zinc
Eo Zinc = Eoreference - Eocell = 0.00 V - 0.76 V = - 0.76 V
2/18/2015
48
Electrochemistry


Relative Strength of Oxidizing and Reducing Agents
Cu 2+ (aq) + 2e- = Cu(s)
Eo = 0.34 V
2H+ (aq) + 2e- = H2 (g)
Eo = 0.00 V
Zn 2+ (aq) + 2e- = Zn(s)
Eo = - 0.76 V
The more positive the Eo value, the more readily the
reaction occurs

Strength of Oxidizing Agents – Cu2+ > H+ > Zn2+

Strength of Reducing Agents – Zn
> H2 > Cu

Oxidizing agents decrease in strength as the value of Eo
decreases, while the strength of the reducing agents
increases as the value of Eo decreases

Cu2+ is the stronger Oxidizing agent

Zn metal (not the ion) is the stronger Reducing agent
2/18/2015
49
Electrochemistry

Table of Standard Electrode Potentials
(The emf Series)
Selected Standard Electrode Potentials (298oC)
 All Values are relative to the “standard hydrogen (reference) electrode
 All reactions are written as “reductions”
2/18/2015
50
Electrochemistry

EMF Series
 All Values are relative to the “standard hydrogen
(reference) electrode
 All reactions are written as “reductions”
 F2 is the strongest oxidizing agent (high, positive Eo)
 Fluorine is very electronegative with a high ionization
potential (easily reduced by gaining electrons)
 By gaining e- it forms the weakest reducing agent,
F- , which is very reluctant to lose electrons)
 Li metal is strongest reducing agent (low, more
negative Eo)
 Lithium has a Low ionization energy and is easily
oxidized by losing electrons
 By losing e- , Lithium forms the weakest oxidizing
agent, Li+
2/18/2015
51
Electrochemistry
Similarities – Acid/Base vs Redox
 Acid Strength vs Base Strength using Ka & Kb values
 Redox (Oxidizing agent vs Reducing agent) using
Standard Electrode Potential (Eo) values
 Appendix D (Table of Standard Electrode Potentials)
 The stronger oxidizing agent (species on left side of
table) has a half-reaction with a larger more positive
(less negative) Eo than a species lower in the list
 The stronger reducing agent (species on the right
side of table) has a half-reaction with a smaller (less
positive) Eo value than a species higher in the list
 A spontaneous reaction between a metal, acting as a
reducing agent by losing electrons (oxidized) to be
gained by the ion of another element (reduced)
would occur if the Eo potential of the metal is less
than that of the metal ion.
52
2/18/2015

Electrochemistry

Writing Spontaneous Redox Reactions

A spontaneous reaction (Eocell > 0) will occur between
an oxidizing agent and any reducing agent that lies
below it in the table

Zn(s) (reducing agent (Eo = -0.76 V) will react
spontaneously with Cu2+ (oxidizing agent) (Eo = +0.34
V), which lies above Zn in the table

The Reduction of Cu2+ to Cu will occur at the Cathode

The Oxidation of Zn to Zn2+ will occur at the Anode

2/18/2015
A spontaneous reaction will occur when the Eo of the
reaction at the Cathode (reduction) minus the Eo of the
reaction at the Anode (oxidation) is > 0 (Positive)
Eocell = (Eocathode (reduction) - Eoanode (oxidation) ) > 0
53
Electrochemistry
Writing Spontaneous Redox Reactions
 Combing half-cell reactions to form a “Net” reaction
 In the Standard Electrode Potential table
(Appendix D), both reactions are written as
“reductions” (e- gain)
 One of the reactions will have to be reversed so that
the applicable reactants are on the left side of the
net equation and the products on the right side
 The sign of Eo for the reversed reaction need not be
reversed
 Which reaction to reverse?
 Scenario #1 - Net reaction is supplied
 Balanced equation clearly indicates which
species is oxidized and which species is
reduced
 Consult table to see which half-cell reaction
(Con’t)
was
reversed
54
2/18/2015

Electrochemistry

Writing Spontaneous Redox Reactions (con’t)

Combing half-cell reactions to form a “Net” reaction
 Net reaction example
Zn(s) + Cu 2+ (aq) 
Stronger
Reducing
Agent
Stronger
Oxidizaing
Agent
Zn 2+ (aq) + Cu(s)
Weaker
Oxidizing
Agent
Weaker
Reducing
Agent
Cu2+ is reduced at Cathode - Cu2+ + 2e-  Cu(s) Eo = + 0.34V
Zn(s) is oxidized at Anode - Zn  s   Zn2+ + 2e-
Eo = - 0.76 V
Zinc metal is the stronger reducing agent (least positive Eo value)
Zn(s) reaction was reversed
Eocell = (Eocathode (reduction) - Eoanode (oxidation) )
Eocell = + 0.34 V - (-0.76V) = + 1.10 V > 0
2/18/2015
Reaction Spontaneous
55
Electrochemistry

Writing Spontaneous Redox Reactions (con’t)
 Combing half-cell reactions to form a “Net” reaction
 Scenario #2 - Net reaction to be determined from
half-cell reactions
Ag + (aq) + e-
= Ag(s)
Sn 2+ (aq) + 2e- = Sn(s)
Eo = + 0.80 V
Eo = - 0.14 V
 Since the net reaction is not known, it is not clear
which reaction occurs at a particular electrode
 For a spontaneous reaction, the cathode
[reduction] potential minus the anode [oxidation]
potential must be “positive” (Eocell > 0)
 To ensure this, the Anode Oxidation term in the
Eocell equation must have an Eo value less than
the Cathode Reduction term
2/18/2015
56
Electrochemistry

Writing Spontaneous Redox Reactions (con’t)
 Combing half-cell reactions to form a “Net” reaction
 Scenario #2 - Net reaction to be determined from
half-cell reactions (con’t)
 Since -0.14 < +0.80, the Sn2+ + 2 e- reaction
must be reversed (converted to the Anode
Oxidation term)
 Balance equations to account for reaction
coefficients (e- lost = e- gained)
2Ag + (aq) + 2eSn(s) =
= 2Ag(s)
Sn 2+ (aq) + 2e-
Sn(s) + 2Ag + (aq) 
Eo = + 0.80 V (Cathode Reduction)
Eo = - 0.14 V (Anode Oxidaton)
Sn 2+ (aq) + 2Ag(s)
(overall reaction)
Eocell = Eocathode reduction(Ag) - Eoanode oxidation(Sn)
2/18/2015
Eocell = 0.80 - (-0.14) = + 0.94 V
(Spontaneous)
57
Practice Problem
Consult the table of standard electrode potentials in your
textbook in order to decide which one of the following
reagents is capable of reducing I2(s) to I- (aq, 1 M)
I2(s) + 2e-  2I-(aq)
+0.53V
Reduction Form
a. Br-(aq)
Br2(l)
b. Ag(s)
Ag+(aq) + e-  Ag(s)
+0.80V
c. Sn(s)
Sn2+(aq) + 2e-  Sn(s)
-0.14V
d. Zn2+ (aq, 1 M)
Zn2+ (aq) + 2e-  Zn(s)
-0.76V
e. Sn4+ (aq,1 M)
Sn4+ (aq) + 2e-  Sn2+ (aq) +0.13V
+ 2e-  2Br-(aq)
+1.07V
(con’t)
2/18/2015
58
Practice Problem (con’t)
The I2 is being reduced – gaining electrons – at Cathode (Red Cat)
The other solution must be capable of being oxidized at Anode (AnOx)
Only answers a, b, c reflect oxidizable elements, i.e., can lose electrons
a. The Br- solution undergoes oxidation (lose e-) at the Anode
Thus
Eocell = Eocathode - Eoanode
+ 0.53 - (+1.07) = - 0.54Eo
Neg Eocell  Bromide (Br - ) will not reduce I 2
b. The Silver (Ag(s) undergoes oxidation (loses e-) to form Ag+
Thus:
Eocell = Eocathode - Eoanode
+ 0.53 - (+0.80) = - 0.27o
Neg Eocell  Ag(s) will not reduce I 2
2/18/2015
(con’t)
59
Practice Problem (con’t)
c. The Sn(s) will undergo oxidation to form Sn2+
Eocell = Eocathode - Eoanode
+ 0.53 - (-0.14) = + 0.67o
Pos Eocell  Sn(s) will reduce I 2
d. The Zn2+ solution is already oxidized (will not lose any
more e-)
Thus:
Zn2+ will not reduce I 2
e. The Sn4+ solution is already oxidized
Thus:
Sn4 will not reduce I 2
2/18/2015
60
Practice Problem

Combine the following 3 half-cell reactions into 3 balanced
spontaneous reactions; Calculate Eocell for each; rank the
relative strengths of the oxidizing & reducing agents
(A)
NO3- (aq) + 4H+ (aq) + 3 e-  NO(g) + 2H2O(l)
N2 (g) + 5H+ + 4e-  N2 H5+
(B)
(C) MnO2 (s) + 4H+ (aq) + 2e-  Mn2+ (aq) + 2H2o (l)
Eo = 0.96 V
Eo = - 0.23 V
Eo = 1.23V
1. Combine & Balance half-reactions A & B
Eo in (A) is more positive than Eo in (B)
Reverse B
(A – cathode reduction; B – anode oxidation)
+
(A) 4NO3 (aq) + 16H (aq) + 12 e  4NO(g) + 8H2O(l)
(B)
3N2 H5
+
 3N2 (g)  15H+ + 12e-
Eo = 0.96 V
Eo = -0.23 V
3N2H5 +(aq) + 4NO3- (aq) + H + (aq)  3N2 (g) + 4NO(g) + 8H2O(l)
Eocell (A + B) = 0.96 V - (-0.23 V) = 1.19 V
2/18/2015
Con’t
61
Practice Problem (con’t)
2. Combine & Balance half-reactions A & C
Reverse half-reaction (A) (Eo 1.23 > Eo 0.96)
(C – cathode reduction; A – anode oxidation)
(C) 3MnO2 (s) + 12H+ (aq) + 6e-  3Mn2+ (aq) + 6H2o (l)
Eo = 1.23 V
(A) 2NO(g) + 4H2O(l)  2NO3- (aq) + 8H+ (aq) + 6 e3MnO2 (s) + 4H+ (aq) + 2NO(g) 
Eo = 0.96 V
3Mn2+ (aq) + 2H2O(l) + 2NO3
Eocell (A + C) = (1.23 V - 0.96 V) = 0.27 V
3. Combine and Balance half-reactions B & C
Reverse half-reaction B (1.23 > -0.23)
(C – cathode reduction; B – anode oxidation)
(C)
(B)
2MnO2 (s) + 8H+ (aq) + 4e-  2Mn2+ (aq) + 4H2o(l)
N 2H5
+
 N2 (g) + 5H + + 4e-
Eo = 1.23 V
Eo = -0.23 V
N2 H5 (aq) + 2MnO2 (s) + 3H+ (aq)  N2 (g) + 2Mn2+ (aq) + 4H2O(l)
2/18/2015
Eocell (B + C) = 1.23 V - (-0.23) = 1.46 V
Con’t
62
Practice Problem (con’t)
Overall Ranking of Oxidizing & Reducing Agents
(C) MnO2 (s) + 4H+ (aq) + 2e-  Mn2+ (aq) + 2H2o (l)
Eo = 1.23 V
(A) NO3- (aq) + 4H+ (aq) + 3 e -  NO(g) + 2H2O(l)
Eo = 0.96 V
(B) N2 (g)
+ 5H+
+ 4e-  N2H5 
Oxidizing Agents : MnO2
> NO3-
Reducing Agents : N2H5+ > NO
Eo = -0.23 V
> N2
> Mn2+
Rank Oxidizing & Reducing Agents Within Each Equation

Equation 1 (A+B): Oxidizing Agent –

Reducing Agent –

Equation 2 (C+A): Oxidizing Agent –

Reducing Agent –

Equation 3 (C+B): Oxidizing Agent –

Reducing Agent –
2/18/2015
NO3- > N2
N2H5+ > NO
MnO2 > NO3NO
> Mn+2
MnO2 > N2
N2H5+ > Mn+2
63
Electrochemistry

Relative Reactivities of Metals
 Metals that displace H2 from acid
 If the Eocell for the reaction of H+ is more positive for
metal A than it is for metal B, metal A is a stronger
reducing agent than metal B and a more active metal
2H + (aq) + 2e-  H 2 (g)
Eo = 0.00 V
(cathode - reduction)
Fe(s)  Fe+2 (aq) + 2eEo = - 0.44 V (anode - oxidation)
Fe(s) + 2H + (aq)  H 2 (g) + Fe 2+ (aq)
Eocell = 0.00 - (-0.44) = 0.44 V
 Metals Li through Pb (includes Fe) in the standard
electrode potential list (appendix D) lie below H+ and
give positive Eocell when reducing H+ to H2, i.e.,
Hydrogen gas is released
2/18/2015
64
Electrochemistry

Relative Reactivities of Metals
 Metals that cannot displace H2 from acid
 Metals that lie above the standard hydrogen
reference half-reaction cannot reduce H+ from acids
 The Eocell for the reversed metal half-reaction is
negative and the reaction does not occur
2H + (aq) + 2e-  H 2 (g)
Eo = 0.00 V (cathode - reduction)
2 Ag(s)
2Ag + (aq) + 2e- Eo = 0.80 V (anode - oxidation)
2Ag(s) + 2H + (aq)  H 2 (g) + 2Ag + (aq)
Eocell = 0.00 V - 0.80 V = - 0.80 V (not spontaneous)


2/18/2015
The higher the metal in the list, the more negative is
its Eocell for the reduction of H+ to H2, thus its
reducing strength (and reactivity) is less
Thus, Gold (Au3+, Eo = +1.5V) is less active than
Silver (Ag+, Eo = +0.8V) and does not release H2 gas
65
Electrochemistry

Relative Reactivities of Metals
 Metals that displace H2 from water
 Metals that lie below the half-cell reaction potential
for water can displace H2 from water
 In the reaction below the E value for water is not the
standard state value listed in the table because in
pure water, [OH-] is 1.0 x 10-7 M, not the standard
state value of 1 M (-0.83 V)
2H 2O(l) + 2e-  H 2 (g) + 2OH -
E = -0.42V
2Na(s) 
Eo = -2.71 V
2Na+ (aq) + 2e-
2Na(s) + 2H 2O(l)  2Na + (aq) + H 2 + 2OH -
Eocell = - 0.42 V - (-2.71 V) = + 2.29 V
Ecell > 0  Sodium displaces Hydrogen from water
2/18/2015
66
Electrochemistry

Relative Reactivities of Metals
 Metals that can displace other metals from solution
 Any metal that is lower in the standard electrode
half-cell list can reduce the ion of a metal that is
higher in the list, thus displacing that metal from
solution (See next slide and slide #48)
Fe2+ (aq) + 2e-  Fe(s)
Eo = -0.44V (cathod; reduction)
Zn(s)  Zn 2+ (aq) + 2e-
Eo = -0.76V (anode; oxidation)
Zn(s) + Fe2+  Zn 2+ (aq) + Fe(s)
Eocell = - 0.44 V - (-0.76 V) = 0.32 V

Ecell > 0 Zinc is the stronger reducing agent reducing
Fe2+ to Fe and displacing it from solution
2/18/2015
67
Electrochemistry
2/18/2015
68
Electrochemistry

Free Energy and Electrical Work

Electrical Work
 Potential (Ecell, in volts) times the charge
Work(w) = Potential difference (E)  Charge
w = Ecell  charge






2/18/2015
Ecell measured with no current flowing
No energy lost to heating
Ecell voltage is maximum possible for cell
Work is maximum possible
Only reversible process can do maximum work
Reversible process with no current flow:
 Forward reaction if opposing potential is smaller
 Reverse reaction if opposing potential is larger
69
Electrochemistry



Spontaneous Reaction – G < 0
Spontaneous Reaction – Ecell > 0
ΔG  - Ecell
The voltaic cell loses energy as it does work on the
surroundings; thus the work term (wmax) is negative
w max = - Ecell  charge
ΔG = wmax
Recall Slide 70 from Chapter 20
w max = - Ecell  charge = G
G = w max = - Ecell  charge
G = w max = - Ecell  nF
ΔG o = - Eocell  n F
Recall :
(Recall slide # 39)
(components in standard states)
n = moles
Coulombs
J
4
F = 9.65  10
= 9.65  10
mole
V • mol e 4
2/18/2015
70
Electrochemistry
Electrical Work (con’t)


Relate standard cell potential to equilibrium
constant (K) of the redox reaction
ΔG o = - RT lnK
(Slide 82 from Chapter 20)
ΔG o = - Eocell  n F
-Eocell  n F = - RT lnK
J
 298.15 K
RT
mol rxn  K
=
lnK =
 2.303(log K)
n mol e
J
nF
(9.65  104
)
mol rxn
V • mol e
8.314
Eocell
E
2/18/2015
o
cell
0.0592 V
=
 log K
n
nEocell
log K =
0.0592 V
(at 298.15K)
71
Electrochemistry
Summary Relationship between
Go
2/18/2015
Eocell
K
72
Electrochemistry

Effect of Concentration on Cell Potential
 Most cells do not start with concentrations in their
“standard” states
(Slide 86 from Chapter 20)
ΔG = ΔGo + RTlnQ
 Recall:
G = - nF  Ecell
o
ΔG o = - nF × Ecell
(Standard State)
o
ΔG = - nF × Ecell = ΔGo + RTlnQ = - n F × Ecell
+ RTlnQ
o
-nF × Ecell = - n F × Ecell
+ RT lnQ
nF
RTlnQ
o
Ecell =
 Ecell
(Nernst Equation)
nF
nF
J
8.314
 298.15 K
o
mol rxn  K
Ecell = Ecell
 2.303 log Q
n mol eJ
(9.65  104
)
mol rxn
V • mol e
o
Ecell = Ecell
-
2/18/2015
0.0592 V
× log Q
n
(at 298.15 K)
73
Electrochemistry

Changes in Potential During Cell Operation

The potential of a cell changes as the concentration of
the cell components change
Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s)
2+
[Zn
]
o
Ecell
= 1.10 V
Q =
[Cu 2+ ]
Ecell = E
o
cell
0.0592 V
× log Q
n
Stage 1 When Q < 1
Ecell > Eocell
Stage 2 When Q = 1
o
Ecell = Ecell
Stage 3 When Q > 1
o
Ecell < Ecell
(at 298.15 K)
Stage 4 When Q = K Ecell = 0 (Equilibrium)
If Q / K < 1, Ecell is positive (+) (cell does work)
If Q / K = 1, Ecell = 0
2/18/2015
(equilibrium - cell no longer does work)
If Q / K > 1, Ecell is negative (-) (reaction reverses until Q / K = 1)
74
Electrochemistry
If Q / K < 1, Ecell is positive (+)
(cell does work)
If Q / K = 1, Ecell = 0
(cell no longer does work)
If Q / K > 1, Ecell is negative (-)
(reaction reverses until Q / K = 1)
Ecell = E
2/18/2015
o
cell
0.0592 V
× log Q
n
75
Electrochemistry

Concentration Cells

In a cell composed of the same substance, but differing
concentrations in the two half-cells, the two
concentrations move to equilibrate producing electrical
energy

The cell reaction is the “sum” of identical half-cell
reactions written in opposite directions


2/18/2015
The Standard Electrode Potentials (Eocell) are both
based on a 1 M solution (standard conditions), so they
“cancel” each other, i. e., Eocell = 0
The non-standard cell potential, Ecell, depends on the
ratio of the two concentrations [A]dil / [A]conc = Q
76
Electrochemistry


How the Concentration Cell Works
The dilute solution is in the Anode compartment
(oxidation) and the concentrated solution is in the
Cathode compartment (Reduction)
Cu(s)  Cu 2+ (aq;0.10M) + 2e- (Oxidation)
Cu 2+( aq;1.0M) + 2e- 
Cu(s)
 Reduction 
In the Anode (dilute) half-cell, Cu atoms give up 2
electrons and the resulting Cu2+ ions enter the
solution and make it more concentrated
 In the Cathode (conc) half-cell, Cu2+ ions gain 2
electrons and the resulting Cu atoms plate out on the
electrode, making the solution less concentrated
 In this type of Voltaic cell, the dilution continues until
equilibrium is attained, i.e.,
Ecell decreases until Ecell = 0 (Q = K)

2/18/2015
77
Electrochemistry

Alkaline Battery
MnO2 (s) + 2H2O(l) + 2e-  Mn(OH)2 (s) + 2OH- (aq)
Zn(s) + 2OH- (aq)  ZnO(s) + H2O(l) + 2e-
Zn(s) + MnO2 (s) + H 2O(l)  ZnO(s) + Mn(OH)2 (s)

Ecell = 1.5 V
[Anode (Oxidation]
Overall Cell Reaction
Nickel-Metal Hydride (Ni-MH) Battery
NiO(OH)(s) + H2O(l) + e-  Ni(OH)2 (s) + OH- (aq)
MH(s) + OH- (aq)  M(s) + H2O(l) + e-
MH(s) + NIO(OH)(s)  M(s) + Ni(OH)2 (s)

[Cathode (Reduction]
Ecell = 1.4 V
[Cathode (Reduction]
[Anode (Oxidation]
Overall Cell Reaction
Lithium Ion Battery
Li1-xMn2O4 (s)(s) + xLi + + xe-  LiMn2O4 (s) + OH- (aq)
Li xC6
 xLi +
+ xe- + C6 (s)
Li xC6 + Li1-xMn 2O4 (s)  LiMn 2O4 (s) + C6(s)
2/18/2015
[Cathode (Reduction]
[Anode (Oxidation]
Ecell = 3.7 V
Overall Cell Reaction
78
Practice Problem
Calculate Ecell for a voltaic cell containing the following halfcells:
Zn(s) I Zn2+ (aq) II H+ I H (g)
2
[Zn2+] = 0.010 M
Construct Eo
[H+] = 2.5 M
PH2 = 0.30 atm
cell
2H+ (aq) + 2e-  H2 (g)
Zn(s)  Zn 2+
(Eo = 0.00 V) (red; cathode)
+ 2e -
(Eo = - 0.76 V) (ox; anode)
2H+ (aq) + Zn(s)  H2 (g) + Zn2+ (aq)
Eocell = 0.00 V - (-0.76 V) = 0.76 V
2+
Calculate Q Q = PH2  [Zn ] = 0.30  0.010 = 4.8  10-4
+ 2
2
[H ]
Ecell = Eocell - 2.303 ×
2/18/2015
2.5
RT
0.0592 V
log Q = Eocell log Q
nF
n
 0.0592 V

Ecell = 0.76V - 
 log(4.8  10-4 )  = 0.76V - (-0.0982 V = 0.86 V
2


79
Practice Problem
What is the equilibrium constant for the following reaction
[Eo(Ce4+/Ce3+) = 1.72 V
Eo(Cl2/Cl-) = 1.36 V
2 Cl-(aq) + 2 Ce4+(aq)  Cl2(g) + 2 Ce3+(aq)
Cl - (aq) I Cl 2 (g) II Ce4+ (aq) I Ce3+ (aq)
2Ce4+ (aq) + 2e-
2Cl - (aq)  Cl 2
 2Ce3+ (aq)
(Eo = 1.36 V) (ox; anode)
+ 2e -
2Ce4+ (aq) + 2Cl -
(Eo = 1.72 V) (red; cathode)
 Cl 2 (g) + 2Ce3+ (aq)
Eocell = 1.72V - 1.36 V = 0.36 V
E
o
cell
0.0592 V
=
 log K
n
nEocell
2  0.36 V
log K =
=
= 12.2
0.0592 V 0.0592 V
2/18/2015
(at 298.15 K)
80
Practice Problem
Write out the overall equation for the cell reaction and
determine the standard cell potential for the following
galvanic cell.
[Eo (Ag+/Ag) = 0.80
Eo (Ni2+/Ni) = -0.26]
Ni(s)|Ni2+(aq)║Ag+(aq)|Ag(s)
2Ag + (aq) + 2e-
 2Ag + (aq)
Ni(aq)  Ni 2+
(Eo = 0.80 V) (red; cathode)
+ 2e -
(Eo = - 0.26 V) (ox; anode)
2Ag + (aq) + Ni(aq)  Ni 2+ (aq) + 2Ag + (aq)
Eocell = 0.80V - (- 0.26) = 1.06 V
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81
Practice Problem
What is the maximum work you can obtain from 15.0 g of Ni
in the galvanic cell shown in the previous problem when the
Ecell is 0.97 V?
[Eo (Ag+/Ag) = 0.80
Eo (Ni2+/Ni) = -0.26]
Ni(s)|Ni2+(aq)║Ag+(aq)|Ag(s)
2Ag + (aq) + Ni(aq)  Ni 2+ (aq) + 2Ag + (aq)
w max = - Ecell  charge
w max
charge = nF
F =
96,485 C
J
=
96,485
mol eV • mol e -
mol eJ
5
= -Ecell  n  F = -0.97 V × 2
× 96,485
=
1.87

10
J / mol
mol Ni
V • mol e
Mass
15.0g
mol(Ni) =
=
= 0.256 mol
Mol Wgt
58.69 g / mol
wmax = 0.256 mol ×(-1.87×105 J / mol = -4.78×104 J
2/18/2015
wmax = - 4.78 ×104 J = - 4.78 kJ
82
Practice Problem
What is the cell voltage (Ecell) for the following galvanic cell?
Cd(s)|Cd2+(0.026 M)║Ni2+(0.00420 M)|Ni(s)
Ni 2+ (aq) + 2e-  Ni(s)
Cd(s)  Cd 2+ + 2e-
Cd(s) + Ni 2+ (aq) 
Eo = - 0.25 V
Eo = - 0.40 V
(Reduction)
(Oxidation)
Cd 2+ + Ni(s)
Eocell = - 0.25 V - (-0.40 V) = 0.15 V
[Cd 2+ ]
0.026M
Q =
=
= 6.19
2+
0.00420M
[Ni ]
o
Ecell = Ecell
-
Ecell = 0.15 V 2/18/2015
0.0592 V
× log Q
n
(at 298.15 K)
0.0592 V
× log 6.19 = 0.15 V - (0.0296  0.79) = 0.13 V
2
83
Practice Problem
Construct a Voltaic Cell to determine the pH of an Unknown
Solution
Compartment #1 – Cathode consisting of Standard
Hydrogen Electrode based on H2/H+
half-cell reaction at standard
conditions (H+ – 1 M; H2 – 1 atm)
Compartment #2 – Anode consisting of same apparatus
but dipping into a solution of
unknown H+(pH)
Although Eocell = 0, the individual half-cells differ in [H+]
and Ecell is not = 0
2H+ (aq; 1 M) +
2e-  H 2 (g; 1 atm)
[cathode; reduction]
H2 (g; 1atm)  2H+ (aq; unknown) + 2e- [anode; oxidation]
2H + (aq; 1M)  2H + (aq; unknown)
2/18/2015
Ecell = ?
Con’t
84
Practice Problem (Con’t)
Ecell = E
o
cell
2
0.0592 V
0.0592 V
[H + ]unknown
o
× log Q = Ecell × log + 2
n
n
]H ]standard
Substituting 1M for [H+ ]standard and 0 V for Eocell gives :
Ecell
2
0.0592 V
[H + ]unknown
0.0592 V
+ 2
= 0V × log
=

log[H
]unknown
2
2
1
2
0.0592 V
× 2 log [H + ]unknown = - 0.0592V  log[H + ]unknown
2
Ecell
-log[H + ]unknown =
0.0592
Ecell = -
Since pH = - log[H + ]
pH =
Ecell
0.0592
Measure the cell potential with a Voltmeter and calculate pH
2/18/2015
85
Practice Problem
What is the pH of the test solution when Ecell = 0.612 V at
25 oC?
Pt|H2(g)(1 atm)|H+(test sol’n)║AgCl(s),Ag(s)|Cl-(2.80 M)
AgCl(s) + e-  Ag(s) + Cl - (aq)
Eo = 0.22V
[cathode; reduction]
H2 (g; 1atm)  2H+ (aq; unknown) + 2e-
Eo = 0.0V
[anode; oxidation]
H2 (g;1atm) + 2AgCl(s)  2Ag(s) + 2Cl - (aq) + 2H+ (aq, unknown)
[Cl - ]2 [H + ]2
Q =
H 2 (g;1atm)
Ecell = E
o
cell
Eocell = 0.22 V - 0.0 V = 0.22 V
0.0592 V
0.0592 V
[Cl - ]2 [H + ]2
o
× log Q = Ecell × log
n
2
H 2 (g;1atm)
0.0592 V
(2.80)2 [H + ]2
0.612 V = 0.22 V × log
= 0.22 V - 0.0296(log(7.84) + 2log[H + ])
2
1
-0.0296  2log[H + ] = 0.612 - 0.22 + 0.0296  0.894
-log[H + ] = pH = 7.07
2/18/2015
86
Summary Equations
charge
Charge = moles of e 
mol eCharge = nF
F - Faraday Constant
-
F =
96, 485 C
mol e-
(C - coulomb, SI unit of charge)
Eocell = Eocathode (reduction) - Eoanode (oxidation)
2H + (aq) + 2e-  H 2 (g)
Eo = 0.00 V (cathode - reduction)
Ag(s)  Ag + (aq) + e- Eo = 0.80 V (anode - oxidation)
2Ag(s) + 2H + (aq)  H 2 (g) + 2Ag + (aq)
Eocell = 0.00 V - 0.80 V = - 0.80 V
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87
Summary Equations
G = w max = - Ecell  charge
G = w max = - Ecell  nF
o
ΔG o = - Ecell
 nF
(components in standard states)
ΔG o = - RT lnK
-Eocell  n F = - RT lnK
ΔG = ΔGo + RTlnQ
-nF  Ecell = - n F  Eocell + RTlnQ
o
Ecell = Ecell
-
o
Ecell = Ecell
-
2/18/2015
RTlnQ
nF
(Nernst Equation)
0.0592 V
× log Q
n
(at 298.15 K)
88