Voltaic Cell Measurements Guide

Voltaic Cell Measurements Pre-Lab Activities: After reading the lab, complete items a, b, c, and d (title, purpose, chemicals and equipment, and summary of procedure) as described on page 10 of Exp. 1 on an 8 1/2 x 11 sheet of paper. Answer the following questions on 8 1/2 x 11 sheet of paper or in your laboratory notebook if one is required by your instructor: 1. Write the oxidation and reduction half reactions; a. Fe2+, Fe3+ ∥ Br2, Brb. Mg , Mg2+ ∥ Cr3+, Cr2+ 2. If E°Hg2+,Hg = +0.80V and E°Co2+,Co = -0.28V, then what is the Ecell for the reaction Co + HgCl2  CoCl2 + Hg? 3. If you assume the E°Ag+,Ag = 0.00 volts, then what is E°Cu2+,Cu when the voltage measured across the electrochemical cell below is +0.46 volts? Ecell = +0.46V for Cu,Cu2+ ∥ Ag+,Ag 4. Using the information above, what is E°Cl2,Cl- given the system below? Ecell = +1.02V for Cu,Cu2+ ∥ Cl2,Cl5. A student measures the potential of a cell made up with 1 M CuSO4 in one solution and 1 M AgNO3 in the other. There is a Cu electrode in the CuSO4 and an Ag electrode in the AgNO3, and the cell is set up as in Figure 6-1. She finds that the potential, or voltage, of the cell, Ecell, is 0.45V, and that the Cu electrode is negative. a. b. c. d. At which electrode is oxidation occurring? Write the equation for the oxidation reaction Write the equation for the reduction reaction. If the potential of silver, silver ion electrode, E°Ag+,Ag is taken to be 0.000 V in oxidation or reduction, what is the value of the potential for the oxidation reaction: E°Cu,Cu2+oxid? E°cell = E°oxid + E°red e. If E°Ag+,Ag red equals 0.80 V, as in standard tables of electrode potentials, what is the value of the potential of the oxidation reaction of copper, E°Cu,Cu2+ oxid? f. Write the net ionic equation for the spontaneous reaction that occurs in the cell that the student studied. g. The student adds 6 M NH3 to be CuSO4 solution until the Cu2+ ion is essentially all converted to Cu(NH3)42+ ion. The voltage of the cell, Ecell, goes up to 0.92V and the Cu electrode is still negative. Find the residual concentration of Cu2+ ion in the cell (Use Eq. 3). Lab Activities: Go over prelab questions with your lab instructor Complete lab and fill in data sheets. Objectives To calculate the reduction potentials (Ered) for seven electrode systems. To learn how to measure electrical potentials in a voltaic cell using a voltmeter. To learn to set up a simple electrochemical cell. Introduction: Many chemical reactions can be classified as oxidation-reduction reactions, because they involve the oxidation of one species and the reduction of another. Such reactions can conveniently be considered as the result of two half-reactions, one of oxidation and the other reduction. In the case of the oxidation-reduction reaction Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s) that would occur if a piece of metallic zinc were put into a solution of copper sulfate, the two reactions would be: Zn(s)  Zn2+(aq) + 2e- oxidation 2 e- + Cu2+(aq)  Cu(s) reduction The tendency for an oxidation-reduction reaction to occur can be measured if the two reactions are made to occur in separate regions connected by a barrier that is porous to ion movement. An apparatus called a voltaic cell in which this reaction might be carried out under this condition is shown in Figure 9 -1. If we connect a voltmeter between the two electrodes we will find that there is a voltage, or potential, between them. The magnitude of the potential is a direct measure of the driving force or thermodynamic tendency of the spontaneous oxidation-reduction reaction to occur. If we study several oxidation-reduction reactions we find that the voltage of each associated voltaic cell can be considered to be the sum of a potential for the oxidation reaction and a potential for the reduction reaction. In the Zn,Zn2+ ∥ Cu2+,Cu cell we have been discussing, [Red Probe] [Black Probe] Figure 9-1 For example, Ecell = EZn,Zn2+oxidation reaction + ECu2+,Cu reduction reaction [1] By convention, the negative electrode in a voltaic cell is taken to be the one from which electrons are emitted (i.e., where oxidation occurs). The negative electrode is the one that is connected to the minus pole of the voltmeter when the voltage is measured. Since any cell potential is the sum of two electrode potentials, it is not possible, by measuring cell potentials, to determine individual absolute electrode potentials. However, if a value of potential is arbitrarily assigned to one electrode reaction, other electrode potentials can be given definite values, based on the assigned value. The usual procedure is to assign a value of 0.0000 volts to the standard potential for the electrode reaction: 2H+(aq)  2e-  H2(g); E°H+,H2 red = 0.0000 V For the Zn,Zn2+ ∥H+,H2 cell, the measured potential is 0.76 V, and the zinc electrode is negative. Zinc metal is therefore oxidized, and the cell reaction must be: Zn(s) + 2H+(aq)  Zn2+(aq) + H2(g); Ecell = 0.76 V Given this information, one can readily find the potential for the oxidation of Zn to Zn2+. E°cell + E°Zn,Zn2+ oxid + E°H+,H2 red 0.76 V = E°Zn,Zn2+ oxid + 0.000 V; E°Zn,Zn2+ oxid = +0.76 V If the potential for a half-reaction is known, the potential for the reverse reaction can be obtained by changing the sign. For example: if E°Zn,Zn2+ oxid = +0.76 V, then E°Zn2+,Zn red = -0.76 V if E°Pb2+,Pb red = +Y V, then E°Pb,Pb2+ oxid = -Y V In the first part of this experiment you will measure the voltages of several different cells. By arbitrarily assigning the potential of a particular half-reaction to be 0.00V, you will be able to calculate the potentials corresponding to all of the various half-reactions that occurred in all of the cells. In our discussion so far we have not considered the possible effects of such system variables as temperature, potential at the liquid-liquid junction, size of metal electrodes, and concentrations of solute species. Although temperature and liquid junctions do have a definite effect on cell potentials, taking account of their influence involves rather complex thermodynamic concepts and is usually not of concern in any elementary course. The size of a metal electrode has no appreciable effect on electrode potential, although it does relate directly to the capacity of the cell to produce useful electrical energy. In this experiment we will operate the cells so that they deliver essentially no energy but exert their maximum potentials. The effect of solute ion concentrations is important and can be described relatively easily. For the cell reaction at 25C: aA(s) + bB+(aq)  cC(s) + dD2+(aq) Ecell = Ecell 0 − 0.06 n x log (conc. D2+ ) d (conc. B+ )b [2] Where Ecell is a constant for a given reaction and is called the standard cell potential and n is the number of electrons in either electrode reaction. By Equation 2 you can see that the measured cell potential, Ecell, will equal the standard cell potential if the molarities of D2+ and B+ are both unity, or, if d equals b, if the molarities are simply equal to each other. We will carry out experiments under such conditions that the cell potentials you observe will be very close to the standard potentials given in the tables in your chemistry text. Considering the Cu,Cu2+ ∥ Ag+,Ag cell as a specific example, the observed cell reaction would be: Cu(s) + 2Ag+(aq)  Cu2+(aq) + 2Ag(s) For this cell, Equation 2 takes the form: Ecell = Ecell0 − 0.06 2 (conc. Cu2+ ) x log (conc. Ag+ )2 In the equation n = 2 because in the cell reaction, two electrons are transferred in each of the two half-reactions. E would be the cell potential when the copper and silver salt solutions are both 1 M, since then the logarithm term is equal to zero. If we decrease the Cu2+ concentration, keeping that of Ag+ at 1 M, the potential of the cell will go up by about 0.03 volts for every factor of ten by which we decrease conc. Cu2+. Ordinarily it is not convenient to change concentrations of an ion by several orders of magnitude, so in general, concentration effects in cells are relatively small. Procedure: WEAR YOUR SAFETY GLASSES WHILE PERFORMING THIS EXPERIMENT You may work in pairs in this experiment. A. Cell Potentials. In this experiment you will be working with these seven electrode systems. Ag+, Ag(s) Cu2+, Cu(s) Zn2+, Zn(s) * Br2(l), Br-, Pt * Cl2(g, 1atm), Cl-, Pt * I2(s), I-, Pt * Fe3+, Fe2+, Pt Your purpose will be to measure enough voltaic cell potentials to allow you to determine the electrode potentials of each electrode by comparing it with an arbitrarily chosen electrode potential. Using the apparatus shown in Figure 9-1, set up a voltaic cell involving any two of the electrodes in the list. About 10 mL of each solution should be enough for making the cell. The solute ion concentrations may be assumed to be one molar and all other species may be assumed to be at unit activity, so that the potentials of the cells you set up will be essentially the standard potentials. Measure the cell potential with the Lab Quest and voltmeter attachment and record it along with which electrode has negative polarity. You will read the potential, in volts, on the LabQuest screen. Write down your readings on a separate paper. [3] If you observe an unexpected negative or positive value, reverse the RED and BLACK probe clips, in order to observe its opposite sign (+/-), still having the same value. In a similar manner set up and measure the cell voltage and polarities of the cells listed on the data table (A.1). Do not combine the silver electrode system with any of the halogen electrode systems (halogens are Br, Cl & I), because a precipitate will form. The data from this part of the experiment should be entered in the first three columns of the data table (A.1). DISPOSAL OF REAGENTS: As you finish each voltage measurement, pour the two solutions into a beaker. At the end of the experiment, pour the contents of the beaker into the labeled waste container. *Graphite electrodes will be substituted for platinum (Pt) Why are the values of Ered on the two bases related to each other in such a simple way? (Answer this question in the Discussion for this experiment.) Data: Table A.1: Cell Potentials Cell Potential E0cell (volts) Negative Electrode Oxidation Reaction E0oxidation in volts Reduction Reaction 1. Zn, Zn2+ || Cu, Cu2+ ____________ __________ ______________________ ____________ ______________________ _______ 2. Cu, Cu2+ || Pt, Fe2+, Fe3+ ____________ __________ ______________________ ____________ ______________________ _______ Electrode System used in Each Cell 3. Pt, Fe2+, Fe3+ || Pt, Br-, Br2 E0reduct vol ____________ __________ ______________________ ____________ ______________________ _______ 4. Pt, Br-, Br2 || Pt, I-, I2 ____________ __________ ______________________ ____________ ______________________ _______ 5. Zn, Zn2+ || Pt, I-, I2 ____________ __________ ______________________ ____________ ______________________ _______ || Ag, Ag+ ____________ __________ ______________________ ____________ ______________________ _______ 7. Zn, Zn2+ || Ag, Ag+ ____________ __________ ______________________ ____________ ______________________ _______ 6. Cu, Cu2+ Calculations: A. Noting that oxidation occurs at the negative pole in a cell, write the oxidation reaction in each of the cells. The other electrode system must undergo reduction; write the reduction that occurs in each cell. B. Assume that E°Ag+,Ag = 0.00 volts (whether in reduction or oxidation). Enter that value in the table for all of the silver electrode systems you used in your cells. Since Ecell = Eoxidation + Ereduction, you can calculate E values for all the electrode systems in which the Ag,Ag+ system was involved. Enter those values in the table. C. Using the values and relations in (B) and taking advantage of the fact that for any given electrode system, Eoxidation = -Ereduction, complete the table of Evalues. The best way to do this is to use one of the E values you found in (B) in another cell with that electrode system. That potential, along with Ecell, will allow you to find the potential of the other electrode. Continue this process with other cells until all the potentials have been determined. Table A.2: Electrode Potentials In Table A.1., you should have a value for Ered or Eoxid for each of the electrode systems you have studied. Remembering that for any electrode system, Ered = -Eoxid, you can find the value for Ered for each system. List those potentials in the left column of the table below in order of decreasing value. E0reduction (E0Ag+, Ag = 0.00V) Electrode Reaction in Reduction E0reduction (E°H+, H2 = 0.00V) ____________________ ____________________ ____________________ ____________________ ____________________ ____________________ ____________________ ____________________ ____________________ ____________________ ____________________ ____________________ ____________________ ____________________ ____________________ ____________________ ____________________ ____________________ ____________________ ____________________ ____________________ The electrode potentials you have determined are based on E°Ag+,Ag = 0.00 V. The usual assumption is that E°H+,H2 red = 0.00 volts, under which conditions E°Ag+,Ag red = 0.80 V. Convert from one base to the other by adding 0.80 volts to each of the electrode potentials, and enter these values in the third column of the table. Why are the values of Ered on the two bases related to each other in such a simple way? (Answer this question in the Discussion for this experiment.)