Download ELECTROCHEMISTRY Chapter 21

CHAPTER 20
PRINCIPLES OF REACTIVITY:
ELECTRON TRANSFER
REACTIONS
1-11 + all bold numbered problems
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ELECTROCHEMISTRY
Electric automobile
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Why Study Electrochemistry?
• Batteries
• Corrosion
• Industrial
production of
chemicals such
as Cl2, NaOH,
F2 and Al
A rusted car.
• Biological
redox reactions
The heme group
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20.1 OXIDATION-REDUCTION
REACTIONS
• Redox reactions make up
electrochemical cells.
• Those that produce energy are called
batteries or galvanic cells (voltaic).
• Those that require energy to operate are
called electrolysis cells.
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TRANSFER REACTIONS
Atom transfer
HOAc + H2O ---> OAc- + H3O+
no change in oxidation state of any of the elements
Electron transfer
Cu(s) + 2 Ag+(aq) ---> Cu2+(aq) + 2 Ag(s)
Cu looses 2 eOIL
and Au+ gains 1 eRIG
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Electron Transfer Reactions
• Electron transfer reactions are
oxidation-reduction or redox
reactions.
• Redox reactions can result in the
generation of an electric current
or be caused by imposing an
electric current.
• Therefore, this field of chemistry
is often called
ELECTROCHEMISTRY.
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CHAPTER OVERVIEW
• This chapter examines electron
transfer reactions,
redox
Reduction/Oxidation
Note: Oil Rig =
Oxidation is Loss of an electron
Reduction is Gain of an electron
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Review of Terminology for
Redox Reactions
• OXIDATION—loss of electron(s) by
a species; increase in oxidation
number.
• REDUCTION—gain of electron(s);
decrease in oxidation number.
• OXIDIZING AGENT—electron
acceptor; species is reduced.
• REDUCING AGENT—electron
donor; species is oxidized.
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Solid interacting with aqueous ions
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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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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
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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.
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CHEMICAL CHANGE -->
ELECTRIC CURRENT
Zn metal
2+ ions
Cu
Cu2+
ions
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) + 2eCu2+ is reduced and is the oxidizing agent
Cu2+(aq) + 2e- ---> Cu(s)
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CHEMICAL CHANGE -->
ELECTRIC CURRENT
Zn metal
2+ ions
Cu
Cu2+
ions
Electrons are
transferred from Zn to
Cu2+, but there is no
useful electric current.
Oxidation: Zn(s) ---> Zn2+(aq) + 2eReduction: Cu2+(aq) + 2e- ---> Cu(s)
-------------------------------------------------------Cu2+(aq) + Zn(s) ---> Zn2+(aq) + Cu(s)
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CHEMICAL CHANGE -->
ELECTRIC CURRENT
•To obtain a useful current,
we separate the oxidizing
wire
e le c t ro ns
and reducing agents so
salt
Zn
bridge
that electron transfer
occurs thru an external
wire.
Zn 2+ ions
•This is accomplished in a
GALVANIC or VOLTAIC
cell.
•A group of such cells is
called a battery.
Cu
Cu2+ ions
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•Electrons travel thru external wire.
•Salt bridge allows anions and cations to move
between electrode compartments.
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•This maintains electrical neutrality.
Electrons move
from anode to
cathode in the wire.
Anions & cations
move thru the salt
bridge.
Electrochemical
Cell or Battery
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Balancing Redox Equations
OXIDATION-REDUCTION REACTIONS
• To be useful, the reactions for these
cells must be balanced.
• The simplest method of balancing is
the ion-electron or half-reaction
method.
• We learned this method in the first
half of General Chemistry, but a quick
review is in order.
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OXIDATION-REDUCTION REACTIONS
• Oxidation is the loss of electron,
and reducing agents are oxidized.
• Reduction is the gain of
electrons, and oxidizing agents
are reduced.
• The simple steps in balancing a
redox equation are:
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BALANCING REDOX REACTIONS
1. Separate the reaction into half reactions.
2. Perform a mass balance for each element
beginning with the element being oxidized
or reduced. Next balance the oxygen
atoms with water and the hydrogen atoms
with hydrogen ions.
3. Perform a charge balance by adding
electrons to balance the charge.
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BALANCING REDOX REACTIONS
4. If the reaction is in basic
solution, add as many hydroxide
ions to both sides as there are
hydrogen ions and simplify by
forming water. Eliminate the
excess water, if any occurs on
both sides of the reaction.
5. Multiply each half-reaction by
the appropriate factor so the
electrons are eliminated when the
half-reactions are added.
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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
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Balancing Equations
Consider the reduction of
Ag+ ions with copper metal.
Cu + Ag+
-- give --> Cu2+ + Ag
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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 for mass. Already
done in this case.
Step 3: Balance each half-reaction for
charge by adding electrons.
Ox
Red
Cu ---> Cu2+ + 2eAg+ + e- ---> Ag
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Balancing Equations
Step 4: Multiply each half-reaction by a
factor that means 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+
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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
VO2+ ---> VO2+ + H2O
2 H+
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+ + 2e-
Red
e- + 2 H+ + VO2+ ---> VO2+ + H2O
Step 4:
Multiply by an appropriate factor.
Ox
Zn ---> Zn2+ + 2e-
Red
2e- + 4 H+ + 2 VO2+
---> 2 VO2+ + 2 H2O
Step 5:
Add 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.
MORE
PRACTICE!!29
20.2 CHEMICAL CHANGES LEADING
TO ELECTRIC CURRENT
• An electrochemical cell has two
electrodes, the anode where oxidation
occurs, and the cathode where
reduction occurs.
• The compartment containing the anode
is called the anodic compartment
• The compartment containing the
cathode is called the cathodic
compartment.
• These two half-cells are frequently
connected by a porous barrier or a salt
bridge.
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CHEMICAL CHANGES LEADING
TO ELECTRIC CURRENT
• The electrical current is carried by
electrons in the metals present, and
by ions in the solution.
• Ions carrying charge are called
electrolytes.
• Electrons travel from the anode (-) to
the cathode (+)
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Reduction
Oxidation
Cathodic
compartment
Anodic
compartment
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Illustrates a general cell
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Abbreviated Cell Notation
Anode | Anodic sol’n || Cathodic sol’n | Cathode
This simple notation is very efficient
2Fe3+(aq) + H2(g)  2Fe2+(aq) + 2H+(aq)
Pt|H2(1atm)|H+(1M)||Fe2+(1M),Fe3+(1M)|Pt
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Anode | Anodic sol’n || Cathodic sol’n | Cathode
Cu|Cu2+(1.0 M)||Ag+(1.0 M)|Ag
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20.3 ELECTROCHEMICAL
CELLS AND POTENTIALS
• The electrical potential is called the
electromotive force or emf, and is given
the symbol E.
• The unit of charge is called the coulomb
• emf is measured in joules/coulomb or
volts, V
• Under standard conditions, 1.0 M
for solutes, 1.0 bar for gases, and
298.15 K, the emf is given the symbol
Eo, and called the standard potential.
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Eo vs. DGo
• The electrical work is nFE, where F is
Faraday's constant, 96, 485 coulombs/
mole electrons or J/V mole, and n is
the moles of electrons in the balanced
equation.
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Eo vs. DGo
This leads to the following equation
between E and DG: DG = - nFE.
• This can be derived from:
G = H -TS = E + PDV - TDS
• Since DE = q + w and w = - PDV – wext
• If the process is done reversibly at
constant temperature and pressure,
q = TDS.
 DG = q - PDV – wext + PDV - TDS
• This leads to DG = - wext.
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Eo vs. DGo
In the standard state, DGo = - nFEo.
• Now we can use E and Eo to predict
reaction spontaneity, i.e. whether a
reaction is product-favored.
• Calculations of DGo from Eo and visa
versa are algebraic.
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Calculating the Potential Eo of
an Electrochemical Cell
• All standard potentials are bases
on a relative scale where the
hydrogen half-cell is assigned a
value to 0.00 V at all temperature,
and is called the standard
hydrogen electrode, SHE.
• All electrode potentials are written
as reductions and derive their
values from direct or indirect
measurements vs. the SHE.
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Calculating the Potential Eo of
an Electrochemical Cell
• Demonstrates this measure for the zinc
electrode potential.
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Demonstrates
this measure for
the copper
electrode
potential.
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CELL POTENTIAL, E
Zn and Zn2+,
anode
4
3
Cu and Cu2+,
cathode
• Electrons are “driven” from anode to cathode by
an electromotive force or emf.
• For Zn/Cu cell, this is indicated by a voltage of
1.10 V at 25 C and when [Zn2+] and [Cu2+] = 1.0 M.
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CELL
POTENTIAL, E
• For Zn/Cu cell, voltage 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.
2 I- ---> I2 + 2e-
2 H2O + 2e- ---> 2 OH- + H2
------------------------------------------------2 I- + 2 H2O --> I2 + 2 OH- + H2
If we know Eo for each half-reaction,
we could get Eo for net reaction.
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CELL POTENTIALS,
o
E
• Can’t measure 1/2 reaction Eo
directly. Therefore, measure it
relative to a STANDARD HALF CELL,
SHE.
2 H+(aq, 1 M) + 2e- --> H2(g, 1 atm)
Eo = 0.0 V
Typically made of Pt or Au, are inert, and important
because SHE’s can act as either an anode or
cathode, difficult to maintain, not used often 
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Other Reference Electrodes, know Eo, easy to make
Self-contained, only make once, sealed, last a longtime
||Hg2Cl2(sat),KCl(xM)|Hg
||AgCl(sat), KCl(xM)|Ag
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||TlCl(sat),KCl(xM)|Tl
Zn/Zn2+ half-cell hooked to a SHE.
Eo for the cell = +0.76 V
Remember Eo
for SHE = 0.0 V
Volts
Zn
-
+
Salt Bridge
H2
Zn2+
Zn
Zn2+ + 2eOXIDATION
ANODE
H+
2 H+ + 2eH2
REDUCTION
CATHODE
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-
Zn
Volts
Salt Bridge
+
H2
H+
Zn 2+
2+
Zn
Zn + 2 e- 2 H + + 2 eOXIDATION
REDUCTION
ANODE
CATHODE
Remember Eo
for SHE = 0.0 V
H2
Overall reaction is reduction of H+ by Zn metal.
Zn(s) + 2 H+ (aq) --> Zn2+ + H2(g)
Eo = + 0.76 V
Therefore, Eo for Zn ---> Zn2+ (aq) + 2e- is
+ 0.76 V. Reduction potential is - 0.76 V.
Zn is a better reducing agent (Zn gets oxidized)
than) H2.
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Cu/Cu2+ and H2/H+ Cell
Eo = +0.34 V
Volts
Cu
+
Salt Bridge
H2
Cu2+
Cu2+ + 2eCu
REDUCTION
CATHODE
H+
H2
2 H+ + 2eOXIDATION
ANODE
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Cu/Cu2+ and H2/H+ Cell
Volts
Cu
+
Salt Bridge
H2
Cu 2+
Cu 2+ + 2eCu
REDUCTION
CATHODE
•
•
•
•
H+
H2
2 H + + 2eOXIDATION
ANODE
Overall reaction is reduction of Cu2+ by H2 gas.
Cu2+ (aq) + H2(g) ---> Cu(s) + 2 H+(aq)
Measured Eo = + 0.34 V
Therefore, Eo for Cu2+ + 2e- ---> Cu is
+ 0.34 V so Cu is a better Oxidizing agent (Cu gets
reduced) the H+
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COMBINING Voltaic CELLS
• Lets Combine the work we just did
with the Zinc and Copper Cells
and combine them into ONE cell
and see the net effect
Zn2+(aq) + 2e- ---> Zn(s)
Cu2+(aq) + 2e- ---> Cu(s)
Eo = - 0.76 V
Eo = + 0.34 V
Zn is a better reducing agent than H+
Cu is a better Oxidizing agent than H+
From this info we know Zn MUST be oxidized
and that Cu MUST be reduced
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Zn/Cu Electrochemical Cell
wire
elect rons
Anode,
negative,
source of
electrons
(oxidation)
Zn
Zn2+ ions
salt
bridge
Cu
Cu2+ ions
Cathode,
positive,
sink for
electrons
(reduction)
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 (calculated) = + 1.10 V
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Uses of
o
E
Values
These experiments show we can
a) decide on relative ability of
elements to act as reducing
agents (or oxidizing agents)
b) assign a voltage to a halfreaction that reflects this ability.
wire
elect rons
Zn
Zn2+ ions
salt
bridge
Cu
Cu2+ ions
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USING STANDARD POTENTIALS
Eocell = Eocathodic - Eoanodic
Eocell = Eoreduction + Eooxidation
• The first equation uses the values right
out of the table of standard electrode
potentials.
• The table can also give us information
about which reactions will be productfavored (produce current) in the
standard state since these reactions
with have a positive Eocell.
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56
57
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USING STANDARD POTENTIALS
• The following is an
abbreviated table form the
previous tables to give
practice at using them.
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TABLE OF STANDARD
POTENTIALS
Oxidizing
ability of Ion
Eo (V)
Cu2+ + 2e-
Cu
+0.34
2 H+ + 2e
H2
0.00
Zn2+ + 2e-
Zn
-0.76
reducing ability
of element
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Standard Redox Potentials, Eo
oxidizing
ability of ion
Eo (V)
Cu2+ + 2e-
Cu
+0.34
2 H+ + 2e-
H2
0.00
Zn2+ + 2e-
Zn
-0.76
reducing ability
of element
• Any substance on
the right will reduce
any substance
higher than it on the
left.
• Zn can reduce H+
and Cu2+.
• H2 can reduce Cu2+
but not Zn2+
• Cu cannot reduce
H+ or Zn2+.
61
Standard Redox Potentials, Eo
oxidizing
ability of ion
Eo (V)
Cu2+ + 2e-
Cu
+0.34
2 H+ + 2e-
H2
0.00
Zn2+ + 2e-
Zn
-0.76
reducing ability
of element
•
•
•
•
•
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 (calculated) = + 1.10 V
62
Using Standard Potentials, Eo
See next slide
Which is the best oxidizing agent:
O2, H2O2, or Cl2? _________________
• Which is the best reducing agent:
Hg, Al, or Sn? ____________________
• Which is the correct direction for the following
reaction to produce a favorable potential?
2Al(s) + 3 Cl2(aq) ---> AlCl3(aq)
AlCl3(aq) ---> 2Al(s) + 3 Cl2(aq)
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64
o
E
for a Voltaic Cell
Volts
Cd
Fe
Salt Bridge
Cd2+
Cd --> Cd2+ + 2eor
Cd2+ + 2e- --> Cd
Fe2+
Fe --> Fe2+ + 2eor
Fe2+ + 2e- --> Fe
See previous slide
65
o
E
for a Voltaic Cell
Volts
Cd
Fe
Salt Bridge
Cd2+
Fe2+
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 = + 0.04 V
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Examples of Cells that NEED electricity
(opposite of Voltaic or Galvanic cells
Production of
Oxygen and
Hydrogen
67
Electrolysis of SnCl2(aq)
Sn2+(aq) + 2Cl-(aq)  Sn(s) and Cl2(g)
Anode (+)
Cathode (-)
68
NaI
phenolphthalein
More on this
later, industrial
applications
OH- detected
69
o
E
and D
o
G
Eo is related to DGo, the free
energy change for the
reaction.
o
DG
= -nF
o
E
where F = Faraday constant
= 9.6485 x 104 J/V•mol
and n is the number of moles
of electrons transferred
Michael Faraday
1791-1867
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Michael Faraday
1791-1867
Originated the terms anode,
cathode, anion, cation,
electrode.
Discoverer of
• electrolysis
• magnetic props. of matter
• electromagnetic induction
• benzene and other organic
chemicals
Was a popular lecturer.
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o
E
and D
o
G
DGo = - n F Eo
For a product-favored reaction
Reactants ----> Products
DGo < 0 and so Eo > 0
Eo is positive, results in current
For a reactant-favored reaction
Reactants <---- Products
DGo > 0 and so Eo < 0
Eo is negative , needs current
72
ELECTROCHEMICAL CELLS AT
NON STANDARD CONDITIONS
• We learned the relationship between
DG and DGo in chapter 19.
DG
= DGo + RT ln Q.
• This yields: -nFE = -nFEo + RT ln Q
• At 298.15 K, we solve for E
E = Eo - [0.0592 V/n] log Q or
E = Eo - [0.0257 V/n] ln Q
73
20.5 Eo and the Equilibrium Constant
Nernst Equations
• At equilibrium for the cell reaction,
E = 0, (equivalent to saying the cell is
dead) Therefore:
Eo = [0.0592 V/n] log K or
Eo = [0.0257 V/n] ln K
• These equations can be solved for K
using the inverse functions.
• n is the number of moles of e-
74
Calculating Ksp, Kf, and other K's
from cell potentials
• In the laboratory we will be calculating Ksp's and
Kf 's for reactions using electrochemical cells.
• Determine the Kf for AlF6-3 from the cell below:
E = 2.465 V
Al |10.0 mL 0.10 M Al(NO3)3,40.0 mL 1.0 M NaF ||0.10 M CuSO4|Cu
2 Al + 3 Cu+2 ===> 2 Al+3 + 3 Cu
Al+3 + 6F- <===> AlF6-3
Given: [F-] = 0.68 M, [AlF6-3] = 0.020 M
Note: no reaction between Al3+ or Cu2+ and SO4-2
75
2 Al + 3 Cu+2 ===> 2 Al+3 + 3 Cu
Eo = 1.676 + .337 = 2.013 V
Ecell = 2.465 V (given), 0.10 M Cu+2
Ecell = Eo - [0.0257 V/n] ln K
2.465 = 2.013 - [0.0257/6] Ln (Al+3)2/(.1)3
solved for Al+3, note that as soon as Al+3 forms it
immediately reacts with F-, so Al+3 is NOT .1M
Which gives: [Al+3] = 4.0x10-25 M
Given: [F-] = 0.68 M, [AlF6-3] = 0.020 M
Sub these values into Kf
Kf = 5.0 x 1023
76
21.6 BATTERIES AND FUEL CELLS
• There are two types of batteries:
• Primary, which are not
rechargeable
• Secondary, or storage batteries
which are easily recharged.
77
Primary Batteries
• 3 types: Dry, Alkaline, Button (Hg, Li)
• The dry cell produces H2 gas
• The battery has a zinc anode and
reduces ammonium ion to ammonia and
hydrogen gas at a graphite cathode.
• The net-reaction does not contain the
hydrogen gas since MnO2 is added to
oxidize the hydrogen gas to hydrogen
ion.
• This battery has a voltage of 1.5 Volts
when new.
78
Dry Cell Battery
Common
dry cell
79
Anode (-)
Zn ---> Zn2+ + 2eCathode (+)
2 NH4+ + 2e- --->
2 NH3 + H2
1.5 Volts
79
Problem alkaline & button: gasses formed
cathode
2NH + 2e  2NH3 (g ) + H 2 (g )
anode
Zn  Zn
+
4
-
2+
+ 2e
-
The gasses NH3 (g ) + H2 (g ) are removed:
Zn
2+
+ 2NH3 (g ) + 2Cl  Zn(NH3 )2Cl2 ( s)
-
and
2MnO2 ( s) + H 2 (g )  Mn 2O3 (s) + H 2O(l )
80
Primary Batteries: No H2 Gas produced
• The alkaline battery has a zinc anode
and a graphite cathode, the MnO2 is
reduced directly to Mn2O3 in a basic
environment.
• It's voltage is 1.54 Volts.
• The mercury and lithium batteries are
button shaped and have voltages of
1.35 V and 3.0 V respectively.
• 50% stronger then a dry cell
81
Alkaline Battery
Nearly same reactions as in
common dry cell, but under basic
conditions (1.54V)
82
Mercury Battery
Anode:
Zn is reducing agent under basic conditions
Cathode:
HgO + H2O + 2e- ---> Hg + 2 OH-
Primary, Hg (1.35V, Li (3.0V)
83
Note OH-, either KOH or NaOH
cathode
anode
Zn(s) + 2OH-  ZnO + 2H2O(l ) + 2e2MnO2 + H2O(l ) + 2e  MnO2 + 2OH
-
-
Also note that Zn here is the cathode, it is the anode in a dry cell
84
At anode
Primary
85
Primary: MnO replaced with-Hg/Li (1.35V, Li (3.0V)86
Secondary Batteries
• The net-reaction turns lead and
lead(IV)oxide, in the presence of
sulfuric acid, into lead(II)sulfate and
water
• The nickel-cadmium battery is a
rechargeable battery.
Fuel Cells
• Fuel cells convert hydrogen and
oxygen gas into water and electrical
energy without combustion.
87
88
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
89
Ni-Cad Battery
Anode (-)
Cd + 2 OH- ---> Cd(OH)2 + 2eCathode (+)
NiO(OH) + H2O + e- ---> Ni(OH)2 + OH-
90
CORROSION: REDOX REACTIONS
IN THE ENVIRONMENT
• Corrosion of iron produces
iron(III)oxide and other iron oxides
depending on the conditions,
moisture and oxygen levels.
• Other active metals also corrode
and form oxides, but frequently this
is in the form of a stable oxide
coating on the metal surface.
• The oxides of iron flake off easily,
exposing fresh iron surfaces for
additional oxidation.
91
CORROSION: REDOX REACTIONS
IN THE ENVIRONMENT
• These processes are true
electrochemical processes and a cathode
and an anode can be identified.
• The presence of conducting ions
enhances the process. (Salt water spray,
salt water from ice suppression, etc.)
• One protection process is called anodic
inhibition and involves changing the
surface of the metal.
• Paint and other coatings are examples.
92
CORROSION
• A more effective method is
cathodic protection which places
the metal in electrical contact with
a more active metal like
magnesium or zinc.
• The latter with iron is called
galvanization and the former with
iron is called a sacrificial anode.
93
Cathodic Protection
94
Fe corrosion to rust Fe2O3
95
Why Aluminum does not rust
96
20.8 ELECTROLYSIS:
CHEMICAL CHANGE FROM
ELECTRIC ENERGY
• When a reaction is driven by an
external electrical energy source
the process is called electrolysis.
• The anode and cathode of such
cells are the sites of oxidation and
reduction as in a galvanic cell, but
the polarity of the electrodes is
reversed.
• The anode is now (+) and
the cathode (-) .
97
ELECTROLYSIS
• For the electrolysis of molten NaCl, Na+ is
reduced to Na, and Cl- is oxidized to Cl2.
• In aqueous solution, H2O is reduced to H2
and OH- and Cl- is oxidized to Cl2.
• These decisions are made by consulting
the standard electrode potential table to
determine which species are most easily
oxidized and reduced.
98
ELECTROLYSIS
• The oxidation and reduction of water
can be a little tricky since the table is a
Standard electrode potential table
which means the all reactants and
products are present at standard
concentrations, 1M and 1 atm.
• Reduction of water to form H2(g) occurs
at - 0.414 V when concentration is
taken into account, and oxidation of
water to form O2(g) occurs at - 0.816 V.
• Over-voltages for hydrogen and oxygen
gases also complicate matters.
99
Electrolysis of Aqueous NaOH
Electric Energy ----> Chemical Change
Anode (+)
Eo = - 0.40 V
4 OH- ---> O2(g) + 2 H2O + 2eCathode (-)
Eo = - 0.83 V
4 H2O + 4e- ---> 2 H2 + 4 OHEo for cell = - 1.23 V
100
Electrolysis
Electric Energy ---> Chemical Change
electrons
BATTERY
+
Anode
Cathode
Cl-
Na+
Electrolysis of
molten NaCl.
Here a battery
“pumps”
electrons from
Cl- to Na+.
101
Electrolysis of Molten NaCl
Anode (+)
Eo = - 1.36 V
2 Cl- ---> Cl2(g) + 2eEo = - 2.71 V
electrons
Cathode (-)
BATTERY
Na+ + e- ---> Na
+
Anode
Cathode
Cl-
Na+
Eo for cell = - 4.07 V
External energy needed
because Eo is (-).
Note that signs of electrodes
are reversed from batteries.
102
Simpler process for making Na(s)?
• We just saw the molten process of
taking NaCl and by applying a
voltage (electrolysis) we can make
both Na(s) and Cl2(g)
• Wouldn’t it be easier to take seawater, perform electrolysis on the
“brine” to get Na(s) and Cl2(g)
103
What are the possible reactions
At the anode
Na+ + e-  Na(s)
2H2O(l) + 2e-  H2(g) + 2OH-
E = -2.714
E = -0.8277
At the cathode
2Cl-  Cl2 + 2e2H2O(l)  O2(g) + 4H+ + 4e-
E = -1.360
E = -1.229
Once a potential is applied the process
with the lowest potential will start FIRST!
104
Once a potential is applied the process
with the lowest potential will start FIRST!
At the anode
Na+ + e-  Na(s)
2H2O(l) + 2e-  H2(g) + 2OH-
E = -2.714
E = -0.8277
At the cathode
2Cl-  Cl2 + 2e2H2O(l)  O2(g) + 4H+ + 4e-
E = -1.360
E = -1.229
The process with the lowest potential:
4H2O(l) + 4e-  2H2(g) + 4OHE = -0.8277
2H2O(l)  O2(g) + 4H+ + 4eE = -1.229
2H2O(l)  O2(g) + 2H2
Ecell = -2.057V 105
Electrolysis of Aqueous
NaCl
Since Na(s) can not
be made from seawater due to
hydrogen and
oxygen gas are made
another technique is
used on sea-water
106
“Brine Process” for making NaOH
from sea-water
107
Electrolysis of Aqueous NaCl
Cells like these are the source of
NaOH and Cl2.
In 1995
25.1 x 109 lb Cl2
26.1 x 109 lb NaOH
Also the source
of NaOCl for
use in bleach.
108
Electrolysis of Aqueous CuCl2
Anode (+)
Eo = - 1.36 V
2 Cl- ---> Cl2(g) + 2eCathode (-)
Eo = + 0.34 V
Cu2+ + 2e- ---> Cu
electrons
Eo for cell = - 1.02 V
BATTERY
Note that Cu+2 is more
easily reduced than
either H2O or Na+.
The water process potential:
4H2O(l) + 4e-  2H2(g) + 4OH2H2O(l)  O2(g) + 4H+ + 4e2H2O(l)  O2(g) + 2H2
+
Anode
E = -0.8277
E = -1.229
Ecell = -2.057V
Cathode
Cl- Cu2+
H2O
109
THE COMMERCIAL PRODUCTION
OF CHEMICALS BY
ELECTROCHEMICAL METHODS
• Aluminum
• Chlorine, sodium hydroxide,
and hydrogen
• These chemicals are extremely
important to the economy.
• The power required for these
processes can be enormous.
110
111
Producing Aluminum
2 Al2O3 + 3 C ---> 4 Al + 3 CO2
Charles Hall (1863-1914) at age of 19 developed the Al
electrolysis process. Founded Alcoa!
112
COMMERCIAL PRODUCTION
OF CHEMICALS
• Power is measured in watts and
has basic units of
joules/second.
• Power is calculated from the
equation
P = V I, (J/C) (C/S) = J/S
• A kilowatt-hour is an energy
unit and is equal to
3.60x106J:
3600 S/hr times 1000 W/kW 113
20.8 COUNTING ELECTRONS
• Current, I, is measured in amperes and
has basic units of coulombs/second.
• Faraday's Law tells us that the current
times the time can be used to calculate the
mass of material oxidized or reduced.
• F, Faraday's constant, is 9.65x104C/mol e-.
• These unit factor problems change:
current x time ==> coulombs ==>
moles e- ==> moles oxidized or
reduced species ==> mass
114
COUNTING ELECTRONS
• The balanced half-reaction
provides the conversion from
moles electrons to moles of
species sought.
• Calculate the grams of aluminum
formed at the cathode if a current
of 4.40 A flows for 1.50 hours.
–Let’s do an example first.
115
Quantitative Aspects of
Electrochemistry
Consider electrolysis of aqueous
silver ion.
Ag+ (aq) + e- ---> Ag(s)
1 mol e---> 1 mol Ag
If we could measure the moles of
e-, we could know the quantity of
Ag formed.
But how to measure moles of e-?
116
Quantitative Aspects of
Electrochemistry
Consider electrolysis of aqueous silver ion.
Ag+ (aq) + e- ---> Ag(s)
1 mol e---> 1 mol Ag
If we could measure the moles of e-, we
could know the quantity of Ag formed.
But how to measure moles of e-?
charge passing
Current =
time
I (amps) =
coulombs
seconds
117
Quantitative Aspects of
Electrochemistry
Current
=
charge passing
time
I (amps) =
coulombs
seconds
But how is charge related to moles of
electrons?
Charge on 1 mol of e- =
(1.60 x 10-19 C/e-)(6.02 x 1023 e-/mol)
= 96,500
C/mol e- = 1 Faraday
118
Quantitative Aspects of Electrochemistry
I (amps) =
coulombs
seconds
1.50 amps flow thru a Ag+(aq)
solution for 15.0 min. What mass
of Ag metal is deposited?
Solution
(a) Calculate the charge
Coulombs = amps x time
= (1.5 amps)(15.0 min)(60 s/min)
= 1350 C
119
Quantitative Aspects of Electrochemistry
1.50 amps flow thru a Ag+(aq) solution for 15.0
min. What mass of Ag metal is deposited?
Solution
(a) Charge = 1350 C
(b) Calculate moles of e- used
1 mol e1350 C •
96,500 C
=
0.0140 mol e-
(c) Calculate quantity of Ag
0.0140 mol e-
•
1 mol Ag
= 0.0140 mol Ag or 1.51 g Ag
1 mol e
120
Quantitative Aspects of Electrochemistry
The anode reaction in a lead storage battery is
Pb(s) + HSO4-(aq) ---> PbSO4(s) + H+(aq) + 2eIf a battery delivers 1.50 amp, and you have
454 g of Pb, how long will the battery last?
Solution
a)
454 g Pb = 2.19 mol Pb
b)
Calculate moles of e-
c)
2 mol e2.19molPb •
= 4.38 mol e1 mol Pb
Calculate charge
4.38 mol e- • 96,500 C/mol e- = 423,000 C
121
Quantitative Aspects of Electrochemistry
a)
b)
c)
Solution
454 g Pb = 2.19 mol Pb
Mol of e- = 4.38 mol
Charge = 423,000 C
d) Calculate time
Time(s)
=
Charge (C)
I (amps)
423,000 C
Time(s) =
282,000 s
=
1.50 amp
About 78 hours
122
COUNTING ELECTRONS
• Remember:
–The balanced half-reaction
provides the conversion
from moles electrons to
moles of species sought.
Now solve the problem:
• Calculate the grams of aluminum
formed at the cathode if a current
of 4.40 A flows for 1.50 hours.
123
BALANCING REDOX
EQUATIONS
The Half-Reaction Method
• Separate the equation into halfreactions.
• Balance the half-reactions.
• Combine the half-reactions to form a
balanced equation containing no
electrons.
124
Balancing Half-Reactions
• First balance the element changing
oxidation state.
• Balance the oxygen atoms with water.
• Balance the hydrogen atoms with H+.
• Balance the charge with electrons.
After combining the half-reactions,
check for mass and charge balance.
125
Practice Problems
Balance the following equation:
MnO4- + H2SO3 ----> Mn+2 + SO4-2
1. Separate into half reactions:
MnO4- ----> Mn+2
H2SO3 ----> SO4-2
2. Balance atoms:
8 H+ + MnO4- ----> Mn+2 + 4 H2O
H2O + H2SO3 ----> SO4-2 + 4 H+
126
Practice Problems
3. Balance charges:
5 e- + 8 H+ + MnO4-----> Mn+2 + 4 H2O
H2O + H2SO3 ----> SO4-2 + 4 H+ + 2 e4. Equal electrons gained and lost:
10
16
2
2
8
2(5 e- + 8 H+ + MnO4- ----> Mn+2 + 4 H2O)
5
5
5
20
10
5(H2O + H2SO3 ----> SO4-2+ 4 H++ 2 e-)
127
Practice Problems
5. Simplify and Add:
10
16
2
2
8 3
2(5 e- + 8 H+ + MnO4- ----> Mn+2 + 4 H2O)
5
5
5
20 4
10
5(H2O + H2SO3 ----> SO4-2 + 4 H++2 e-)
2MnO4- + 5 H2SO3
---->
2 Mn+2 + 5 SO4-2 + 4 H+ + 3 H2O
128
Practice Problems
Balance the following equation:
Al + NO3- ---> Al(OH)4- + NH3
1. Separate into half reactions:
Al ----> Al(OH)4NO3- ----> NH3
2. Balance atoms:
4 H2O + Al ----> Al(OH)4- + 4 H+
9 H+ + NO3- ----> NH3 + 3 H2O
129
Practice Problems
3. Balance charges:
4 H2O + Al ----> Al(OH)4- + 4
+ +- 3 e+
H
8 e + 9 H + NO3 ----> NH3 + 3
H2O 4. Equal electrons gained and lost:
32
8
8
32
24
8(4 H2O + Al ----> Al(OH)4- + 4 H+
-)
+
3
e
24
27
3
3
9
3(8 e- + 9 H+ + NO3- ----> NH3 + 3
H2O)
130
Practice Problems
5. Simplify and Add:
32 23
8
8(4 H2O + Al
24
27
3
H+ +
3(8 e- + 9
H2O)
23
H2O + 8 Al
32 5
24
----> Al(OH)4- + 4 H+
+ 3 e3-)
9
NO3- ----> NH3 + 3
+ 3 NO3- ---->
8 Al(OH)4- +
8
5 H+ + 3 NH3
131
Practice Problems
18
6. Change to basic solution:
23 H2O + 8 Al + 3 NO3- ---->
+ - + 3 NH
+55 H
OH
3
8 Al(OH)4- +
+ 5 OH-
5 H2O
18 H2O + 5 OH- + 8 Al + 3 NO3- ---->8
Al(OH)4- + 3 NH3
132
Practice Problems
1. MnO2
+
HBr
-->
Br2
+
MnBr2
2e- + 4 H+ + MnO2 ----> Mn2+ + 2
H2O
2 Br- ----> Br2 + 2e4 H+ + MnO2 + 2 Br- ----> Mn2+ + 2
H2O + Br2
133
Practice Problems
2. Cl2
+
NaBr --> NaCl
+
Br2
2e- + Cl2 --> 2 Cl2 Br --> Br2 + 2eCl2 + 2 Br- ----> 2 Cl- +
Br2
134
Practice Problems
3. H2S
+
3
3(H2S
6
82
HNO3 --> S
3
+
6
NO
6
----> S + 2 H+ + 2 e-)
2
H+ +
2
NO3-
4
2(3 e- + 4
----> NO + 2
3 HH
+ 2 H+ + 2 NO3- ----> 3 S + 2 NO + 4 H2O
2S2O)
135
Practice Problems
4. PbO2
6
6
+ Sb --> PbO + NaSbO2
(base)
3
3
3
3(2 e- + 2H+ + PbO2 ----> PbO +
4 1
2 H2O)2
82
6
2(2 H2O + Sb ----> SbO2- + 4
+ + 3 e-)
H
3 PbO2 + H2O + 2 Sb --->3 PbO + 2
1
+- 2 OH- +
+
2
OH
2 H2 O
SbO2 + 2 H
3PbO2 + 2 OH- + 2Sb --->3PbO +
-
136