Electrochemical Cells
Manatee Community College
Introduction
If we place a strip of Iron metal in copper(II) sulfate solution we can see evidence
that a chemical reaction is occurring. The blue solution caused by the copper(II) sulfate
begins to fade and the iron metal has a deposit of metallic copper on it. The reaction that
is taking place can be seen in Equation 1.
(1) Fe(s) + Cu2+(aq)  Fe2+(aq) + Cu(s)
In Equation 1 the iron has lost 2 electrons and the copper has gained 2 electrons.
These types of reactions are called oxidation-reduction or redox reactions. In redox
reactions we can break up the total reaction into two parts, the oxidation half and the
reduction half. Equation 2 shows the oxidation half and Equation 3 shows the reduction
half.
(2) Fe(s)  Fe2+(aq) + 2 electrons
(3) Cu2+(aq) + 2 electrons  Cu(s)
In the oxidation half reaction (Equation 2) Fe(s) loses 2 electrons to become Fe2+(aq).
Oxidation results in the oxidation number of the substance increasing. When we set up a
half reaction we have to balance for the number of electrons for the reactant side with
respect to the product side. If we look at the periodic table we see that Fe(s) has 26
electrons, this means that Fe2+(aq) has only 24 electrons. To balance the number of
electrons we must add 2 electrons to the product side. You can only add electrons, never
subtract electrons!
In the reduction half reaction (Equation 3) Cu2+(aq) gains 2 electrons to become Cu(s).
Reduction results in a decrease in the oxidation number of a substance. When we set up
the half reaction, the electrons on the reactant side and the product side have to be equal
to each other. If we look at the periodic table Cu(s) has 29 electrons which means that
Cu2+ has 27 electrons. To get the reactant and product side to have equal number of
electrons we must add 2 electrons to the reactant side.
If we place a strip of copper in Iron sulfate solution (FeSO4) we observe no evidence
of a reaction. This shows that the reverse of the reaction in Equation 1 does not occur
spontaneously. We may conclude that Cu2+ ion has a greater tendency to accept electrons
from Fe than Fe2+ ion does to accept electrons from Cu. In other words, Cu2+ ion has a
greater reduction potential than Fe2+ ion.
Table 1 ranks selected metal ions by standard reduction potentials, determined under
standard conditions: 25C, 1 atm, 1 mol/L ion concentration. The standard reduction
potential, Ered, measured in volts (V) provides a quantitative measure of a metal ion’s
tendency to accept electrons. The greater the reduction potential the greater the tendency
for reduction.
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Table 2 Selected standard reduction potentials for half-reactions
Reduction Half-Equation
Ered, V
+
Ag (aq) + 1 electron  Ag(s)
.80
Cu2+(aq) + 2 electrons  Cu(s)
.34
+
2 H (aq) + 2 electrons  H2(g)
.00
Pb2+(aq) + 2 electrons  Pb(s)
-.13
2+
Sn (aq) + 2 electrons  Sn(s)
-.14
Ni2+(aq) + 2 electrons  Ni(s)
-.26
2+
Fe (aq) + 2 electrons  Fe(s)
-.44
2+
Zn (aq) + 2 electrons  Zn(s)
-.76
Cr2+(aq) + 2 electrons  Cr(s)
-.91
3+
Al (aq) + 3 electrons  Al(s)
-1.66
Mg2+(aq) + 2 electrons  Mg(s)
-2.37
The half-reactions that represent the oxidation of the metals in Table 1 to their
respective metal ions are the reverse of the half-reactions in Table 1. The standard
oxidation potential, Eox, of an oxidation half-reaction has the same numerical value, but
the opposite sign. An example of this with respect to Tin can be seen in Equation 4 & 5.
(4) Sn2+(aq) + 2 electrons  Sn(s)
Ered = -.14 V
(5) Sn(s)  Sn2+(aq) + 2 electrons
Eox = .14 V
We can use the standard oxidation and reduction potentials to predict if a reaction
will be spontaneous or non-spontaneous. Let’s take a look at the reaction of Zn2+(aq) with
Ni(s). First let’s set up a general reaction, which can be seen in Equation 6.
(6) Zn2+(aq) + Ni(s)  Zn(s) + Ni2+(aq)
Now we must separate the whole reaction into two half-reactions, one for
oxidation (Equation 7) and one for reduction (Equation 8). When we set these up we
need to include their respective standard oxidation and standard reduction potentials.
(7) Ni(s)  Ni2+(aq) + 2 electrons
Eox = .26 V
(8) Zn2+(aq) + 2 electrons  Zn(s)
Ered = -.76
(9) Ni(s) + Zn2+(aq)  Ni2+(aq) + Zn(s)
Enet = .26+(-.76) = -.50 V
To obtain the net equation we have to add the two half-reactions (see Equation 9).
This results in the electrons canceling out and Enet = -.50 V. Because the standard net
potential (Enet), or sum of the standard reduction and oxidation potentials, is a negative
number we can predict that this reaction would not occur. Remember that in a galvanic
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cell/redox reaction there must be a positive voltage. Having a negative voltage means
that we would have to supply a current to get this reaction to proceed.
Galvanic Cells
Oxidation-reduction reactions can occur between reactants in separate containers.
In this situation electrons are transferred from the reactants being oxidized to the reactant
being reduced through a wire connecting the two containers.
For example, suppose we place a Ni strip in an aqueous Ni2+ salt solution. Then
we place a Cu strip in an aqueous Cu2+ salt solution in a separate container. (the metal
strips are called electrodes) Next, we connect one end of a wire to the Ni strip and the
other end to the Cu strip. This wire allows electrons to pass from the Ni strip to the Cu
strip, provided that we make an additional connection between the containers to complete
the electrical circuit. This second connection is called a salt bridge. This assembly is
called a galvanic or voltaic cell. A galvanic cell is a system that utilizes a spontaneous
oxidation-reduction reaction to pump electrons through an electrical circuit. In this cell
we use a voltmeter to measure the net cell potential. This is the same as the Standard net
potential except that it’s reactants are not necessarily in a standard state condition.
Electrolytic Cells
A second type of electrochemical cell is the electrolytic cell. This is one in which an
external electric current is used to drive a non-spontaneous chemical reaction (Figure 1).
This forms the basis for all electroplating processes.
Figure 1
The layout of this cell is the same as a galvanic cell except the anode is now positive and
the cathode negative. Since the power source is driving electrons into the cathode, the
metal cations in solution will be reduced to form the metal on the cathode. As cations are
taken out of solution by this process additional metal atoms on the anode are oxidized to
fresh cations. The net result is that the cathode gains mass and the anode loses mass.
In this lab you will plate copper from a solution of copper sulfate onto a copper cathode.
The overall reaction is:
(10) Cu+2(aq) + 2e-  Cu(s)
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Michael Faraday showed the quantitative relationship between the number of electrons
flowing into the cathode and the amount of substance produced. It requires two “moles”
of electrons to produce one mole of copper. There are three quantities you will measure:
the current flow, time and the mass of copper. Current flow, amperes or amps, is a
measure of the charge passing through the system. The units are Coulombs per second.
So multiplying amps by time gives coulombs.
(11) Coulombs = amps x time (seconds)
Faraday’s Constant is the total charge on a mole of electrons.
(12) F = 96,500 Coulombs/mole
So we can get moles of electrons by dividing Coulombs by the Faraday Constant. But
since it takes two moles of electrons to make a mole of copper we divide the resultant by
two:
(13) Moles Copper = amps x time
2F
Procedure
I. Oxidation-Reduction Reactions
1.
The Zn-Cu System
Obtain and place a Zinc strip in a small test tube.
a) Add enough 0.1 M CuSO4 solution into the zinc test tube to
completely cover the zinc.
b) Wait 5 minutes, record your observations.
c) Take the zinc strip out of the solution and discard the zinc,
keep the solution for the next step.
2.
The Pb-Cu System
Obtain a lead strip and place it in the CuSO4 solution from step 1
a) Wait 5 minutes, record your observations
b) Place the metal strip and the solution into the designated waste
containers.
3.
The Zn-Pb System
Obtain and place a zinc strip in a small test tube
a) Add enough 0.1M Pb(NO3)2 solution into the zinc test tube to
completely cover the zinc.
b) Wait 5 minutes, record your observations.
c) Place the metal strip and the solution in the designated waste
containers.
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II.
Galvanic Cells
1.
The Zn-Cu Galvanic Cell
a) Label a 20 mL beaker “Zn” and place 15 mL of 0.1 M ZnSO4 in it
b) Label a 20 mL beaker “Cu” and place 15 mL of 0.1 M CuSO4 in it
c) Fold a piece of filter paper to about ½ inch in width, fold this piece in
half and place ½ in the copper solution and ½ in the zinc solution
d) Using the KNO3 solution completely soak the filter paper (this is the
salt bridge)
e) Connect one end of the voltmeter, via the alligator clamps, to a Zinc
strip
f) Connect the other end of the voltmeter to a copper strip
g) Make sure the voltmeter is on and set to the right setting
h) Place the copper strip in the copper solution and the zinc strip in the
zinc solution, Make sure to hold them both steady and not let the
alligator clips to touch the solution
i) Record the voltage on the voltmeter
j) Disconnect the alligator clips from the metal strips and dispose of all
metal and solution in the appropriate waste containers
2.
The Fe-Cu Galvanic cell
a) Follow the same procedure given in the previous cell while changing
the metal strips and the solutions used
b) Use 15 mL of 0.1 M FeSO4 and 15 mL of 0.1 M CuSO4
c) Place the Iron strip in the FeSO4 solution and the Copper strip in the
CuSO4 solution
d) Take the voltage
e) Disconnect the alligator clips from the metal strips and dispose of all
metal and solution in the appropriate waste containers
III. The Copper Electrolytic Cell
1. Obtain 2 copper strips and remove the oxide from the surface with sandpaper
or steel wool.
2. Label one strip A for anode and the other C for cathode and weigh them.
3. Place the strips in a 50 mL beaker connected to a power supply and ammeter
as shown in the schematic on the next page.
4. Add 40 mL of 1.0M CuSO4/0.1M H2SO4 and a magnetic stirrer. Stir the
solution gently.
5. Raise the voltage dial on the power supply until you get a reading of about
0.25 amps on the ammeter (approximately 4 volts).
6. Time the process for ten minutes and carefully note the current flow.
7. Turn off the power, remove the copper strips and rinse them in a beaker of
water. Dry them carefully and weigh the strips.
8. Determine the mass of copper plated, the moles of copper plated and compare
the results with the expected value from the current flow and time.
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Electrolytic Cell Schematic
+
-
Power Supply
Cu Anode (+)
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Ammeter
(-) Cu Cathode
6
Data Sheet
I.
II.
Oxidation-Reduction Reactions
1.
The Zn-Cu System
Observations:
2.
The Pb-Cu System
Observations:
3.
The Zn-Pb System
Observations:
Galvanic Cells
1. The Zn-Cu Galvanic Cell
Cell voltage:
___
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2. The Fe-Cu Galvanic Cell
Cell voltage:
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Copper Electrolytic Cell Data Table
Current (amps):______________
Time( seconds):______________
Coulombs:_____________
Calculated moles of copper produced (Show calculations):_______________
Initial Anode Mass:___________
Initial Cathode Mass:__________
Final Anode Mass:___________
Final Cathode Mass:___________
Mass of Cu Lost:____________
Mass of Cu Formed:___________
Average Copper Mass Produced:______________
Moles of Copper Produced:_________________
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Post-Lab Questions
I.
Oxidation-Reduction Reactions
1.
The Zn-Cu System
Oxidation ½ reaction:
Eoxid = ____
Reduction ½ reaction:
Ered = ____
Net Equation:
Enet = _____
Prediction of spontaneity:
Comparison of predicted spontaneity with experimental observations:
2.
The Pb-Cu System
Oxidation ½ reaction:
Eoxid = ____
Reduction ½ reaction:
Ered =_____
Net Equation:
Enet = ___
Prediction of spontaneity:
.
Comparison of predicted spontaneity with experimental observations:
3.
The Zn-Pb System
Oxidation ½ reaction:
Eoxid = ____
Reduction ½ reaction:
Ered =_____
Net Equation:
Enet =_____
Prediction of spontaneity:
Comparison of predicted spontaneity with experimental observations:
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II.
Galvanic Cells
1.
The Zn-Cu Galvanic Cell
Cu ½ reaction:
E = ______
Zn ½ reaction:
E = _______
Net cell reaction:
Enet = ______
Comparison of observed and calculated cell voltages:
_
2.
The Fe-Cu Galvanic Cell
Cu ½ reaction:
E = _______
Fe ½ reaction:
E = _______
Net cell reaction:
Enet = _______
Comparison of observed and calculated cell voltages:
III Electrolytic Cells
1. Aluminum is manufactured by the Hall-Heroult Process which produces
aluminum from aluminum oxide in an electrolytic cell. The net reaction is:
Al+3 + 3e-  Al(s)
You need to design a cell that will produce Al at the rate of 1 kg per hour. How
many amps must your power supply produce?
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Pre-Lab Questions:
1.
Answer the following questions about a galvanic cell that would have the
following net reaction
Cu(s) + 2 Ag+(aq)  Cu2+(aq) + 2 Ag(s)
a) Write the oxidation and reduction half reactions, net reaction, include
the cell potential of the oxidation, reduction, and net cell potential
Oxidation ½:
Eoxid =_____
Reduction ½:
Ered =_____
Net reaction:
Enet =_____
b) Is this reaction spontaneous? Why or why not?
2.
Rank the following metals by ease of reduction, starting with the easiest to
reduce
Ag+, Ni2+, Al3+, Pb2+, Mg2+
3.
Rank the following metals by ease of oxidation, starting with the easiest to
oxidize
Mg, Ag, Fe, Cu, Cr
4.
For the following reactions state if the reactions are favored in the forward
or reverse direction
a) Ag+(aq) + Cu(s)  Cu2+(aq) + Ag(s)
b) Ni2+(aq) + Sn(s)  Ni(s) + Sn2+(aq)
c) Mg(s) + Fe2+(aq)  Mg2+(aq) + Fe(s)
d) Pb(s) + Zn2+(aq)  Zn(s) + Pb2+(aq)
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