CHEM122L
General Chemistry Lab
Revision 2.0
Construction of Electrochemical Cells
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To learn about Electrochemical Cells; Voltaic and Electrolytic Cells.
To learn about Half-Reactions and Half-Cells.
To learn about how Concentration affects Electromotive Force.
In this laboratory exercise, we will construct a number of Voltaic Cells and measure the
Electrochemical Potential (cell) of each using a Digital Voltmeter (DVM). We will then
compare our measurements with the expected results. We will also construct a Concentration
Cell and again compare the measured Cell Potential with its expected result. Finally, we will
construct an Electrolytic Cell that can be used to electrolyze Water and examine the results of the
electrolysis.
The first Electrochemical Cell constructed was invented by Alessandro Volta, which he
described in a communication on 20 March 1800 to Sir Joseph Banks, President of the Royal
Society of London. His Cell was a Pile of pairs of Zinc and Silver disks separated by paper disks
soaked in salt water. With a tall Pile, he could detect a weak electric shock when he touched its
two ends.
(Volta’s Pile Schematics)
(A Voltaic Pile)
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We wish to understand how it was Volta's Pile was able to produce an electric shock and more
generally, how electrochemical cells work.
Electrochemical Cells are built using the chemistry of Oxidation-Reduction reactions. Batteries,
or Voltaic Cells, use the spontaneity of the underlying Cell Reaction (G < 0) to “drive” the Cell
so that electrons can be shunted through an external device (lightbulb, etc.) to do useful work.
On the other hand, the non-spontaneous Cell Reaction (G > 0) of an Electrolytic Cell must be
“driven” by a generator, primary cell (battery) or other source of electromotive force.
As a first step in understanding the operation of a Voltaic Cell, consider an oxidation-reduction
reaction that occurs spontaneously. A classic example is provided by the resulting chemical
reaction when a strip of Zinc metal is dipped into an aqueous solution of Cupric Sulfate, CuSO4.
Over time, we observe the disappearance of the bluish color of the Cu2+ ion and note the
appearance of a dark coppery platting on the Zinc metal. The Zinc metal also appears to
dissolve. Hence, we have the following chemical reaction occurring:
Zn(s) + CuSO4(aq)
Cu(s) + ZnSO4(aq)
(Eq. 1)
Cu(s) + Zn2+(aq)
(Eq. 2)
In Net Ionic form, this is written as:
Zn(s) + Cu2+(aq)
Here we note Zinc is being Oxidized while the Cupric Ion is being Reduced. Electrons are being
transferred from the Zinc metal to the Cupric Ion and the system behaves like a shorted battery.
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So, this chemical reaction can be used to construct a Voltaic Cell because, first, the reaction
proceeds spontaneously and, second, it involves a transfer of electrons which can potentially be
shunted through an external device.
In order to construct a Voltail Cell using this Redox Reaction we must first split the reaction into
two Half-Reactions; one half-reaction associated with the Oxidation and the other the Reduction.
Oxidation:
Zn(s)
Reduction:
Cu2+(aq) + 2 e-
Zn2+(aq) + 2 eCu(s)
(Eq. 3)
(Eq. 4)
Then we must arrange things physically so the two Half-Reactions occur in separate regions of
the Cell; two separate Half-Cells for the present case. This is done such that the electrons are
forced to travel through an external device in order for the Cell Reaction to progress forward. In
the present example, we can do this by separating the two reactions into two different beakers.
A Zinc electrode is used in the beaker on the left. Over time this electrode will dissolve and
produce Zinc ions. Copper metal is used as an electrode on which to plate the Cupric ions in the
beaker on the right. Any inert metal that is less Active than Copper, such as Platinum (Pt), Silver
(Ag) or Gold (Au), can be used as this electrode.
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The Zn electrode on the left is referred to as an Anode; it involves the Cells's Oxidation HalfReaction and produces electrons. The Cu electrode on the right is the Cathode and involves the
Cells's Reduction Half-Reaction, consuming electrons. This can be generalized:
Anode:
Electrode supplying electrons; where oxidation occurs.
Cathode:
Electrode consuming electrons; where reduction occurs.
A prob now em with our Cell construction now presents itself. The cathodic half-cell is
producing an excess of positive ions and the anodic half-cell is depleting them. This imbalance
within both half-cells is rectified by adding a Salt-Bridge to the Cell. A Salt-Bridge can be
constructed using a glass tube filled with an inert salt such as KCl in high concentration that
employs a porous plug at both ends, a saline soaked piece of paper or cardboard, or a frittedglass disk. In any case, the salt-bridge is arranged so that inert ions can leak into the half-cells,
but gross mixing of the half-cell solutions is prevented. For example, using a glass-tube
construction, where the tube is filled with KCl, in our Cell, Cl- will leak from the Salt-Bridge
into the anodic half-cell so as to counter the build-up of positive ions.
As a point of Cell nomenclature, the two electrodes of our battery are marked (+) or (-) according
to:
Cathode ===> (+)
Anode ===> (-)
This is so we can view our Cell in a more abstract way:
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This “+” and “-“ convention is reversed in an Electrolytic Cell.
Our sketch of the Cell above can be shortened into a Cell Diagram, where phase boundaries are
denoted with a | and the salt-bridge is denoted by ||. The anode, by convention, is placed on the
left-hand side of the Diagram and the cathode on the right. For our simple battery, the Cell
Diagram would be:
Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)
In operational terms, the useful Work which can be derived from a battery is related to its
Electrochemical Potential (). This is defined as the Voltage difference between the two
electrodes when the battery is operated in an Open-Circuit configuration; the device is
effectively removed and the electrodes are not connected. In practical terms, the
Electrochemical Potential can be measured with a Voltmeter having a very high internal
impedance. If the impedance is high enough, the current flow through the meter is sufficiently
low that the Cell is effectively operating as an Open-Circuit. Thus, we have:
 = Voltage across the Meter when Current ~ 0
(Eq. 5)
Because the Gibbs Free Energy change for the Cell reaction is the maximum useful work
obtainable, we have:
G = - n F 
(Eq. 6)
where n represents the number of moles of electrons shunted per mole of reaction and F is the
Faraday Constant; 96485 C/mole.
If the aqueous solutions in our Zn|Zn2+||Cu2+|Cu Cell are prepared such that the electrolytic
solutions are in their Standard State (1M concentration and Ideal behavior), then the
electrochemical potential is found by measurement to be 1.10 Volts and is generally denoted as
Cello.
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In a similar manner, each Half-Cell can be compared to a Standard Hydrogen Electrode (SHE)
and the associated Electrochemical Potentials are measured.
Zn(s) | Zn2+(1M,Ideal) || SHE
Cello = 0.762 Volt
SHE || Cu2+(1M,Ideal) | Cu(s)
Cello = 0.339 Volt
Here, SHE is short-hand for a Half-Cell built on the following Half-Reaction:
2 H+(1M,Ideal) + 2 e-
H2(1 atm,Ideal)
By convention, we take SHE = 0 Volts. If that is done, the Electrochemical Potential for each
Half-Cell measured relative to the SHE is:
Zn(s)
Zn2+(aq) + 2 e-
Cu2+(aq) + 2 e-
Zn|Zn2+,oxo = 0.762 Volt
Cu|Cu2+,redo = 0.339 Volt
Cu(s)
The Electrochemical Potential for any Cell is then simply the sum of the Electrochemical
Potentials for the two Half-Cells. For our example:
cello = Zn|Zn2+,oxo + Cu|Cu2+,redo = 0.762 V + 0.339 V = 1.10 Volt
Half-Cell Electrochemical Potentials are always Tabulated as Cathodic Reactions. So, the
Electrochemical Potential for the Zn|Zn2+ electrode would be listed as:
Zn2+(aq) + 2 e-
Zn(s)
Zn|Zn2+,redo = - 0.762 Volt
Note the change in sign. The Electrochemical Potential for the Cu|Cu2+ half-cell is already listed
in Cathodic form.
If the concentrations of the electrolyte solutions used in the Cell are not at 1 Molar, then the Cell
Potential will change slightly. This adjustment is given by the Nernst Equation:
Cell ~ Cello - (RT/nF) lnQ
(Eq. 7)
where for a Cell Reaction of the form:
aA + bB
cC + dD
we have:
Q = [C]c [D]d / [A]a [B]b
(Eq. 8)
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A note concerning Standard States is useful here. Gases are in their
standard state if they are at a pressure of Po = 1 barr and behave
ideally. Solutes in solution are in their standard state if they are at a
concentration of co = 1M and behave ideally. Because charged ions
in an aqueous solution interact very strongly, electrolyte solutions
are usually very non-ideal and hence the approximation of Equation
7.
Walther Nernst
(https://en.wikipedia.org/wiki/Walther_Nernst)
Activity Coefficients () are used to correct each solute concentration for the effects of nonidealities:
Q = (C [C])c (D [D])d / (A [A])a ) (B [B])b
(Eq. 9)
If activity coefficients are included, Equation 7 becomes exact. Typically, activity coefficients
must be measured experimentally. However, for dilute electrolytic solutions, they can be
estimated using the theory of Peter Debye and Erich Huckel:
log i = - 0509 zi2 I1/2
(Eq. 10)
where i is the activity coefficient of ion "i" whose charge is zi and which is in an aqueous
solution at 25oC with an ionic strength of I.
Peter Debye
(https://en.wikipedia.org/wiki/Peter_Debye)
Erich Huckel
(https://en.wikipedia.org/wiki/Erich_H%C3%BCckel)
The Ionic Strength of a solution is calculated according to:
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I = ½  ci zi2
(Eq. 11)
where ci is the molar concentration of ion "i" and the sum is over all the ions in solution.
Example
Determine the activity coefficient of the Cu2+ ion, Cu2+, in an aqueous solution that is 0.001 M
Cu(NO3)2.
First determine the ionic strength of the solution:
I = ½ (cCu2+ zCu2+2 + cNO3 zNO32) = ½ ((0.001M) (+2)2 + (0.002M) (-1)2) = 0.003M
Then estimate Cu2+:
log Cu2+ = - 0.509 zCu2+2 I1/2 = - 0.509 (+2)2 (0.003)1/2 = - 0.112
Cu2+ = 10-0.112 = 0.773
It should be noted from this example that these "correction factors" can be quite significant.
Turning our attention now to Electrolytic Cells, as mentioned above, electrolytic cells must be
electrically driven. They frequently involve an electrolytic break-down of the Cell system. For
instance, the electrolysis of molten NaCl in a Down's Cell generates liquid Na and gaseous Cl2.
2 NaCl(l)
2 Na(l) + Cl2(g)
Here, the underlying half-reactions are:
Na+ + e2 Cl-
Na
Cl2 + 2 e-
(Cathode is a Steel Electrode)
(Anode is a Graphite Electrode)
This is useful if the electrolytic breakdown products are commercially desirable.
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This experiment will involve the construction of several electrochemical cells. In the first case,
we will construct an electrolytic cell for the electrolysis of Water and observe what happens as
the electrolysis proceeds.
2 H2O
2 H2(g) + O2(g)
(Eq. 12)
At each electrode we expect the following chemistry:
Anode:
2 H2O
Cathode:
2 H2O + 2 e-
O2(g) + 4 H+(aq) + 4 eH2(g) + 2 OH-(aq)
Next we will construct the following Voltaic Cells and measure their Cell Potentials.
Zn | Zn2+(aq, 0.001 M) || Pb2+(aq, 0.001 M) | Pb
Zn | Zn2+(aq, 0.001 M) || Cu2+(aq, 0.001 M) | Cu
Mg | Mg2+(aq, 0.001 M) || Zn2+(aq, 0.001 M) | Zn
(Eq. 13)
(Eq. 14)
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Finally, we will construct the following Concentration Cell and measure its Cell Potential.
Cu(s) | Cu2+(aq, 0.001M) || Cu2+(aq, 0.01M) | Cu(s)
The measured Potentials for each of the above Voltaic cells will be compoared with that
predicted by Nernst's formulation and deviations will be considered.
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Pre-Lab Questions
1.
If I wish to construct a Voltaic Cell using Zinc-Zinc Ion and Silver-Silver Ion HalfCells, which metal will serve as the Anode and which the Cathode? Justify your
answer quantitatively.
Data
Zn|Zn2+,redo = - 0.762 V
Ag|Ag+,redo
= + 0.80 V
2.
Write a Cell Reaction for the above Cell. Write appropriate Half-Reactions.
Identify the Oxidation and Reduction.
3.
Draw a Cell Diagram for this Cell.
4.
Determine Cello if 0.001 M ZnSO4 and 0.001 M Ag2SO4 are used as electrolytes.
i) First calculate Cello.
ii) Then calculate the ionic strength I (Eq. 11) for the ZnSO4 and Ag2SO4 electrolyte
solutions.
iii) Use the Debye-Huckel equation (Eq. 10) to determine the activity coefficients Zn2+ and
Ag+.
iv) Finally determine Cell using the Nernst Equation (Eqs. 7 and 9).
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Procedure
Electrolytic Cell for the Electrolysis of Water
1.
We will use a Brownlee Electrolysis Apparatus and Battery Jar as our Electrolytic
Cell for the electrolysis of Water. The Brownlee Apparatus consists of two Pt
electrodes which can be connected to a DC power source and inserted into gas
collection tubes. Obtain the Brownlee Apparatus, the Battery Jar, two gas
collection tubes, a 1500 mL beaker and a 250 mL beaker.
Brownlee Apparatus
Battery Jar
Gas Collection Tubes
2.
As Water is a poor conductor, we will use 0.5 M H2SO4 as the electrolyte solution
to be electrolyzed. Prepare enough 0.5 M H2SO4 to fill the Battery Jar about 2/3full
and to fill the Gas Collection Tubes from the 6 M H2SO4 stock provided. Add a
few drops of Bromthymol Blue indicator to the solution; pKa ~ 7. (The acidic form
of Bromthymol Blue is Yellow and the basic form is Blue.)
3.
Fill the Battery Jar 2/3 full with this solution.
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4.
Fill the Gas Collection Tubes with the same solution and place them in the Battery
Jar. Do this by closing the filled test tube with a thumb while wearing gloves. Turn
the Tube upside down and lower it into the solution in the Battery Jar. Allow the
test Tube to rest on the bottom of the Jar. Make sure the Tube is completely filled
with solution. Repeat this process with the other Gas Collection Tube.
5.
Place the Brownlee Apparatus support across the top of the Jar. Slip the Collection
Tubes over the electrodes and fasten them into the clips.
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6.
Connect the binding posts of the Apparatus to a Power Supply Connection Adapter
which is attached to a 3-12 V AC-DC power converter. This converter will act as
our power source for the electrolysis. Note the immediate formation of Hydrogen
and Oxygen gas. After a moment note the color of the Bromothymol Blue indicator
around each electrode.
7.
Allow the gas generation to continue until the tube collecting the Hydrogen gas is
full. At this point disconnect the power supply. Note the relative volumes of the
Hydrogen and Oxygen gas collected.
8.
Carefully and without losing any collected gas remove each Collection Tube from
the Apparatus. Apply a burning splint to the Tube containing the Hydrogen gas. A
popping sound is indicative of combusting Hydrogen. Apply a glowing splint to the
tube containing the Oxygen gas. The splint should flare in the presence of Oxygen.
Voltaic Cells
1.
We will construct our Voltaic Cells using an H-Cell apparatus. This apparatus uses
a porous fritted glass disk as its Salt-Bridge. (These Cells are very expensive, so
treat them with care.) Obtain an H-Cell from the stockroom and drain out the
0.1 M HNO3 storage solution. Clamp it into place using a ring stand.
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2.
Using successive dilutions, prepare about 100 mL of 0.001 M of each of the
following electrolyte solutions from the 0.1 M stock solutions provided:
ZnSO4
Pb(NO3)2
Mg(NO3)2
CuSO4
2.
Construct a Zn | Zn2+(aq, 0.001 M) || Pb2+(aq, 0.001 M) | Pb Voltaic Cell.
i)
Fill each Half-Cell of the H-Cell with the appropriate electrolyte solution such
that it rises to just above the fritted disk.
ii)
Use a piece Steel-Wool to clean the Metal Strips before using them in the
Cell. Use the bottom of a large weighing dish as a surface on which to
sand down the metal strip. Do not use the Steel-Wool directly on the
countertop. Wear gloves while doing this. If needed, briefly dip the metal
strips in ~1M HNO3 and then rinse them with Deionized Water.
iii)
Place the appropriate metal electrode into each Half-Cell and crimp it over the
lip of the H-Cell.
iv)
Connect the Cathode to the (+) terminal on the Voltmeter and the Anode to
the (-) terminal using alligator clips.
v)
Measure cell.
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vi)
3.
After making the Cell Potential measurement, dispose of the electrolyte
solutions in the labeled waste container. Rinse the H-Cell thoroughly with
distilled Water.
Make the same Cell Potential measurement for the following Cells:
Zn | Zn2+(aq, 0.001 M) || Cu2+(aq, 0.001 M) | Cu
Mg | Mg2+(aq, 0.001 M) || Zn2+(aq, 0.001 M) | Zn
Concentration Cell
1.
Construct the following Concentration Cell and measure its Cell Potential as above.
Cu(s) | Cu2+(aq, 0.001M) || Cu2+(aq, 0.01M) | Cu(s)
2.
When finished, rinse and fill the H-Cell with 0.1 M HNO3 storage solution.
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Data Analysis
Electrolytic Cell for the Electrolysis of Water
1.
For the Electrolysis of Water:
Examine the Half-Reactions (Equations 13 and 14) for this Cell and confirm that your
observations are indeed consistent with them. Your comments should include mention
of:
Relative Volumes of Gases Produced
Color of the Acid-Base Indicator
Results of the Splint Tests
Voltaic Cells
1.
For each Voltaic Cell constructed:
a) Write the Half-Reactions and identify the Anode and Cathode.
b) Determine Cell using the Nernst Equation (Eq. 7). Be sure to correct for nonidealities using the Debye-Huckel theory (Eqs. 9, 10, 11). Calculate the
Percentage Difference between your measurement and the theoretical
prediction. Tabulated values of o for each Half-Cell can be found in your
textbook. Explain any discrepancies.
Concentration Cell
1.
For the Concentration Cell:
Cu(s) | Cu2+(0.001M) || Cu2+(1M) | Cu(s)
Calculate Cell using the Nernst Equation; as above. Compare this (percentage difference)
with your measured value. Explain any discrepancy.
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Post Lab Questions
1.
Arrange the following metals according to their placement in the Activity Series:
Zinc
Magnesium
Lead
Copper
How does this arrangement compare to the Std. Electrochemical Potentials for the HalfCells prepared from these metals as tabulated in your textbook?
2.
The Cell reaction for a Mercury Battery, used in watches and hearing aids, is:
Zn(s) + HgO(s) + H2O
Identify the Oxidation and Reduction.
Zn(OH)2(s) + Hg(l)