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Homework answers / question archive / Experiment 8 Establishing a Table of Reduction Potentials The main objective of this experiment is to establish the reduction potentials of four metals relative to an arbitrarily chosen metal, and then arrange them into a table that has the form of a table of standard reduction potentials
Experiment 8
Establishing a Table of Reduction Potentials
The main objective of this experiment is to establish the reduction potentials of four metals relative to an arbitrarily chosen metal, and then arrange them into a table that has the form of a table of standard reduction potentials. This will be done by measuring the voltage difference between various pairs of half-cells made from the metals and suitable ionic solutions. Note: The words “voltage” and “potential” mean the same thing—historical usage determines when one or the other word is used.
A voltaic cell utilizes a spontaneous oxidation-reduction reaction to produce an electrical current and a voltage. (Note: The term “voltaic cell” means the same thing as the word “battery.”) Half-cells are normally produced by placing a piece of metal into a solution containing the cation of the metal (e.g., Cu metal in a solution of CuSO4 which ionizes to give Cu2+). In this version of a voltaic cell, the half cell will be a small piece of metal placed onto 3 drops of solution which is on a piece of filter paper. The solution contains the cation of the solid metal. Figure 1 shows the arrangement of half-cells on a piece of filter paper. The two half-reactions are normally separated by a porous barrier or a salt bridge. Here, the salt bridge will be several drops of aqueous NaNO3 solution placed on the filter paper between the two half cells. Using the computer as a voltmeter, the red lead makes contact with one metal and the black lead makes contact with another metal.
By comparing the voltages obtained for four pairs of half-cells, and by recording which metal was the (+) or (–) electrode, you can establish a table of reduction potentials for the five metals in this lab.
Windows PC |
one piece of filter paper, 11.0 cm diameter |
Vernier LabPro interface |
1 cm X 1 cm sheets of metals Cu, Zn, Pb, Ag, Fe |
Logger Pro |
1 M NaNO3 |
Vernier Voltage Probe |
1 M solutions of Cu2+, Zn2+, Pb2+, Ag+, Fe2+ |
one Petri dish |
steel wool |
forceps
Chemistry with Computers. Then open the experiment file called “Exp 28 LabPro Voltage.” The sensor that will be automatically set is called “Voltage (–10 to +10).”
The calibration is “Volts_10V.” The computer is now set to monitor voltage, in volts.
You can read the voltage in the Meter window when the voltage probe leads are connected to a cell. When the leads are not in contact with a cell (or each other), a meaningless default voltage may be displayed. If you touch the two leads together, the voltage will drop to 0.00 V.
Figure 1
Voltaic Cell (metals used) |
Measured Volt
#1 #2 |
age (V) #3 |
Metal at the (+) Termi nal |
Metal at the (–) Termi nal |
Half reaction that goes as a reduction (this is for the metal at the (+) terminal)* |
Half reaction that goes as an oxidation (this is for the metal at the (–) terminal)* |
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Cu – Zn |
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Cu – Pb |
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Cu – Ag |
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Cu – Fe
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TO HELP YOU DECIDE THE CHARGE OF EACH TERMINAL OF A HALF CELL.
An oxidation gives up electrons. Electrons are negative. So the half cell
where an oxidation is taking place is going to be the negative terminal because the negative electrons are coming out of this half cell.
A reduction takes electrons. Taking of electrons is essentially making
the half cell positive because the negative electrons are going into the half cell. So the half cell that goes as a reduction is going to be the positive electrode.
Voltaic Cell (metals used) |
Average Measured Voltage from the Data Table (V) |
Metal at the (+) Terminal |
Metal at the (–) Terminal |
Reduction potential of metal reacting with Cu (assumes that Cu has a reduction voltage of 0.00 V)1 |
Adjusted reduction potential of metal reacting with Cu (assumes that Cu has a reduction voltage of 0.34 V)2 |
Cu – Zn |
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Cu – Pb |
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Cu – Ag |
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Cu – Fe |
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Footnote #1: If Cu is the (+) terminal and the other metal (M) is the (–) terminal, the Cu is going as a reduction and M is going as an oxidation. By definition, the reduction potential of M is (–). If Cu is (–) and M is (+), then Cu is going as an oxidation and other metal is going as a reduction. By definition, the reduction potential of M is (+). Essentially, this means that the sign of each voltage in the first column is to be reversed when entered in this column.
Footnote #2: In the previous column, Cu was arbitrarily assigned a voltage of 0.00 V. However, the actual standard reduction potential of Cu is 0.34 V. Thus, to convert the reduction potential in the previous column to actual reduction potentials, add 0.34 V to each entry.
Reduction potential in descending order from most (+) to most (–) reordered from the last column in Results Table 1. (Don’t forget to insert Cu at the +0.34 V position in this table) |
Metal |
Actual reduction potential of each metal from the textbook (Zumdahl, 6th ed, p 833)* |
Percent error between your measured reduction voltage (column #1) and the actual reduction voltage (column #3). (Cu will, of course, have a 0% error)**
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* Be careful to chose the proper half reactions. These involve the metal and the ions listed in the materials section on page L–1.
**Students in the past have found that iron gives the largest percent error. It turns out that the “iron” we are using is really mild steel and is alloyed with other metals, so a large error is to be expected. We have not found a source of pure iron at this time.
Is the order of the metals in Results Table 2 the same as that in the table of Standard Reduction Potentials on p 833 in Zumdahl’s text?
Yes No (circle one)
If you circled No, check all your calculations. The order should come out the same as in the standard table.
The formula for percent error:
% error=yourvalue-book valuex 100% book value
The vertical lines stand for absolute value, meaning that the sign of the number between the vertical lines is always positive.
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