21-1
ELECTROCHEMISTRY I
• VOLTAIC OR GALVANIC CELLS
• STANDARD ELECTRODE POTENTIALS
• NONSTANDARD CELLS
• Chapter 18.1-18.10 (McM)
• Chapter 21.2-21.5 (Silberberg)

21-2
Chapter 21
Electrochemistry:
Chemical Change and Electrical Work

21-3
Electrochemistry:
Chemical Change and Electrical Work
21.3 Cell Potential: Output of a Voltaic Cell
21.4 Free Energy and Electrical Work
21.5 Electrochemical Processes in Batteries

21-4
Goals & Objectives
• See the following Learning Objectives on
pages 894.
• Understand these Concepts:
• 21.4-12.
• Master these Skills:
• 21.2-7.

21-5
Electrochemical Cells
A voltaic cell uses a spontaneous redox reaction
(DG < 0) to generate electrical energy.
- The system does work on the surroundings.
A electrolytic cell uses electrical energy to drive a
nonspontaneous reaction (DG > 0).
- The surroundings do work on the system.
Both types of cell are constructed using two electrodes
placed in an electrolyte solution.
The anode is the electrode at which oxidation occurs.
The cathode is the electrode at which reduction occurs.

21-6
Figure 21.2 General characteristics of (A) voltaic and (B) electrolytic
cells.

21-7
Spontaneous Redox Reactions
A strip of zinc metal in a solution of Cu2+ ions will react
spontaneously:
Cu2+(aq) + 2e- → Cu(s) [reduction]
Zn(s) → Zn2+(aq) + 2e- [oxidation]
Cu2+(aq) + Zn(s) → Zn2+(aq) + Cu(s)
Zn is oxidized, and loses electrons to Cu2+.
Although e- are being transferred, electrical energy is not
generated because the reacting substances are in the
same container.

21-8
The Zinc-Copper Cell
• The cell consists of a strip of copper in a
1.0M CuSO4 solution and a strip of zinc in a
1.0M ZnSO4 solution. A wire and salt
bridge complete the circuit.
• The copper electrode gains mass and the
[Cu2+] decreases while the zinc electrode
loses mass and the [Zn2+] increases as the
cell operates.
• The initial voltage is 1.10 volts(v).

21-9
Figure 21.3 The spontaneous reaction between zinc and
copper(II) ion.

21-10
The Zinc-Copper Cell

21-11
The Zinc-Copper Cell

21-12
The Zinc-Copper Cell
• Anode reaction:
– Zn --> Zn2+ + 2e-
• Cathode reaction:
– Cu2+ + 2e- --> Cu
• Overall reaction:
– Zn + Cu2+ --> Zn2+ + Cu

21-13
The Zinc-Copper Cell
• Shorthand Notation
– Zn / Zn2+(1.0M) // Cu2+(1.0M) / Cu
• or
– Zn / Zn2+ // Cu2+ / Cu

21-14
Construction of a Voltaic Cell
Each half-reaction takes place in its own half-cell, so
that the reactions are physically separate.
Each half-cell consists of an electrode in an electrolyte
solution.
The half-cells are connected by the external circuit.
A salt bridge completes the electrical circuit.

21-15
Operation of the Voltaic Cell
Oxidation (loss of e-) occurs at the anode, which is
therefore the source of e-.
Zn(s) → Zn2+(aq) + 2e-
Over time, the Zn(s) anode decreases in mass and the
[Zn2+] in the electrolyte solution increases.
Reduction (gain of e-) occurs at the cathode, where the e-
are used up.
Cu2+(aq) + 2e- → Cu(s)
Over time, the [Cu2+] in this half-cell decreases and the
mass of the Cu(s) cathode increases.

21-16
Charges of the Electrodes
The anode produces e- by the oxidation of Zn(s). The
anode is the negative electrode in a voltaic cell.
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
Electrons flow through the external wire from the anode
to the cathode, where they are used to reduce Cu2+ ions.
The cathode is the positive electrode in a voltaic cell.

21-17
Figure 21.4A A voltaic cell based on the zinc-copper reaction.
Oxidation half-reaction
Zn(s) → Zn2+(aq) + 2e-
Reduction half-reaction
Cu2+(aq) + 2e- → Cu(s)
Overall (cell) reaction
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

21-18
Figure 21.4B A voltaic cell based on the zinc-copper reaction.
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
Oxidation half-reaction
Zn(s) → Zn2+(aq) + 2e-
After several hours, the
Zn anode weighs less as
Zn is oxidized to Zn2+.
Reduction half-reaction
Cu2+(aq) + 2e- → Cu(s)
The Cu cathode gains
mass over time as Cu2+
ions are reduced to Cu.

21-19
The Salt Bridge
The salt bridge completes the electrical circuit and allows
ions to flow through both half-cells.
As Zn is oxidized at the anode, Zn2+ ions are formed and
enter the solution.
Cu2+ ions leave solution to be reduced at the cathode.
The salt bridge maintains electrical neutrality by allowing
excess Zn2+ ions to enter from the anode, and excess
negative ions to enter from the cathode.
A salt bridge contains nonreacting cations and anions,
often K+ and NO3
-, dissolved in a gel.

21-20
Flow of Charge in a Voltaic Cell
Zn(s) → Zn2+(aq) + 2e- Cu2+(aq) + 2e- → Cu(s)
Electrons flow through the wire from anode to cathode.
Cations move through the salt
bridge from the anode solution
to the cathode solution.
Zn2+
Anions move through the salt
bridge from the cathode solution
to the anode solution.
SO4
2-
By convention, a voltaic cell is shown with the anode on
the left and the cathode on the right.

21-21
Active and Inactive Electrodes
An inactive electrode provides a surface for the reaction
and completes the circuit. It does not participate actively
in the overall reaction.
- Inactive electrodes are necessary when none of the reaction
components can be used as an electrode.
An active electrode is an active component in its half-
cell and is a reactant or product in the overall reaction.
Inactive electrodes are usually unreactive substances such
as graphite or platinum.

21-22
Figure 21.5 A voltaic cell using inactive electrodes.
Reduction half-reaction
MnO4
-(aq) + 8H+(aq) + 5e- → Mn2+(aq) + 4H2O(l)
Oxidation half-reaction
2I-(aq) → I2(s) + 2e-
Overall (cell) reaction
2MnO4
-(aq) + 16H+(aq) + 10I-(aq) → 2Mn2+(aq) + 5I2(s) + 8H2O(l)

21-23
Notation for a Voltaic Cell
Zn(s)│Zn2+(aq)║Cu2+(aq) │Cu(s)
The anode components
are written on the left.
The cathode components
are written on the right.
The single line shows a phase
boundary between the
components of a half-cell.
The double line shows that the half-
cells are physically separated.
The components of each half-cell are written in the same
order as in their half-reactions.
If needed, concentrations of dissolved components
are given in parentheses. (If not stated, it is assumed
that they are 1 M.)

21-24
graphite I-(aq)│I2(s)║MnO4
-(aq), H+(aq), Mn2+(aq) │graphite
Notation for a Voltaic Cell
The inert electrode is specified.
A comma is used to show components
that are in the same phase.

21-25
Sample Problem 21.2 Describing a Voltaic Cell with Diagram
and Notation
PROBLEM: Draw a diagram, show balanced equations, and write the
notation for a voltaic cell that consists of one half-cell with
a Cr bar in a Cr(NO3)3 solution, another half-cell with an
Ag bar in an AgNO3 solution, and a KNO3 salt bridge.
Measurement indicates that the Cr electrode is negative
relative to the Ag electrode.
PLAN: From the given contents of the half-cells, we write the half-
reactions. To determine which is the anode compartment
(oxidation) and which is the cathode (reduction), we note the
relative electrode charges. Electrons are released into the
anode during oxidation, so it has a negative charge. Since Cr
is negative, it must be the anode, and Ag is the cathode.

21-26
SOLUTION:
Sample Problem 21.2
Ag+(aq) + e- → Ag(s) [reduction; cathode]
Cr(s) → Cr3+(aq) + 3e- [oxidation; anode]
3Ag+ + Cr(s) → 3Ag(s) + Cr3+(aq)
The half-reactions are:
The balanced overall equation is:
The cell notation is given by:
Cr(s)│Cr3+(aq)║Ag+(aq)│Ag(s)
The cell diagram shows the anode on
the left and the cathode on the right.

21-27
The Copper-Silver Cell
• The cell consists of a strip of copper in a
1.0M CuSO4 solution and a strip of silver in
a 1.0M AgNO3 solution. A wire and salt
bridge complete the circuit.
• The copper electrode loses mass and the
[Cu2+] increases while the silver electrode
gains mass and the [Ag+] decreases as the
cell operates.
• The initial voltage is 0.46 volts(v).

21-28
The Copper-Silver Cell
• Anode reaction:
– Cu --> Cu2+ + 2e-
• Cathode reaction:
– e- + Ag+ --> Ag
– 2(e- + Ag+ --> Ag)
• Overall reaction:
– Cu + 2Ag+ --> Cu2+ + 2Ag
• Shorthand notation:
– Cu / Cu2+ // Ag+ / Ag

21-29
The Standard Hydrogen Electrode
Half-cell potentials are measured relative to a standard
reference half-cell.
The standard hydrogen electrode has a standard
electrode potential defined as zero (E°reference = 0.00 V).
This standard electrode consists of a Pt electrode with H2
gas at 1 atm bubbling through it. The Pt electrode is
immersed in 1 M strong acid.
2H+(aq; 1 M) + 2e- H2(g; 1 atm) E°ref = 0.00V

21-30
The Zinc-SHE Electrode
• The cell consists of a SHE and a strip of
Zn in a 1.0M ZnSO4 solution solution. A
wire and salt bridge complete the circuit.
• The zinc electrode loses mass and the
[Zn2+] increases while the [H+] decreases
in the SHE and hydrogen gas is produced
as the cell operates.
• The initial voltage is 0.763 volts(v).

21-31
Figure 21.7 Determining an unknown E°half-cell with the standard
reference (hydrogen) electrode.
Oxidation half-reaction
Zn(s) → Zn2+(aq) + 2e−
Reduction half-reaction
2H3O+(aq) + 2e- → H2(g) + 2H2O(l)
Overall (cell) reaction
Zn(s) + 2H3O+(aq) → Zn2+(aq) + H2(g) + 2H2O(l)

21-32
The Zinc-SHE Electrode
• Anode:
– Zn --> Zn2+ + 2e-
• Cathode:
– 2e- + 2H+ --> H2
• Overall:
– Zn + 2H+ --> Zn2+ + H2
• Zn / Zn2+ // H+ / H2

21-33
The Copper-SHE Cell

21-34
The Copper-SHE Cell
• Anode:
– H2 --> 2H+ + 2e-
• Cathode:
– 2e- + Cu2+ --> Cu
• Overall:
– H2 + Cu2+ --> 2H+ + Cu
• Cell voltage = 0.337v

21-35
Table 21.2 Selected Standard Electrode Potentials (298 K)
Half-Reaction E°(V)
+2.87
−3.05
+1.36
+1.23
+0.96
+0.80
+0.77
+0.40
+0.34
0.00
−0.23
−0.44
−0.83
−2.71
strengthofreducingagent
strengthofoxidizingagent
F2(g) + 2e− 2F−(aq)
Cl2(g) + 2e− 2Cl−(aq)
MnO2(g) + 4H+(aq) + 2e− Mn2+(aq) + 2H2O(l)
NO3
-(aq) + 4H+(aq) + 3e− NO(g) + 2H2O(l)
Ag+(aq) + e− Ag(s)
Fe3+(g) + e− Fe2+(aq)
O2(g) + 2H2O(l) + 4e− 4OH−(aq)
Cu2+(aq) + 2e− Cu(s)
N2(g) + 5H+(aq) + 4e− N2H5
+(aq)
Fe2+(aq) + 2e− Fe(s)
2H2O(l) + 2e− H2(g) + 2OH−(aq)
Na+(aq) + e− Na(s)
Li+(aq) + e− Li(s)
2H+(aq) + 2e− H2(g)

21-36
Comparing E°half-cell values
Standard electrode potentials refer to the half-reaction as
a reduction.
E° values therefore reflect the ability of the reactant to act
as an oxidizing agent.
The more positive the E° value, the more readily the
reactant will act as an oxidizing agent.
The more negative the E° value, the more readily the
product will act as a reducing agent.

21-37
Electrical Potential and the Voltaic Cell
When the switch is closed and no reaction is occurring,
each half-cell is in an equilibrium state:
Zn(s) Zn2+(aq) + 2e- (in Zn metal)
Cu(s) Cu2+(aq) + 2e- (in Cu metal)
Zn is a stronger reducing agent than Cu, so the position of
the Zn equilibrium lies farther to the right.
Zn has a higher electrical potential than Cu. When the
switch is closed, e- flow from Zn to Cu to equalize the
difference in electrical potential
The spontaneous reaction occurs as a result of the different
abilities of these metals to give up their electrons.

21-38
Cell Potential
A voltaic cell converts the DG of a spontaneous redox
reaction into the kinetic energy of electrons.
The cell potential (Ecell) of a voltaic cell depends on the
difference in electrical potential between the two
electrodes.
Cell potential is also called the voltage of the cell or the
electromotive forces (emf).
Ecell > 0 for a spontaneous process.

21-39
Standard Electrode Potentials
The standard electrode potential (E°half-cell) is the potential
of a given half-reaction when all components are in their
standard states.
By convention, all standard electrode potentials refer to
the half-reaction written as a reduction.
The standard cell potential depends on the difference
between the abilities of the two electrodes to act as
reducing agents.
E°cell = E°cathode (reduction) - E°anode (oxidation)

21-40
Sample Problem 21.3 Calculating an Unknown E°half-cell from E°cell
PROBLEM: A voltaic cell houses the reaction between aqueous bromine
and zinc metal:
Br2(aq) + Zn(s) → Zn2+(aq) + 2Br-(aq) E°cell = 1.83 V.
Calculate E°bromine, given that E°zInc = -0.76 V
PLAN: E°cell is positive, so the reaction is spontaneous as
written. By dividing the reaction into half-reactions, we
see that Br2 is reduced and Zn is oxidized; thus, the zinc
half-cell contains the anode. We can use the equation for
E°cell to calculate E°bromine.
SOLUTION:
Br2(aq) + 2e- → 2Br-(aq) [reduction; cathode]
Zn(s) → Zn2+(aq) + 2e- [oxidation; anode] E°zinc = -0.76 V

21-41
Sample Problem 21.3
E°cell = E°cathode − E°anode
1.83 = E°bromine – (-0.76)
1.83 + 0.76 = E°bromine
E°bromine = 1.07 V

21-42
Writing Spontaneous Redox Reactions
Each half-reaction contains both a reducing agent and an
oxidizing agent.
The stronger oxidizing and reducing agents react
spontaneously to form the weaker ones.
A spontaneous redox reaction (E°cell > 0) will occur
between an oxidizing agent and any reducing agent that
lies below it in the emf series (i.e., one that has a less
positive value for E°).
The oxidizing agent is the reactant from the half-reaction
with the more positive E°half-cell.

21-43
Using half-reactions to write a spontaneous redox reaction:
Sn2+(aq) + 2e- → Sn(s) E°tin = -0.14 V
Ag+(aq) + e- → Ag(s) E°silver = 0.80 V
Step 1: Reverse one of the half-reactions into an oxidation step
so that the difference between the E° values will be positive.
Here the Ag+/Ag half-reaction has the more positive E° value, so it
must be the reduction. This half-reaction remains as written.
We reverse the Sn2+/Sn half-reaction, but we do not reverse the sign:
Sn(s) → Sn2+(aq) + 2e- E°tin = -0.14 V

21-44
Step 2: Multiply the half-reactions if necessary so that the number
of e- lost is equal to the number or e- gained.
2Ag+(aq) + 2e- → 2Ag(s) E°silver = 0.80 V
Note that we multiply the equation but not the value for E°.
Sn(s) → Sn2+(aq) + 2e- E°tin = -0.14 V
2Ag+(aq) + 2e- → 2Ag(s) E°silver = 0.80 V
Step 3: Add the reactions together, cancelling common species.
Calculate E°cell = E°cathode - E°anode.
Sn(s) + 2Ag+(aq) → 2Ag(s) + Sn2+(aq) E°cell = 0.94 V
E°cell = E°silver – E°tin = 0.80 – (-0.14) = 0.94 V

21-45
Writing Spontaneous Cell
Reactions
• Write the spontaneous cell reaction and
Eo
cell expected when a zinc-silver cell is
set up in the usual fashion.
• Write the shorthand notation for this cell.
• Draw the cell indicating the anode,
cathode, salt bridge, direction of electron
flow, and direction of ion flow from the
salt bridge.

21-46

21-47

21-48
Writing Spontaneous Cell
Reactions
• Write the spontaneous cell reaction and
Eo
cell expected when an Fe3+/Fe2+-MnO4
-
/Mn2+ cell is set up in the usual fashion.
• Fe3+ (0.771v) Fe2+ MnO4
- (1.51v) Mn2+
• Write the shorthand notation for this cell.

21-49

21-50
Writing Spontaneous Cell
Reactions
• Write the spontaneous cell reaction and
Eo
cell expected when a cell using the
following half-reactions is set up in the
usual fashion.
• NO3
- (0.96v) NO
• H3AsO4
(0.58v) H3AsO3
• Write the shorthand notation for this cell.

21-51

21-52

21-53
Electrochemical Processes in Batteries
A primary battery cannot be recharged. The battery is
“dead” when the cell reaction has reached equilibrium.
A secondary battery is rechargeable. Once it has run
down, electrical energy is supplied to reverse the cell
reaction and form more reactant.
A battery consists of self-contained voltaic cells arranged
in series, so their individual voltages are added.

21-54
Alkaline battery.Figure 21.15
Anode (oxidation): Zn(s) + 2OH-(aq) → ZnO(s) + H2O(l) + 2e-
Cathode (reduction): MnO2(s) + 2H2O(l) + 2e- → Mn(OH)2(s) + 2OH-(aq)
Overall (cell) reaction:
Zn(s) + MnO2(s) + H2O(l) → ZnO(s) + Mn(OH)2(s) Ecell = 1.5 V

21-55
Silver button battery.Figure 21.16
Anode (oxidation): Zn(s) + 2OH-(aq) → ZnO(s) + H2O(l) + 2e-
Cathode (reduction): Ag2O(s) + H2O(l) + 2e- → 2Ag(s) + 2OH-(aq)
Overall (cell) reaction: Zn(s) + Ag2O(s) → ZnO(s) + 2Ag(s)
Ecell = 1.6 V
The mercury battery uses HgO as the oxidizing agent instead of
Ag2O and has cell potential of 1.3 V.

21-56
Figure 21.17 Lithium battery.
Anode (oxidation):
3.5Li(s) → 3.5Li+ + 3.5e-
Cathode (reduction):
AgV2O5.5 + 3.5Li- + 3.5e- → Li3.5V2O5.5
Overall (cell) reaction:
AgV2O5.5 + 3.5Li(s) → Li3.5V2O5.5
The primary lithium battery is widely used
in watches, implanted medical devices,
and remote-control devices.

21-57
Lead-acid battery.Figure 21.18
The lead-acid car battery is a secondary battery and is rechargeable.

21-58
Anode (oxidation): Pb(s) + HSO4
-(aq) → PbSO4(s) + H+(aq) + 2e-
Cathode (reduction):
PbO2(s) + 3H+(aq) + HSO4
-(aq) + 2e- → PbSO4(s) + 2H2O(l)
Overall (cell) reaction (discharge):
PbO2(s) + Pb(s) + H2SO4(aq) → 2PbSO4(s) + 2H2O(l) Ecell = 2.1 V
The reactions in a lead-acid battery:
Overall (cell) reaction (recharge):
2PbSO4(s) + 2H2O(l) → PbO2(s) + Pb(s) + H2SO4(aq)
The cell generates electrical energy when it discharges as a voltaic cell.

21-59
Nickel-metal hydride batteryFigure 21.19
Anode (oxidation): MH(s) + OH-(aq) → M(s) + H2O(l) + e-
Cathode (reduction): NiO(OH)(s) + H2O(l) + e- → Ni(OH)2(s) + OH-(aq)
Overall (cell) reaction:
MH(s) + NiO(OH)(s) → M(s) + Ni(OH)2(s) Ecell = 1.4 V

21-60
Lithium-ion battery.Figure 21.20
Anode (oxidation):
LixC6(s) → xLi+ + xe- + C6(s)
Cathode (reduction):
Li1-xMn2O4(s) + xLi+ + xe- → LiMn2O4(s)
Overall (cell) reaction:
LixC6(s) + Li1-xMn2O4(s) → LiMn2O4(s)
Ecell = 3.7 V
The secondary (rechargeable) lithium-ion battery is used to power laptop
computers, cell phones, and camcorders.

21-61
In a fuel cell, also called a flow cell, reactants enter the
cell and products leave, generating electricity through
controlled combustion.
Fuel Cells
Reaction rates are lower in fuel cells than in other
batteries, so an electrocatalyst is used to decrease
the activation energy.

21-62
Figure 21.21 Hydrogen fuel cell.
Anode (oxidation): 2H2(g) → 4H+(aq) + 4e-
Cathode (reduction): O2(g) + 4H+(aq) + 4e- → 2H2O(g)
Overall (cell) reaction: 2H2(g) + O2(g) → 2H2O(g) Ecell = 1.2 V

21-63
Walther Hermann Nernst
• Walther Hermann Nernst
(June 25, 1864 – November 18,
1941) was a German physicist
who is known for his theories
behind the calculation of
chemical
• affinity as embodied in the
third law of thermodynamics,
for which he won the 1920
Nobel Prize in chemistry. He
is also known for developing
the Nernst equation.

21-64
Nonstandard Cells
• The Nernst equation
– E = Eo - (0.0592/n)logQ
– where n = moles of electrons transferred
– and Q = reaction quotient
• For Cu2+ + e- --> Cu+ Eo = +0.153
• E = Eo - (0.0592/1)log([Cu+]/[Cu2+])
– n = 1
– Q = [Cu+]/[Cu2+]

21-65
The Nernst Equation
• For the above cell when
[Cu2+]=[Cu+]=1.0M
• E = Eo - (0.0592/1)log(1)
• E = Eo

21-66
Nonstandard Cells
• Calculate the potential for the Cu2+/Cu+
cell at 25oC when [Cu2+] = 2.0x10-3M
and [Cu+]=6.0x10-3M. Eo = 0.153v

21-67

21-68

21-69
Nonstandard Cells
• Determine the electrode potential for the
hydrogen electrode when
• [H+] = 1.0x10-3M and
• the p(hydrogen gas) = 0.50atm.

21-70

21-71
Nonstandard Cells
• Determine the initial potential for the
following cell reaction when
• the [Fe3+]=(1.0x10-2),
• the [Fe2+] = 0.10M,
• the [Sn4+]=1.0M and
• the [Sn2+]=0.10M
• Fe3+ (0.771v) Fe2+ Sn4+ (0.15v) Sn2+

21-72

21-73