CHEM122L

General Chemistry Lab

Revision 2.0

Construction of Electrochemical Cells

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)

In Net Ionic form, this is written as:

Zn(s) + Cu2+

(aq) Cu(s) + Zn2+

(aq) (Eq. 2)

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) Zn2+

(aq) + 2 e- (Eq. 3)

Reduction: Cu2+

(aq) + 2 e- Cu(s) (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 Half-

Reaction 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 fritted-

glass 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:

H2(1 atm,Ideal) 2 H+(1M,Ideal) + 2 e

-

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- Zn|Zn2+,ox

o = 0.762 Volt

Cu2+

(aq) + 2 e- Cu(s) Cu|Cu2+,red

o = 0.339 Volt

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+,ox

o + Cu|Cu2+,red

o = 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+,red

o = - 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:

a A + b B c C + d D

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 non-

idealities:

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 I

1/2 (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 Erich Huckel

(https://en.wikipedia.org/wiki/Peter_Debye) (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 zNO3

2) = ½ ((0.001M) (+2)

2 + (0.002M) (-1)

2) = 0.003M

Then estimate Cu2+:

log Cu2+ = - 0.509 zCu2+2 I

1/2 = - 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+ + e

- Na (Cathode is a Steel Electrode)

2 Cl- Cl2 + 2 e

- (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 O2(g) + 4 H+(aq) + 4 e

- (Eq. 13)

Cathode:

2 H2O + 2 e- H2(g) + 2 OH

-(aq) (Eq. 14)

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

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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 Half-

Cells, 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) After making the Cell Potential measurement, dispose of the electrolyte

solutions in the labeled waste container. Rinse the H-Cell thoroughly with

distilled Water.

3. 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 non-

idealities 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 Half-

Cells 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 Zn(OH)2(s) + Hg(l)

Identify the Oxidation and Reduction.