Prof. T.L. Heise, CHE 116
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Chapter Twenty
Copyright © Tyna L. Heise 2001 - 2002
All Rights Reserved
Prof. T.L. Heise, CHE 116
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Electrochemistry
Redox reactions
Oxidation is the loss of electrons
Reduction is the gain of electrons
The transfer of electrons that occurs is
exothermic and can be used to produce
electricity
Prof. T.L. Heise, CHE 116
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Electrochemistry
At other times the reactions you want
to occur may be nonspontaneous and
require the addition of electricity to
drive the reaction.
Electrochemistry is the branch of
chemistry that deals with the
relationships between electricity and
chemical reactions
Prof. T.L. Heise, CHE 116
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Oxidation - Reductions
Reaction
Determination of oxidation reduction
reactions occurs by examining the
following:
assign oxidation numbers
look for the change in oxidation
states
don’t forget, something that
oxidizes is also the reducing agent,
something that reduces is also the
oxidizing agent
Prof. T.L. Heise, CHE 116
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Oxidation - Reductions
Reaction
Sample Exercise: Identify the oxidizing
and reducing agents in the following
oxidation-reduction equation:
2H2O(l) + Al(s) + MnO4-(aq) 
Al(OH)4-(aq) +
MnO2(s)
Prof. T.L. Heise, CHE 116
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Oxidation - Reductions
Reaction
Sample Exercise: Identify the oxidizing
and reducing agents in the following
oxidation-reduction equation:
+1 -2
0
+7 -2
2H2O(l) + Al(s) + MnO4-(aq) 
+3 -2 +1
+4
-2
Al(OH)4-(aq) + MnO2(s)
Prof. T.L. Heise, CHE 116
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Oxidation - Reductions
Reaction
Sample Exercise: Identify the oxidizing
and reducing agents in the following
oxidation-reduction equation:
+1 -2
0
+7 -2
2H2O(l) + Al(s) + MnO4-(aq) 
+3 -2 +1
+4
-2
Al(OH)4-(aq) + MnO2(s)
Prof. T.L. Heise, CHE 116
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Balancing Oxidation Reductions Reaction
Whenever we balance a chemical
equation, we must obey the conservation
of mass.
As we balance redox reactions we will see
an additional rule: the balancing of charge
as well.
Prof. T.L. Heise, CHE 116
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Balancing Oxidation Reductions Reaction
1. Assign oxidation numbers to all
elements
2. Bridge the elements that are changing
in oxidation state
3. Identify the number of electrons being
lost and gained and balance them.
4. Balance all remaining elements besides
H and O
5. Balance O by adding H2O to the
deficient side
Prof. T.L. Heise, CHE 116
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Balancing Oxidation Reductions Reaction
6. Balance H by adding H+ to the deficient
side
7. Check to make sure your charge for
each side is the same
Prof. T.L. Heise, CHE 116
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Balancing Oxidation Reductions Reaction
Sample Exercise: Complete and balance
the following oxidation-reduction
equations in acidic solutions.
A) Cu(s) + NO3-(aq)  Cu2+(aq) + NO2(g)
Prof. T.L. Heise, CHE 116
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Balancing Oxidation Reductions Reaction
Sample Exercise: Complete and balance
the following oxidation-reduction
equations in acidic solutions.
A) 0
+5 -2
+2
+4 -2
Cu(s) + NO3-(aq)  Cu2+(aq) + NO2(g)
Prof. T.L. Heise, CHE 116
Balancing Oxidation Reductions Reaction
13
Sample Exercise: Complete and balance
the following oxidation-reduction
equations in acidic solutions.
Leo = RA 2e A)
0
+5 -2
+2
+4 -2
Cu(s) + NO3-(aq)  Cu2+(aq) + NO2(g)
Ger = OA 1e -
Prof. T.L. Heise, CHE 116
Balancing Oxidation Reductions Reaction
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Sample Exercise: Complete and balance
the following oxidation-reduction
equations in acidic solutions.
Leo = RA 2e A)
0
+5 -2
+2
+4 -2
Cu(s) + 2NO3-(aq)Cu2+(aq) +
2NO2(g)
Ger = OA 2(1e -)
Prof. T.L. Heise, CHE 116
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Balancing Oxidation Reductions Reaction
Sample Exercise: Complete and balance the
following oxidation-reduction equations in
acidic solutions.
Leo = RA 2e A)
0
+5 -2
+2
+4 -2
Cu(s)+2NO3-(aq)Cu2+(aq)+2NO2(g)+2H2O
Ger = OA 2(1e -)
Prof. T.L. Heise, CHE 116
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Balancing Oxidation Reductions Reaction
Sample Exercise: Complete and balance the
following oxidation-reduction equations in
acidic solutions.
Leo = RA 2e A)
0
+5 -2
+2
+4 -2
Cu +2NO3(aq)+4H+Cu2+(aq)+2NO2(g)+2H2O
Ger = OA 2(1e -)
Prof. T.L. Heise, CHE 116
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Balancing Oxidation Reductions Reaction
Sample Exercise: Complete and balance
the following oxidation-reduction
equations in acidic solutions.
B) Mn2+ + NaBiO3  Bi3+ + MnO4-
Prof. T.L. Heise, CHE 116
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Balancing Oxidation Reductions Reaction
Sample Exercise: Complete and balance
the following oxidation-reduction
equations in acidic solutions.
B) +2
+5 -2
+3
+7 -2
Mn2+ + BiO3-  Bi3+ + MnO4-
Prof. T.L. Heise, CHE 116
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Balancing Oxidation Reductions Reaction
Sample Exercise: Complete and balance
the following oxidation-reduction
equations in acidic solutions.
Leo = RA 5e B)
+2
+5 -2
+3
+7
-2
Mn2+ + BiO3 -  Bi3+ + MnO4Ger = OA 2e -
Prof. T.L. Heise, CHE 116
Balancing Oxidation Reductions Reaction
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Sample Exercise: Complete and balance
the following oxidation-reduction
equations in acidic solutions.
Leo = RA 2(5e -)
B)
+2
+5 -2
+3
+7
-2
2Mn2+ + BiO3 - Bi3+ + 2MnO4Ger = OA 2e -
Prof. T.L. Heise, CHE 116
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Balancing Oxidation Reductions Reaction
Sample Exercise: Complete and balance
the following oxidation-reduction
equations in acidic solutions.
Leo = RA 2(5e -)
B)
+2
+5 -2
+3
+7
-2
2Mn2+ + 5BiO3 - 5Bi3+ + 2MnO4Ger = OA 5(2e -)
Prof. T.L. Heise, CHE 116
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Balancing Oxidation Reductions Reaction
Sample Exercise: Complete and balance
the following oxidation-reduction
equations in acidic solutions.
B)
2Mn2+ + 5BiO3- + 14H+
5Bi3+ + 2MnO4- + 7H2O
Prof. T.L. Heise, CHE 116
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Balancing Oxidation Reductions Reaction
At times the solution in which a redox
reaction occurs is a basic one. The
balancing of an equation in a basic
solution is slightly different. Instead of
balancing with H2O and H+, you must
balance with H2O and OH-.
Prof. T.L. Heise, CHE 116
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Balancing Oxidation Reductions Reaction
Sample exercise: Complete and balance
the following equations for oxidation and
reduction reactions that occur in basic
solutions.
A) NO2- + Al  NH3 + Al(OH)4-
Prof. T.L. Heise, CHE 116
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Balancing Oxidation Reductions Reaction
Sample exercise: Complete and balance
the following equations for oxidation and
reduction reactions that occur in basic
solutions.
A) +3 -2
0
-3 +1
+3 -2 +1
NO2- + Al  NH3 + Al(OH)4-
Prof. T.L. Heise, CHE 116
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Balancing Oxidation Reductions Reaction
Sample exercise: Complete and balance
the following equations for oxidation and
reduction reactions that occur in basic
solutions.
A) Ger = OA 6e +3 -2
0
-3 +1
+3 -2 +1
NO2- + Al  NH3 + Al(OH)4Leo = RA 3e -
Prof. T.L. Heise, CHE 116
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Balancing Oxidation Reductions Reaction
Sample exercise: Complete and balance
the following equations for oxidation and
reduction reactions that occur in basic
solutions.
A) Ger = OA 6e +3 -2
0
-3 +1
+3 -2 +1
NO2- + 2Al  NH3 + 2Al(OH)4Leo = RA 2(3e -)
Prof. T.L. Heise, CHE 116
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Balancing Oxidation Reductions Reaction
Sample exercise: Complete and balance
the following equations for oxidation and
reduction reactions that occur in basic
solutions.
A)
NO2- + 2Al + 5H2O + OH- 
NH3 + 2Al(OH)4-
Prof. T.L. Heise, CHE 116
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Balancing Oxidation Reductions Reaction
Sample exercise: Complete and balance
the following equations for oxidation and
reduction reactions that occur in basic
solutions.
B) Cr(OH)3 + ClO-  CrO42- + Cl2
Prof. T.L. Heise, CHE 116
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Balancing Oxidation Reductions Reaction
Sample exercise: Complete and balance
the following equations for oxidation and
reduction reactions that occur in basic
solutions.
B) +3 -2 +1
+1 -2
+6 -2
0
Cr(OH)3 + ClO-  CrO42- + Cl2
Prof. T.L. Heise, CHE 116
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Balancing Oxidation Reductions Reaction
Sample exercise: Complete and balance
the following equations for oxidation and
reduction reactions that occur in basic
solutions.
Leo = RA 3e B)
+3 -2 +1
+1 -2
+6 -2
0
Cr(OH)3 + ClO-  CrO42- + Cl2
Ger = OA 1e -
Prof. T.L. Heise, CHE 116
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Balancing Oxidation Reductions Reaction
Sample exercise: Complete and balance
the following equations for oxidation and
reduction reactions that occur in basic
solutions.
Leo = RA 3e B)
+3 -2 +1
+1 -2
+6 -2
0
Cr(OH)3 + 6ClO-  CrO42- + 3Cl2
Ger = OA 6(1e -)
Prof. T.L. Heise, CHE 116
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Balancing Oxidation Reductions Reaction
Sample exercise: Complete and balance
the following equations for oxidation and
reduction reactions that occur in basic
solutions.
Leo = RA 2(3e -)
B)
+3 -2 +1
+1 -2
+6 -2
0
2Cr(OH)3 + 6ClO-  2CrO42- + 3Cl2
Ger = OA 6(1e -)
Prof. T.L. Heise, CHE 116
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Balancing Oxidation Reductions Reaction
Sample exercise: Complete and balance
the following equations for oxidation and
reduction reactions that occur in basic
solutions.
B) 2Cr(OH)3 + 6ClO- 
2CrO42- + 3Cl2 + 2H2O + 2OH-
Prof. T.L. Heise, CHE 116
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Voltaic Cells
The energy released in a spontaneous
redox reaction can be used to perform
electrical work.
This task is accomplished in a voltaic cell,
a device in which the transfer of electrons
takes place through an external pathway
rather than directly between reactants.
Prof. T.L. Heise, CHE 116
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Voltaic Cells
The
spontaneous
reaction
occurs in a
single cell,
but splitting
the cell into
two halves
allows for
the external
pathway of
electrons to
be created.
Prof. T.L. Heise, CHE 116
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Voltaic Cells
Prof. T.L. Heise, CHE 116
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Voltaic Cells
Prof. T.L. Heise, CHE 116
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Voltaic Cells
Parts of Voltaic Cell:
the two solid metals connected by the
external circuit are called electrodes.
the electrode at which oxidation occurs
is called the anode
the electrode at which reduction occurs
is called the cathode.
Prof. T.L. Heise, CHE 116
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Voltaic Cells
Parts of Voltaic Cell:
the anode metal is the substance that
undergoes oxidation to form more
positive ions.
the positive metallic ion in the beaker
that contains the cathode is the substance
that undergoes reduction to form more of
the metal electrode.
Prof. T.L. Heise, CHE 116
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Voltaic Cells
Parts of Voltaic Cell:
As the oxidation half cell gets a larger
concentration of positive ions, negative
ions flow through the semipermeable
membrane that forms the salt bridge to
balance the charges in the beaker.
As the reduction half cell gets a smaller
concentration of positive ions, positive
ions flow through the membrane to
replace the loss.
Prof. T.L. Heise, CHE 116
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Voltaic Cells
Parts of Voltaic Cell:
in any voltaic cell the electrons flow
from the anode (-) to the cathode (+)
through the external circuit
The collection of lost electrons at the
anode causes it to be negative, and the
cathode is positive so it can attract the
electrons and pull them through the wire.
Prof. T.L. Heise, CHE 116
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Voltaic Cells
Prof. T.L. Heise, CHE 116
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Voltaic Cells
Sample exercise: The two half-reactions in a
voltaic cell are
Zn(s)  Zn2+(aq) + 2eClO3-(aq) + 6H+ + 6e-  Cl-(aq) +
3H2O(l)
a) Indicate which reaction occurs at the
anode and the cathode
Prof. T.L. Heise, CHE 116
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Voltaic Cells
Sample exercise: The two half-reactions in a
voltaic cell are
Zn(s)  Zn2+(aq) + 2eClO3-(aq) + 6H+ + 6e-  Cl-(aq) +
3H2O(l)
Prof. T.L. Heise, CHE 116
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Voltaic Cells
Sample exercise: The two half-reactions in a
voltaic cell are
Zn(s)  Zn2+(aq) + 2eClO3-(aq) + 6H+ + 6e-  Cl-(aq) +
3H2O(l)
b) Which electrode is consumed first in the
cell reaction?
Prof. T.L. Heise, CHE 116
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Voltaic Cells
Sample exercise: The two half-reactions in a
voltaic cell are
Zn(s)  Zn2+(aq) + 2eClO3-(aq) + 6H+ + 6e-  Cl-(aq) +
3H2O(l)
b) Which electrode is consumed first in the
cell reaction?
The anode is consumed
Prof. T.L. Heise, CHE 116
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Voltaic Cells
Sample exercise: The two half-reactions in a
voltaic cell are
Zn(s)  Zn2+(aq) + 2eClO3-(aq) + 6H+ + 6e-  Cl-(aq) +
3H2O(l)
c) Which electrode is positive?
Prof. T.L. Heise, CHE 116
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Voltaic Cells
Sample exercise: The two half-reactions in a
voltaic cell are
Zn(s)  Zn2+(aq) + 2eClO3-(aq) + 6H+ + 6e-  Cl-(aq) +
3H2O(l)
c) Which electrode is positive?
The cathode is positive
Prof. T.L. Heise, CHE 116
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Voltaic Cells
An atomic view of what occurs in a single
cell in which both oxidation and reduction
are occurring may give better insight.
Prof. T.L. Heise, CHE 116
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Voltaic Cells
A molecular view allows a better
understanding of what is occurring in each
half cell.
Prof. T.L. Heise, CHE 116
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Cell EMF
The reason that electrons flow
spontaneously is do to the “driving force”
behind certain redox reactions.
The ‘driving force’ is a difference in
potential energy.
The anode has a higher potential
energy, and electrons flow to the cathode
to achieve a lower potential energy.
As in many natural wonders, the flow
from higher concentration to lower
continues
Prof. T.L. Heise, CHE 116
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Cell EMF
The difference in potential energy per
charge is measured in volts.
1 V (volt) is the potential difference
required to impart 1 J (joule) of energy to
a charge of 1 C (coulomb)
the potential difference is called an
electromotive force or emf
emf, or Ecell, is also called the cell
potential
Prof. T.L. Heise, CHE 116
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Cell EMF
The difference in potential energy per
charge is measured in volts.
The emf of a particular cell is
dependent on the reaction that is
occurring there.
We will always assume standard
conditions, and the cell potential should
be denoted as such, E°cell
Prof. T.L. Heise, CHE 116
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Cell EMF
Standard reduction potential:
the cell potential will depend on the
particular cathode and anode used
the cell potential is the difference
between two electrode potentials
Ecell = Ecathode + Eanode
the amount of moles of each DOES
NOT change the potentials of cells
Prof. T.L. Heise, CHE 116
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Cell EMF
Standard reduction potential:
when finding the potentials, a reduction
table will be used.
Finding the potential for cathodes is
straightforward, simply read the table
Finding the potential for anodes means
reading the table BACKWARDS and the
sign of the potential should be switched
Table 20.1 is a standard reduction table
Prof. T.L. Heise, CHE 116
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Cell EMF
Sample exercise: Calculate the standard
emf for a cell that employs the following
overall reaction:
0
Ni(s) +
Ni+2(aq)
+3
2Fe3+(aq)
+2
 2Fe2+(aq) +
+2
Prof. T.L. Heise, CHE 116
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Cell EMF
Sample exercise: Calculate the standard
emf for a cell that employs the following
overall reaction:
0
Ni(s) +
Ni+2(aq)
+3
2Fe3+(aq)
+2
 2Fe2+(aq) +
+2
Prof. T.L. Heise, CHE 116
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Cell EMF
Sample exercise: Calculate the standard
emf for a cell that employs the following
overall reaction: Ger = cathode
0
Ni(s) +
+3
2Fe3+(aq)
+2
+2
 2Fe2+(aq) + Ni+2(aq)
Leo = anode
Prof. T.L. Heise, CHE 116
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Cell EMF
Sample exercise: Calculate the standard
emf for a cell that employs the following
overall reaction: Ger = cathode
0
Ni(s) +
Ni+2(aq)
+3
2Fe3+(aq)
+2
 2Fe2+(aq) +
Leo = anode
Ecell = Ecathode + Eanode
+2
Prof. T.L. Heise, CHE 116
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Cell EMF
Sample exercise: Calculate the standard
emf for a cell that employs the following
overall reaction: Ger = cathode
0
Ni(s) +
+3
2Fe3+(aq)
+2
+2
 2Fe2+(aq) + Ni+2(aq)
Leo = anode
Ecell = Ecathode + Eanode
= +0.77 + +0.28
Prof. T.L. Heise, CHE 116
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Cell EMF
Sample exercise: Calculate the standard
emf for a cell that employs the following
overall reaction:
2Al(s) + 3I2(s)  2Al3+(aq) + 6I-(aq)
Prof. T.L. Heise, CHE 116
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Cell EMF
Sample exercise: Calculate the standard
emf for a cell that employs the following
overall reaction:
0
0
+3
-1
2Al(s) + 3I2(s)  2Al3+(aq) + 6I-(aq)
Prof. T.L. Heise, CHE 116
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Cell EMF
Sample exercise: Calculate the standard
emf for a cell that employs the following
overall reaction:
leo = anode
0
0
+3
-1
2Al(s) + 3I2(s)  2Al3+(aq) + 6I-(aq)
ger = cathode
Prof. T.L. Heise, CHE 116
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Cell EMF
Sample exercise: Calculate the standard
emf for a cell that employs the following
overall reaction:
leo = anode
0
0
+3
-1
2Al(s) + 3I2(s)  2Al3+(aq) + 6I-(aq)
ger = cathode
Ecell = +0.54 + +1.66 = 2.20 V
Prof. T.L. Heise, CHE 116
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Cell EMF
The more positive the potential difference,
the greater the driving force.
Prof. T.L. Heise, CHE 116
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Spontaneity of Redox
Reactions
Any reaction that can occur in a voltaic cell
to produce a positive emf must be
spontaneous.
Ecell = Ered + Eox
a + emf is spontaneous
a - emf is not spontaneous
Prof. T.L. Heise, CHE 116
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Spontaneity of Redox
Reactions
Sample exercise: Using the standard
reduction potentials, determine which of the
following reactions are spontaneous.
A) I2(s) + 5Cu2+(aq) + 6H2O(l) 
2IO3-(aq) + 5Cu(s) + 12H+(aq)
Prof. T.L. Heise, CHE 116
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Spontaneity of Redox
Reactions
Sample exercise: Using the standard
reduction potentials, determine which of the
following reactions are spontaneous.
A) I2(s) + 5Cu2+(aq) + 6H2O(l) 
2IO3-(aq) + 5Cu(s) + 12H+(aq)
Red: Cu2+(aq)  Cu(s)
Ox: I2(s) + 6H2O  2IO3-(aq) + 12H+(aq)
Prof. T.L. Heise, CHE 116
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Spontaneity of Redox
Reactions
Sample exercise: Using the standard
reduction potentials, determine which of the
following reactions are spontaneous.
A) I2(s) + 5Cu2+(aq) + 6H2O(l) 
2IO3-(aq) + 5Cu(s) + 12H+(aq)
Red: Cu2+(aq)  Cu(s)
Ox: I2(s) + 6H2O  2IO3-(aq) + 12H+(aq)
Ecell = +0.34 + -1.195
Prof. T.L. Heise, CHE 116
71
Spontaneity of Redox
Reactions
Sample exercise: Using the standard
reduction potentials, determine which of the
following reactions are spontaneous.
A) I2(s) + 5Cu2+(aq) + 6H2O(l) 
2IO3-(aq) + 5Cu(s) + 12H+(aq)
Red: Cu2+(aq)  Cu(s)
Ox: I2(s) + 6H2O  2IO3-(aq) + 12H+(aq)
Ecell = +0.34 + -1.195 = -0.855 NOT
Prof. T.L. Heise, CHE 116
72
Spontaneity of Redox
Reactions
Sample exercise: Using the standard
reduction potentials, determine which of the
following reactions are spontaneous.
B) Hg2+(aq) + 2I-(aq)  Hg(l) + I2(s)
Prof. T.L. Heise, CHE 116
73
Spontaneity of Redox
Reactions
Sample exercise: Using the standard
reduction potentials, determine which of the
following reactions are spontaneous.
B) Hg2+(aq) + 2I-(aq)  Hg(l) + I2(s)
Red: Hg2+(aq)  Hg(l)
Ox: 2I-(aq)  I2(s)
Prof. T.L. Heise, CHE 116
74
Spontaneity of Redox
Reactions
Sample exercise: Using the standard
reduction potentials, determine which of the
following reactions are spontaneous.
B) Hg2+(aq) + 2I-(aq)  Hg(l) + I2(s)
Red: Hg2+(aq)  Hg(l)
Ox: 2I-(aq)  I2(s)
Ecell = +0.854 + -0.536
Prof. T.L. Heise, CHE 116
75
Spontaneity of Redox
Reactions
Sample exercise: Using the standard
reduction potentials, determine which of the
following reactions are spontaneous.
B) Hg2+(aq) + 2I-(aq)  Hg(l) + I2(s)
Red: Hg2+(aq)  Hg(l)
Ox: 2I-(aq)  I2(s)
Ecell = +0.854 + -0.536 = +0.318 SPON
Prof. T.L. Heise, CHE 116
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Spontaneity of Redox
Reactions
Looking at the standard reduction table, a
general rule can be established:
any metal in the series is able to
spontaneously react with any ion beneath
it
there is also a relationship between emf
and Gibbs Free Energy
(remember a negative DG is
spontaneous)
DG = -nFE
Prof. T.L. Heise, CHE 116
77
Spontaneity of Redox
Reactions
DG = -nFE
n = number of electrons
transferred in reaction
F = Faraday’s constant
1 F = 96,500 J/V-mol
E = emf of cell
units of DG are J/mol
Prof. T.L. Heise, CHE 116
78
Spontaneity of Redox
Reactions
Sample exercise: Consider the following
reaction:
3Ni2+(aq) + 2Cr(OH)3(s) + 10OH- 
3Ni(s) + 2CrO42-(aq) + 8H2O(l)
a) what is the value of n for this reaction?
Prof. T.L. Heise, CHE 116
79
Spontaneity of Redox
Reactions
Sample exercise: Consider the following
reaction:
3Ni2+(aq) + 2Cr(OH)3(s) + 10OH- 
3Ni(s) + 2CrO42-(aq) + 8H2O(l)
a) what is the value of n for this reaction?
Each Ni2+ gains 2e- and there are 3
ions, so 6e- are involved
n=6
Prof. T.L. Heise, CHE 116
80
Spontaneity of Redox
Reactions
Sample exercise: Consider the following
reaction:
3Ni2+(aq) + 2Cr(OH)3(s) + 10OH- 
3Ni(s) + 2CrO42-(aq) + 8H2O(l)
b) Calculate DG
DG = -nFE
Prof. T.L. Heise, CHE 116
81
Spontaneity of Redox
Reactions
Sample exercise: Consider the following
reaction:
3Ni2+(aq) + 2Cr(OH)3(s) + 10OH- 
3Ni(s) + 2CrO42+(aq) + 8H2O(l)
b) Calculate DG
DG = -nFE n = 6
F = 96,500 J/V-mol
E = -0.28 + +0.13= -0.15 V
Prof. T.L. Heise, CHE 116
82
Spontaneity of Redox
Reactions
Sample exercise: Consider the following
reaction:
3Ni2+(aq) + 2Cr(OH)3(s) + 10OH- 
3Ni(s) + 2CrO42+(aq) + 8H2O(l)
b) Calculate DG
DG = -nFE
-(6)(96,500 J/V-mol)(-0.15 V)
+86850 J/mol
NOT Spon
Prof. T.L. Heise, CHE 116
83
Effect of Concentration
on Cell EMF
As a voltaic cell is discharged, the reactants
of the reaction are consumed and the
products are generated, so the
concentrations of these substances change.
The emf progressively drops until E = 0, at
which point we say the cell is dead. At that
point the concentrations in the cell are at
equilibrium.
Prof. T.L. Heise, CHE 116
84
Effect of Concentration
on Cell EMF
The dependence of cell emf on concentration
can be obtained from the dependence of the
free energy change on concentration
Nernst Equation: E = E° - RT (lnQ)
nF
= E° - 0.0592 V (log Q)
n
Prof. T.L. Heise, CHE 116
85
Effect of Concentration
on Cell EMF
The dependence of cell emf on concentration
can be obtained from the dependence of the
free energy change on concentration
Nernst Equation: E = E° - RT (lnQ)
nF
* Q is the reaction
quotient
= E° - 0.0592 V (log Q)
n
Prof. T.L. Heise, CHE 116
86
Effect of Concentration
on Cell EMF
Sample exercise: Calculate the emf
generated by the cell
2Al(s) + 3I2(s)  2Al3+(aq) + 6I-(aq)
when [Al3+] = 4.0 x 10-3M and [I-] = 0.010 M.
Prof. T.L. Heise, CHE 116
87
Effect of Concentration
on Cell EMF
Sample exercise: Calculate the emf
generated by the cell
2Al(s) + 3I2(s)  2Al3+(aq) + 6I-(aq)
when [Al3+] = 4.0 x 10-3M and [I-] = 0.010 M.
E = E° - 0.0592 V (log Q)
n
Prof. T.L. Heise, CHE 116
88
Effect of Concentration
on Cell EMF
Sample exercise: Calculate the emf
generated by the cell
2Al(s) + 3I2(s)  2Al3+(aq) + 6I-(aq)
when [Al3+] = 4.0 x 10-3M and [I-] = 0.010 M.
E = E° - 0.0592 V (log Q)
n
E° = +0.536 + +1.66 = 2.22 V
Prof. T.L. Heise, CHE 116
89
Effect of Concentration
on Cell EMF
Sample exercise: Calculate the emf
generated by the cell
2Al(s) + 3I2(s)  2Al3+(aq) + 6I-(aq)
when [Al3+] = 4.0 x 10-3M and [I-] = 0.010 M.
E = E° - 0.0592 V (log Q)
n
n = 6 e-
Prof. T.L. Heise, CHE 116
90
Effect of Concentration
on Cell EMF
Sample exercise: Calculate the emf
generated by the cell
2Al(s) + 3I2(s)  2Al3+(aq) + 6I-(aq)
when [Al3+] = 4.0 x 10-3M and [I-] = 0.010 M.
E = E° - 0.0592 V (log Q)
n
Q = [Al3+]2[I-]6 = [4.0 x 10-3]2[0.010]6
= 1.6 x 10-17
Prof. T.L. Heise, CHE 116
91
Effect of Concentration
on Cell EMF
Sample exercise: Calculate the emf
generated by the cell
2Al(s) + 3I2(s)  2Al3+(aq) + 6I-(aq)
when [Al3+] = 4.0 x 10-3M and [I-] = 0.010 M.
E = E° - 0.0592 V (log Q)
n
= 2.22 - 0.0592 (log 1.6 x 10-17)
6
Prof. T.L. Heise, CHE 116
92
Effect of Concentration
on Cell EMF
Sample exercise: Calculate the emf
generated by the cell
2Al(s) + 3I2(s)  2Al3+(aq) + 6I-(aq)
when [Al3+] = 4.0 x 10-3M and [I-] = 0.010 M.
E = E° - 0.0592 V (log Q)
n
= 2.22 - 0.0592 (log 1.6 x 10-17)
6
= 2.36 V
Prof. T.L. Heise, CHE 116
93
Effect of Concentration
on Cell EMF
Concentration cells: a voltaic cell with an
emf of ZERO can be constructed if the same
species is used as both the anode and the
cathode if concentrations are the same;
using two different concentrations of the
same species will create a non-zero cell.
A cell based solely on the emf generated
because of a difference in a concentration is
called a concentration cell
Prof. T.L. Heise, CHE 116
94
Effect of Concentration
on Cell EMF
Prof. T.L. Heise, CHE 116
95
Effect of Concentration
on Cell EMF
The standard emf for this cell would have to
zero, but the cell operates due to the
differences in concentration.
The driving force for the cell is provided by
the difference in concentration.
Operation of the cell proceeds to equalize
the concentrations: the more dilute
concentration acts as the anode, the solid
metal electrode will oxidize to turn into
more ions.
Prof. T.L. Heise, CHE 116
96
Effect of Concentration
on Cell EMF
The more concentrated solution acts as the
cathode, where the larger number of ions
reduce to plate the electrode and thereby
reduce there number.
The nernst equation can also be used here to
calculate the Ecell under non-standard
conditions.
Anode: X(s)  X+(dilute) + eCathode: X+(conc.) + e-  X(s)
Prof. T.L. Heise, CHE 116
97
Effect of Concentration
on Cell EMF
Anode: X(s)  X+(dilute) + eCathode: X+(conc.) + e-  X(s)
X+(conc.) + X(s)  X+(dilute) + X(s)
Q = X+(dilute) / X+(conc.)
Prof. T.L. Heise, CHE 116
98
Effect of Concentration
on Cell EMF
Sample exercise: A concentration cell is
constructed with two Zn(s)  Zn2+(aq)
half-cells. The first cell has [Zn2+] = 1.35 M
and the second cell has [Zn2+] = 3.75 x 10-4
M.
a) Which half cell is the anode of the cell?
Prof. T.L. Heise, CHE 116
99
Effect of Concentration
on Cell EMF
Sample exercise: A concentration cell is
constructed with two Zn(s)  Zn2+(aq)
half-cells. The first cell has [Zn2+] = 1.35 M
and the second cell has [Zn2+] = 3.75 x 10-4
M.
a) Which half cell is the anode of the cell?
The cell with [Zn2+] = 3.75 x 10-4 M is
the anode.
Prof. T.L. Heise, CHE 116
100
Effect of Concentration
on Cell EMF
Sample exercise: A concentration cell is
constructed with two Zn(s)  Zn2+(aq)
half-cells. The first cell has [Zn2+] = 1.35 M
and the second cell has [Zn2+] = 3.75 x 10-4
M.
b) What is the emf of the cell?
Prof. T.L. Heise, CHE 116
101
Effect of Concentration
on Cell EMF
Sample exercise: A concentration cell is
constructed with two Zn(s)  Zn2+(aq)
half-cells. The first cell has [Zn2+] = 1.35 M
and the second cell has [Zn2+] = 3.75 x 10-4
M.
b) What is the emf of the cell?
E = E° - 0.0592 V (log Q)
n
Prof. T.L. Heise, CHE 116
102
Effect of Concentration
on Cell EMF
Sample exercise: A concentration cell is
constructed with two Zn(s)  Zn2+(aq)
half-cells. The first cell has [Zn2+] = 1.35 M
and the second cell has [Zn2+] = 3.75 x 10-4
M.
b) What is the emf of the cell?
E = E° - 0.0592 V (log Q)
n
= 0 - 0.0592 (log 3.75x10-4/1.35)
2
Prof. T.L. Heise, CHE 116
103
Effect of Concentration
on Cell EMF
Sample exercise: A concentration cell is
constructed with two Zn(s)  Zn2+(aq)
half-cells. The first cell has [Zn2+] = 1.35 M
and the second cell has [Zn2+] = 3.75 x 10-4
M.
b) What is the emf of the cell?
E = E° - 0.0592 V (log Q)
n
= 0 - 0.0592 (log 3.75x10-4/1.35)
2
Prof. T.L. Heise, CHE 116
104
Effect of Concentration
on Cell EMF
Cell emf and Chemical Equilibrium: the
nernst equation helps us understand why
the emf drops as the cell is discharged: as
the reactants are converted into products,
the value of Q increases, so the value of E
decreases.
The cell emf eventually reaches 0.
When DG = 0, E = 0, and an Ecell = 0 shows a
cell at equilibrium.
Prof. T.L. Heise, CHE 116
105
Effect of Concentration
on Cell EMF
If 0 = E° - 0.0592 V (log Q)
n
Then log K = nE°
0.0592
Prof. T.L. Heise, CHE 116
106
Effect of Concentration
on Cell EMF
Sample exercise: Using standard reduction
potentials, calculate the equilibrium
constant at 25°C for the reaction
Br2(l) + 2Cl-(aq)  Cl2(g) + 2Br-(aq)
Prof. T.L. Heise, CHE 116
107
Effect of Concentration
on Cell EMF
Sample exercise: Using standard reduction
potentials, calculate the equilibrium
constant at 25°C for the reaction
Br2(l) + 2Cl-(aq)  Cl2(g) + 2Br-(aq)
log K = nE°
0.0592
Prof. T.L. Heise, CHE 116
108
Effect of Concentration
on Cell EMF
Sample exercise: Using standard reduction
potentials, calculate the equilibrium
constant at 25°C for the reaction
Br2(l) + 2Cl-(aq)  Cl2(g) + 2Br-(aq)
log K = nE°
0.0592
= 2(+1.065 + -1.359)
0.0592
Prof. T.L. Heise, CHE 116
109
Effect of Concentration
on Cell EMF
Sample exercise: Using standard reduction
potentials, calculate the equilibrium
constant at 25°C for the reaction
Br2(l) + 2Cl-(aq)  Cl2(g) + 2Br-(aq)
log K = nE°
0.0592
= 2(+1.065 + -1.359)
0.0592
= -9.93
Prof. T.L. Heise, CHE 116
110
Effect of Concentration
on Cell EMF
Sample exercise: Using standard reduction
potentials, calculate the equilibrium
constant at 25°C for the reaction
Br2(l) + 2Cl-(aq)  Cl2(g) + 2Br-(aq)
K = 10(-9.93)
= 1.2 x 10-10
Prof. T.L. Heise, CHE 116
111
Batteries
A battery is a portable, self-contained
electrochemical power source that consists
of one or more voltaic cells.
The common 1.5 V battery is a single voltaic
cell, where a 12 V battery uses multiple
voltaic cells in one case.
When cells are connected in series, with the
cathode of one attached to the anode to
another, the battery produces a voltage that
is the sum of the emfs of the single cells.
Prof. T.L. Heise, CHE 116
112
Batteries
Batteries
connected in
series...
Prof. T.L. Heise, CHE 116
113
Batteries
Commerical batteries that has specific
performance characteristics can require
considerable ingenuity.
The emf of a battery is determined by the
substances used as the cathode and anode,
and the usable life of the battery depends on
the quantities of the substances packaged.
Keep in mind the anode and cathode need to
be separated by a porous barrier similar to
a salt bridge.
Prof. T.L. Heise, CHE 116
114
Batteries
The materials used to construct the battery
must be stable under the conditions in which
it is to be used and must be chosen to
minimize health and environmental
concerns upon use and disposal.
Different applications require batteries with
different properties.
Prof. T.L. Heise, CHE 116
115
Batteries
Lead-Acid Battery
Battery used in cars is one of the most
common batteries, with over 100 million
produced annually.
A 12-V battery consists of six voltaic cells in
series, each producing 2-V.
Cathode: lead dioxide packed on a
metal grid
Anode: lead
Prof. T.L. Heise, CHE 116
116
Batteries
Lead-Acid Battery
Both electrodes are immersed in sulfuric
acid.
Both reactants are solids so there is no need
to separate the cell into anode and cathode
compartments.
To keep the electrodes from touching, wood
or fiber glass spacers are placed between
them
Prof. T.L. Heise, CHE 116
117
Batteries
Lead-Acid Battery
Using solids offers another advantage, no
concentration changes occur and the emf
stays mostly constant, fluctuations occuring
with variations in [H2SO4].
One major advantage with this battery is
the ability to be recharged
Prof. T.L. Heise, CHE 116
118
Batteries
Prof. T.L. Heise, CHE 116
119
Batteries
Alkaline Battery:
The most common non-rechargeable battery
is the alkaline battery.
More than 1010 are produced annually.
The anode of this battery consists of
powdered zinc metal immobilized in a gel in
contact with a concentrated solution of
KOH.
The cathode is a mixture of MnO2 and
graphite
Prof. T.L. Heise, CHE 116
120
Batteries
Prof. T.L. Heise, CHE 116
121
Batteries
Nickel-Cadmium, Nickel-Metal-Hydride,
and Lithium-Ion Batteries
All of these batteries are lightweight, easily
rechargeable batteries for cell phones,
notebook computers and video recorders.
Prof. T.L. Heise, CHE 116
122
Batteries
Nickel-Cadmium
Cadmium is oxidized while nickel
oxyhydroxide is reduced
The solid reaction products adhere to the
electrodes, which permits the reactions to be
reversed during charging.
Drawbacks: cadmium is a toxic heavy metal
Prof. T.L. Heise, CHE 116
123
Batteries
Nickel-Metal-Hydride
Nickel oxyhydroxide is still reduced, but
now the anode is a metal alloy that has the
ability to absorb hydrogen atoms.
The hydrogen atoms are released as water
Prof. T.L. Heise, CHE 116
124
Batteries
Lithium-Ion Batteries
Lithium is a very light-weight element , and
the technology is unique.
Li+ ions insert themselves into certain
layered solids; during discharge lithium ions
migrate between two different layered
materials that serve as the anode and
cathode
Prof. T.L. Heise, CHE 116
125
Batteries
Fuel Cells
The direct production of electricity from
fuels by a voltaic cell could, in principle,
yield a higher rate of conversion of the
chemical energy of the reaction.
Voltaic cells that perform this conversion
using conventional fuels are called fuel cells.
Fuel cells are not true batteries because they
are not self-contained.
Prof. T.L. Heise, CHE 116
126
Batteries
Fuel Cells
The most promising
fuel cell system
involves the reaction of
H2 and O2 to form
H2O(l) as the only
product.
Cathode: 4e- + O2 + 2H2O

4OH-
Prof. T.L. Heise, CHE 116
127
Corrosion
Undesirable redox reactions that lead to the
corrosion of metals are the other side to
spontaneous reactions.
For nearly all metals, oxidation is a
thermodynamically favorable process in air
at room temperature.
With certain metals, when oxidation occurs
a protective oxide layer is formed - example:
aluminum
Prof. T.L. Heise, CHE 116
128
Corrosion
Corrosion of iron
From an economic standpoint, this is a
major problem: 20% of the iron produced is
used to replace iron objects that have been
discarded because of rust damage.
The rusting of iron requires both oxygen
and water.
Other factors can accelerate rusting - pH,
salts, contact with metals and stress
Prof. T.L. Heise, CHE 116
129
Corrosion
Corrosion of iron
Corrosion can occur anywhere, but the best
place is the spot where the iron has the most
access to oxygen.
Prof. T.L. Heise, CHE 116
130
Corrosion
Preventing the Corrosion of Iron
Covering with paint or another metal can
protect the iron from corrosion.
Paint offers a layer against the iron coming
into contact
A second metal that is easier to oxidize then
iron will corrode first offering protection.
Prof. T.L. Heise, CHE 116
131
Corrosion
Prof. T.L. Heise, CHE 116
132
Corrosion
Prof. T.L. Heise, CHE 116
133
Electrolysis
It is possible to use electrical energy to
cause nonspontaneous redox reactions to
occur.
Processes which are driven by an outside
source of electrical energy are called
electrolysis reactions and take place in
electrolytic cells.
Prof. T.L. Heise, CHE 116
134
Electrolysis
Electrolytic cells
consists of two electrodes in a molten
salt or a solution.
driven by a battery or some other
source of direct electrical current.
the battery acts as an electron pump,
pushing electrons into one electrode and
pulling them from the other
the electrodes are inert, acting only as
the site for oxidation and reduction
Prof. T.L. Heise, CHE 116
135
Electrolysis
Electrolytic cells
reduction
still occurs
at the
cathode
oxidation
still occurs
at the
anode
Prof. T.L. Heise, CHE 116
136
Electrolysis
Electrolytic cells
Na+
reduces at
cathode to
form Na
Cl- oxidizes
at anode to
form Cl2
Prof. T.L. Heise, CHE 116
137
Electrolysis
Electrolytic cells
the electrode connected to the negative
terminal is now the cathode due to the
battery supplying the electrons for
reduction
the electrode connected to the positive
terminal is now the anode to draw
electrons off the metal and cause the
metal to oxidize and form ions
Prof. T.L. Heise, CHE 116
138
Electrolysis
Electrolysis of Aqueous Solutions
because of the high melting points of
ionic substances, the electrolysis of
molten salts requires very high
temperatures
we can produce ions from soluble salts
at room temperatures by dissolving the
salt in water
the electrolysis of an aqueous solution
is complicated by the presence of water
Prof. T.L. Heise, CHE 116
139
Electrolysis
Electrolysis of Aqueous Solutions
we must consider if the water is
oxidized or reduced
if the water is oxidized, O2 and H+ ions
are formed
if the water is reduced, H2 and OHions are formed
the more positive the value of E°red, the
more favorable the reduction is
Prof. T.L. Heise, CHE 116
140
Electrolysis
Electrolysis with Active Electrodes
several practical applications of
electrochemistry are based on active
electrodes
electroplating involves using
electrolysis to deposit a thin layer of one
metal on top of another in order to
improve beauty or resistance to
corrosion
Prof. T.L. Heise, CHE 116
141
Electrolysis
Electrolysis with Active Electrodes
Prof. T.L. Heise, CHE 116
142
Electrolysis
Quantitative Aspects of Electrolysis
for any half reaction the amount of a
substance that is reduced or oxidized in
an electrolytic cell is directly
proportional to the number of electrons
passed into the cell
Fig 20.31
Prof. T.L. Heise, CHE 116
143
Electrolysis
Sample exercise: The half reaction for
formation of magnesium metal upon
electrolysis of molten MgCl2 is
Mg2+ + 2e- Mg. Calculate the mass of
magnesium formed upon passage of
current of 60.0 A for a period of
4.00 x 103 s.
Prof. T.L. Heise, CHE 116
144
Electrolysis
Sample exercise: Mg2+ + 2e- Mg. Calculate
the mass of magnesium formed upon
passage of current of 60.0 A for a period of
4.00 x 103 s.
Coulombs= A x s
= 60.0 A x 4.00 x 103 s
= 2.4 x 105 C
Prof. T.L. Heise, CHE 116
145
Electrolysis
Sample exercise: Mg2+ + 2e- Mg. Calculate
the mass of magnesium formed upon passage
of current of 60.0 A for a period of 4.00 x 103
s.
2.4 x 105 C 1 mole e- 1 mole Mg 24.305 g Mg
96,500 C 2 mole e1 mole Mg
Prof. T.L. Heise, CHE 116
146
Electrolysis
Sample exercise: Mg2+ + 2e- Mg. Calculate
the mass of magnesium formed upon passage
of current of 60.0 A for a period of 4.00 x 103
s.
2.4 x 105 C 1 mole e- 1 mole Mg 24.305 g Mg
96,500 C 2 mole e1 mole Mg
30.2 g Mg
Prof. T.L. Heise, CHE 116
147
Electrolysis
Work done by electrical current can be
calculated now based on the same formula as
Gibbs Free Energy.
wmax = -nFE * voltaic cells
w = nFEext
*electrolytic cells