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ELECTROCHEMICAL CELLS

Prepared by Ross S. Nord, Masanobu M. Yamauchi, and Stephen E. Schullery, Eastern Michigan University

CELL CONSTRUCTION AND POTENTIALS In principle, any spontaneous redox (oxidation-reduction) reaction can be used to construct an electrochemical cell. Cells will be constructed as shown in Figure 1. First, it is necessary to physically separate the chemicals involved in the oxidation and reduction half-reactions. We will place aqueous solutions of two different metal salts in separate

beakers. Then, suitable electrodes must be found. We will use a piece of each metal. The oxidation and reduction electrodes are commonly referred to as the anode and cathode, respectively. These terms derive from the fact that, when a voltaic cell reaction is running, anions migrate to the anode (to cast off their electrons and make it negative), and cations migrate to the cathode (to grab some electrons and leave it positive).

Figure 1. A sample setup for construction and voltage measurement of electrochemical cells. Next, wires are used to connect the electrodes. We will place a voltmeter in the circuit to measure the cell potential. When using a cell to power an electrical device (motor, light bulb, etc.), the device would be inserted in place of the voltmeter. Finally, provision must be made for ions to flow between the half-cells, to complete the "internal" circuit but yet not allow gross mixing of the reactants. A salt bridge composed of an absorptive material impregnated with an inert electrolyte is commonly used. A strip of filter paper soaked in 0.10 M KNO3 makes a surprisingly adequate salt bridge.

The voltage of a cell (Ecell) is related to the potential energy difference between electrons at the two electrodes, in units of joules/coulomb. This can be thought of as the difference between the two half-cell potentials.

𝐸!"## = 𝐸!"#!!"# − 𝐸!"#$% (1a)

By convention, it was chosen to use reduction potentials for both half-cells. Since oxidation is the opposite of reduction, an oxidation potential is equal in magnitude, but opposite in sign, to the corresponding reduction potential. While reduction is actually occurring at the cathode, oxidation is occurring at the anode. So, the

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reduction potential corresponding to the anode half-reaction is subtracted in equation (1a). HALF-CELL REDUCTION POTENTIALS The voltage that we measure is related to the difference in potential energy between the two half-cells. Thus, we cannot measure the potential of a single half-cell by itself. So, the value for one half-cell is arbitrarily chosen and all other values are measured relative to it. By standard convention, the reduction potential of the standard hydrogen electrode, SHE, is chosen to be 0.00 V.

2  𝐻! 𝑎𝑞, 1  𝑀 + 2  𝑒! → 𝐻! 𝑔, 1  𝑏𝑎𝑟 (2)

So, for example, when a SHE is connected to a half-cell containing Ag+|Ag, the silver electrode is found to be the positive electrode (meaning it is the cathode and that it is easier to reduce the Ag+ than the H+) and the measured cell potential is 0.800 V. Therefore, the Ag+|Ag half-cell is assigned a reduction potential of 0.800 V. Similarly, when the SHE is connected to a Pb2+|Pb half-cell, the SHE is found to be the positive electrode (so it is easier to reduce the H+ than the Pb2+) and the measured cell potential is 0.126 V. So, the Pb2+|Pb electrode is assigned a reduction potential of -0.126 V. We can use the half-cell reduction potentials to predict the cell potential for a Ag/Pb cell.

𝐴𝑔! 𝑎𝑞 + 𝑒! → 𝐴𝑔 𝑠                        𝐸 = 0.800  𝑉

𝑃𝑏!! 𝑎𝑞 + 2  𝑒! → 𝑃𝑏 𝑠              𝐸 = −0.126  𝑉

As we shall see, since the Ag+|Ag half-cell has the greater reduction potential, it is the cathode reaction and the Pb will be oxidized. The complete cell reaction can now be obtained by reversing the Pb half-cell reaction and then adding the two half-reactions together. However, since the number of electrons must cancel, we need to multiply the Ag half-cell reaction by 2 to yield:

                                                 𝑃𝑏 𝑠 → 𝑃𝑏!! 𝑎𝑞 + 2  𝑒!              

   2  𝐴𝑔! 𝑎𝑞 + 2  𝑒! → 2  𝐴𝑔 𝑠

⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯

2  𝐴𝑔! 𝑎𝑞 + 𝑃𝑏 𝑠 → 2  𝐴𝑔 𝑠 + 𝑃𝑏!!(𝑎𝑞)

and  𝐸!"## = 0.800  𝑉 −  −0.126  𝑉 = 0.926  𝑉

Now, let’s say that the Pb2+|Pb half-cell had been chosen to have a reduction potential of 0.00 V (instead of the SHE). Then, all of the other reduction potentials would also increase by 0.126 V (so Ag+|Ag would have a reduction potential of 0.926 V). However, the difference between any two potentials would be unchanged (and still equal to the experimental value). So, we could have chosen any half-cell as the starting point to construct a table of reduction potentials. When a measurement is made under standard conditions (e.g., the concentrations of the ions in each half-cell are exactly 1 M), the result is the standard cell potential

𝐸!"##! = 𝐸!"#!!"#! − 𝐸!"#$%! (1b)

where EO for each half-cell may be obtained from a table of standard reduction potentials.

VOLTAGE AND ΔG The voltage is also related to the Gibbs free energy of the chemical reaction

∆𝐺 = −𝑛𝐹𝐸                                                                                                  (3𝑎)

where n is the number of electrons transferred in the balanced equation (n=2 in our Ag/Pb example), E is the cell's voltage, and F is the Faraday constant (96,500 C/mol or 96.5 kJ/V-mol). If the cell is a standard cell, then

∆𝐺° = −𝑛𝐹𝐸°                                                                                            (3𝑏)

Notice that a negative ΔG, which corresponds to a spontaneous reaction, results in a positive voltage. Thus, the direction of spontaneous reaction can be deduced by noting how the electrodes must be connected to a voltmeter to give a positive reading. VOLTMETER READINGS The available multimeters in lab are capable of measuring resistance, current or voltage. All three functions share one terminal in common, on our meters labeled "COM." By convention, this terminal is to be connected with negative polarity. The other voltage terminal, although

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perhaps labeled only "V," must be the "positive" or "+" terminal. A voltmeter gives a positive reading when the positive terminal is in fact more positive than the common or negative terminal. Thus, we can deduce that a negative terminal must be connected to an electrode where electrons are being produced, i.e., where oxidation occurs. Similarly, a positive terminal must be connected to an electrode from which electrons are being removed, i.e., where reduction spontaneously occurs. Therefore, a positive voltmeter reading means that the half-cell with the higher reduction potential is connected to the positive terminal.

IN THIS EXPERIMENT Five half-cells will be constructed. One of them, the Fe2+|Fe half-cell will be arbitrarily assigned a reduction potential of zero.

𝐹𝑒!! 𝑎𝑞 + 2  𝑒! → 𝐹𝑒 𝑠

The reduction potentials of the other four half-cells will then be measured by connecting them to the Fe2+|Fe half-cell. This will allow for the construction of a table of reduction potentials. Next, six different electrochemical cells will be constructed by pairing all possible combinations of the remaining four half-reactions:

𝑍𝑛!! 𝑎𝑞 + 2  𝑒! → 𝑍𝑛 𝑠

𝑁𝑖!! 𝑎𝑞 + 2  𝑒! → 𝑁𝑖 𝑠

𝐶𝑢!! 𝑎𝑞 + 2  𝑒! → 𝐶𝑢 𝑠

𝐴𝑙!! 𝑎𝑞 + 3  𝑒! → 𝐴𝑙 𝑠

For each pair, one of these reduction half-reactions must be reversed to provide the oxidation necessary for a balanced redox reaction. Although the molecules will decide which half-reaction is to be reversed, you can make a good prediction by comparing the measured reduction potentials. The half-cell with the most positive reduction potential should remain as a reduction. The measured voltage of each cell will be compared with the expected voltage calculated from (1) the measured reduction potentials; and (2) the standard reduction potentials from your textbook. Note that the measured cell voltages are expected to roughly, but not exactly, match voltages calculated using textbook standard half-reaction EO values. The latter correspond to reactions carried out in hypothetical, ideal standard states, which your setups are not. Finally, you will write balanced cell reactions and then calculate ΔG and ΔGO for each reaction.

PRE-LABORATORY PREPARATION

You will need the standard reduction potential for each of the five half-cells you are using,

EXPERIMENTAL SECTION

REAGENTS PROVIDED

0.10 M solutions of KNO3, CuSO4, ZnSO4, NiSO4, and FeSO4

0.050 M Al2(SO4)3. (which is 0.10 M in Al3+) Solid metal electrodes Filter paper for construction of salt bridges

Hazardous Chemicals

NiSO4 – possible carcinogen and poison. Do not swallow it and avoid skin contact. Wash off immediately with copious amounts of water if

your skin comes in contact with it.

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WASTE DISPOSAL The Ni and Cu solutions should be disposed of in the designated waste containers.

PROCEDURE 1. Construct separate half-cells for each of the five half-reactions:

𝑍𝑛!! 𝑎𝑞 + 2  𝑒! → 𝑍𝑛 𝑠

𝐹𝑒!! 𝑎𝑞 + 2  𝑒! → 𝐹𝑒 𝑠

𝐶𝑢!! 𝑎𝑞 + 2  𝑒! → 𝐶𝑢 𝑠

𝑁𝑖!! 𝑎𝑞 + 2  𝑒! → 𝑁𝑖 𝑠

𝐴𝑙!! 𝑎𝑞 + 3  𝑒! → 𝐴𝑙 𝑠 See Figure 1. If necessary, very lightly sandpaper the metal electrodes to remove corrosion products and ensure good contact. Please do NOT bend the electrodes. Make salt bridges by soaking paper strips in 0.10M KNO3. Use a new salt bridge for every voltage measurement. 2. Measure the voltage of each of the four cells it is possible to construct by pairing the Fe2+|Fe half-cell with the other four half-cells: Set the multimeter to the 2V scale. (On some of the voltmeters this is the 2000 mV scale. If you have one of these voltmeters, you will need to divide the mV by 1000 to get the reading in

volts. For example, 347 mV = 0.347 V.) If the voltage is negative, reverse the leads to obtain a positive voltage reading. Do not allow the electrodes to come into contact with each other, or the salt bridge. Watch the voltage for a few seconds and then take a reading. Note which metal electrode was positive. You should determine the technique’s reproducibility by setting up the cells again with fresh solutions and repeating the measurements. Comment on reproducibility in the error analysis section of the report. 3. Measure the voltage of each of the six cells it is possible to construct by pairing the four half-cells other than the Fe2+|Fe half-cell. 4. Determine how the voltage for the cell constructed using the Al3+|Al and Zn2+|Zn changes with time. Construct the cell with fresh solutions and freshly-sanded electrodes. Record the voltage reading immediately after immersing the aluminum electrode in the aluminum solution and completing the circuit. Continue taking voltage readings every 15 seconds for 3 minutes.

DATA ANALYSIS

1. Determine the Measured Reduction Potential for each electrode from the average voltage measured in Step 2 of the procedure.

Remember that we defined Fe2+|Fe to have a reduction potential of zero.

Make sure that each reduction potential is reported with the proper sign. The electrodes that were positive compared to Fe, will have positive reduction potentials, while the electrodes that were negative compared to Fe (i.e., Fe was the positive electrode) will have negative reduction potentials.

2. Compare the Measured Reduction Potential for each half-cell to the corresponding standard reduction potential (as found in your textbook or other source).

Discuss any discrepancies in the error analysis section of your report.

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3. Use equation (1a) and the Measured Reduction Potentials in Step 3 of the procedure to calculate the cell voltage for each cell you constructed.

4. Use equation (1a) and standard reduction potentials to calculate the cell voltage for each cell you constructed in Step 3.

Compare the Measured and standard values and discuss any discrepancies in the error analysis section of your report.

5. Write the balanced (net ionic) equation for the oxidation-reduction reaction occurring in each cell.

6. Discuss the variation in voltage for the Al / Zn cell as a function of time. Al metal forms a stable coating of aluminum oxide that protects the metal underneath so it does not easily oxidize. Explain how sanding the electrodes influences the measured voltage.