Figure 1. A voltaic cell Cu,Cu 2+ Ag +, Ag. gas is, by convention, assigned a reduction potential of 0.00 V.

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1 Voltaic Cells Introduction In this lab you will first prepare a set of simple standard half-cells and then measure the voltage between the half-cells with a voltmeter. From this data you will be able to rank each half-cell by its ability to act as a reducing agent under standard conditions. Secondly, you will explore the effects changes in electrolyte concentration have on cell potential using the Nernst equation. Part A: Oxidation-Reduction and Relative Cell Potentials Many chemical reactions can be classified as oxidation-reduction (or redox) since they involve the oxidation of one species and the simultaneous reduction of another species. Redox reactions can be separated into two half-reactions, one oxidation and one reduction. Consider the reaction between copper and silver ions: Cu(s) + 2Ag + (aq) Cu 2+ (aq) + 2Ag(s) This reaction will occur if a piece of copper is placed in a solution silver ions. The half reactions would be: Cu(s) Cu 2+ (aq) + 2e oxidation 2Ag + (aq) + 2e 2Ag(s) reduction The tendency for a redox reaction to occur can be measured if the two half-reactions are made to occur in separate regions (half-cells) that are connected by a porous barrier that allows ion flow between the half-cells. There are many ways to prepare a voltaic cell. Figure 1 gives an example of one method you will use. If a voltmeter is connected between the two electrodes a voltage, or potential is measured. The magnitude of the potential is a direct measure of the driving force or thermodynamic tendency of the redox reaction to occur. Figure 1. A voltaic cell Cu,Cu 2+ Ag +, Ag gas is, by convention, assigned a reduction potential of 0.00 V. If one studies several redox reactions we find that the potential of the overall cell can be considered the sum of the potential for the individual oxidation and reduction halfreactions. For example, consider the copper-silver cell illustrated in Figure 1 where the cell voltage is shown to be 0.41 V. In terms of the half cell reactions the cell potential, E cell, can be written as: E cell = E red + E ox 0.41 V = E(Ag +,Ag) + E(Cu,Cu 2+ ) By definition, the anode is the electrode where oxidation takes place. The cathode is the electrode where reduction takes place. In our cell Ag is the cathode and Cu is the anode. The absolute values for E red and E ox are not measured with the voltmeter; only their sum of V can be measured. Since we cannot directly measure the voltage of an individual halfreaction the reduction of hydrogen ions to produce hydrogen 2H + (aq) + 2e H 2 (g); V = 0.00 V The hydrogen half-cell thus becomes the reference electrode. All other half-cell potentials are measured relative to the hydrogen electrode half-cell. Refer to your text for a table of half-cell reduction potentials relative to the hydrogen half-cell. For example, the standard reduction potential of Ag + ions to produce Ag metal is given as V. The greater the standard reduction potential, the greater the tendency for that species to become reduced. Ag + ions have a V greater tendency to be reduced than H + ions. In our measurements we will not be using a hydrogen half-cell as a reference electrode. We will use a half-cell of our choice as our reference electrode. This electrode will be assigned a potential of 0.00 V by convention for our measurements. This will not effect the measurements of the overall cell potentials since they are always the difference between two half-cell potentials, and therefore independent of the choice of reference. Voltaic Cells.doc 1 April 18, 2008

2 Making Voltaic Cell Measurements in the Lab The voltmeter will measure only the overall redox potential between the two half-cells. To assign individual half-cell reduction potentials we need to understand how a voltmeter measures a potential and then follow a few conventions of electrochemistry. The voltmeter has two probes, a red and black probe. By design the voltmeter records the voltage difference between the red probe and the black probe. By convention, the potential at the black probe is assigned a voltage of 0.00 V. Therefore, the voltage measured is the potential of the red probe relative to 0.00 V. For example, in Figure 1 the voltmeter reads V with the black probe on the Cu half-cell and the red probe on the Ag half-cell. If we assign a voltage of 0.00 V to the Cu half-cell (black probe) then the Ag half-cell is V higher in potential than the Cu half-cell. This means that the Ag half-cell reduction potential is V above the Cu half-cell reduction potential. If we reverse the probe leads, black on Ag (0.0 V) and red on Cu, we will measure a voltage of 0.41 V for the Cu half-cell. Again the Ag half-cell is V above the Cu half-cell in reduction potential. In summary, the voltmeter gives the reduction potential of the half-cell at the red probe relative to the reduction potential of the half-cell at the black probe. Now we can use this voltage information to assign anode or cathode designations to each half-cell and also rank each halfcell in order of relative reduction potential. From Figure 1 the Ag + /Ag half-cell is at the higher potential and therefore is the cathode. The Cu/Cu 2+ half-cell is at a lower potential and is therefore the anode. If one assigns zero volts to the Cu half-cell reduction potential (black probe) than one must assign V to the Ag half-cell reduction potential (red probe): Ag + (aq) + e > Ag(s) E half-cell = V Cathode Cu 2+ (aq) + 2e > Cu(s) E half-cell = 0.00 V Anode The voltage for any voltaic cell can be calculated from the half-cell reduction potentials at the cathode and anode: E cell = E cathode E anode From Figure 1, E cell = V 0.00 V = V since Ag is at the cathode and Cu at the anode. Using a table of standard reduction potentials, we find the accepted half-cell potentials for Ag + and Cu 2+ are V and V respectively (relative to the hydrogen electrode). The cell potential for the overall redox reaction between Ag + and Cu would be (under standard conditions): E cell = V 0.34 V = V. This is very close to our measurement in Figure 1, even though we have a 0.1 M Ag + concentration instead of 1 M. Measuring other half-cells relative to the Cu half-cell will allow you to rank each half-cell in terms of reduction potential. The more positive the half-cell reduction potential the greater is the potential for the ion to be reduced. Using the Cu half-cell as the reference electrode (black probe) and measuring a Ni/Ni 2+ half-cell (red probe) you will find the voltmeter reads a voltage of about 0.62 V. This means that the Ni 2+ half-cell reduction potential is 0.62 V below the Cu half-cell reduction potential. We can now rank the Ag, Cu, and Ni half-cell potentials relative to each other: Ag + (aq) + e > Ag(s) Cu 2+ (aq) + 2e > Cu(s) Ni 2+ (aq) + 2e _ > Ni(s) E half-cell = V E half-cell = 0.00 V E half-cell = 0.62 V These can be easily converted to half-cell potentials relative to the standard hydrogen electrode. (Think about it: How do you do this conversion?) Reactions are spontaneous when the overall cell potential, E cell, is positive. When the Cu/Cu 2+ and Ni/Ni 2+ half-cells are connected electrons will flow spontaneously to the half-cell with the greatest reduction potential, in this case the Cu/Cu 2+ half-cell. The Cu 2+ ions are reduced while the Ni(s) is oxidized. The overall spontaneous reaction is Cu 2+ (aq) + Ni(s) > Cu(s) + Ni 2+ (aq) E cell = V. Part B: Nonstandard Cell Potentials: The Nernst Equation So far we have not considered the possible effects of potentials at the liquid-liquid junctions, the size of the metal electrodes, temperature and concentration of solutes on the cell potential. Although potentials at the liquid-liquid junctions do have an effect on the cell potential, the effect is minor and taking into account their influence involves complex thermodynamic corrections. The size of the metal electrodes has no effect on the cell voltage but does determine the capacity of the cell to produce a useful current. Temperature and solute concentration will effect the cell potential and their effect is represented in the Nernst equation: E cell = E cell Voltaic Cells.doc 2 April 18, 2008 RT nf ln Q

3 Here, E cell is the cell potential under nonstandard conditions, E cell is the standard cell potential for the redox reaction as written, R is the universal gas constant in units of J/(mol K), T is in Kelvin, n is the number of electrons transferred in the overall redox reaction, and F is Faraday s constant, 96,485 C/mol. Q is the reaction quotient as in our studies of equilibrium, [products]/[reactants], and is derived from the redox reaction as written. Be careful that you always use the sign of E cell that corresponds to how the redox reaction is written. If the reaction is reversed, E cell changes sign and the reaction quotient is inverted. For example, the following reaction is spontaneous under standard conditions: Cu(s) + 2Ag + (aq) Cu 2+ (aq) + 2Ag(s); E cell = V Q is defined as, Q = [Cu 2+ ]. Remember to include the stoichiometric coefficients in the expression for Q. If the above [Ag + 2 ] reaction is reversed, E cell changes sign and Q is inverted: Q is now defined as, Q = [Ag + ] 2. [Cu 2+ ] Cu 2+ (aq) + 2Ag(s) Cu(s) + 2Ag + (aq); E cell = 0.46 V The Nernst equation allows for corrections to the cell potential when concentrations are not 1 M and the temperature is not 25 C. Since R and F are constants the Nernst equation can be simplified to: 8.617x10 5 V E cell ( K ) T = E cell ln Q n At 25 C this further reduces to: E cell = E cell V n For example, assume the concentration of ions in the cell of figure 1 are 0.1 M and M for the Ag + and Cu 2+ ions respectively. The cell potential (at 25 C) would be: V " E cell = E cell ln [Cu2+ ] % n $ #[Ag + ] 2 ' = V V " ln [0.0001] % $ & 2 # [0.1] 2 ' = V & The cell voltage has become more positive, (more spontaneous), because the system is further from equilibrium. If we decrease the copper ion concentration, keeping the silver ion concentration constant at 0.1 M, the cell potential will increase by about 0.03 V for every factor of ten decrease in Cu 2+ concentration. If we should add a complexing agent to the solution containing the Cu 2+ ions, the value of the copper ion concentration would drop drastically, and the voltage change would be appreciable. You will demonstrate this effect by adding ammonia to complex the copper ion. From the measured cell potential, the Nernst equation can be used to calculate the copper ion concentration remaining in solution after the ammonia is added. In a similar experiment you will determine the solubility product constant, K sp, of silver chloride. In this case you will immerse the silver electrode in a solution of known chloride concentration, 1 M Cl, and a small amount of AgNO s (aq) will be added. The Ag + (aq) will precipitate until K sp is satisfied. From the measured cell potential, the Nernst equation can be used to find the Ag + ion concentration remaining in solution. From this K sp for AgCl can be calculated. MATERIALS LabPro interface Metal electrodes: 5x5 mm 2 of Ag, Cu, Pb, Zn, Al, Mg Voltage Probe 250-ml beaker 24-well test plate String steel wool or sand paper plastic Beral pipets Waste beaker 10-mL graduated cylinder Porous crucible 50-mL beaker Digital thermometer Voltaic Cells.doc 3 April 18, 2008 ln Q 0.10 M silver nitrate, AgNO 3 solution 0.10 M copper (II) nitrate, Cu(NO 3 ) 2, solution 0.10 M lead (II) nitrate, Pb(NO 3 ) 2, solution 0.10 M zinc nitrate, Zn(NO 3 ) 2, solution 0.10 M aluminum nitrate, Al(NO 3 ) 3, solution 0.10 M magnesium nitrate, Mg(NO 3 ) 2, solution 1.0 M copper (II) sulfate, CuSO 4, solution 0.5 M potassium nitrate, KNO 3, solution 6 M ammonia, NH 3 solution 1 M potassium chloride, KCl solution

4 Procedure Part A: Voltaic Cell Measurements 1. Obtain and wear goggles. 2. Use a well test (culture) plate to construct your voltaic cell. Use a Beral pipet to transfer a small amount (1 ml) of 0.10 M Cu(NO 3 ) 2 into a center well. Add 0.10 M AgNO 3 solution to a neighboring well in the test plate. CAUTION: Handle these solutions with care. If a spill occurs, ask your instructor how to clean up safely. 3. Obtain one Cu and one Ag metal strip to act as electrodes. Polish each strip with steel wool or sand paper. Connect the Cu strip to the black lead of the voltage probe. Connect the Ag strip in the red lead of the voltage probe. Note: If the metal contacts on the voltage probes appear corroded, polish them using steel wool or sand paper. 4. Make a salt bridge by soaking a short length of string in a 50-mL beaker than contains a small amount of 0.5 M KNO 3 solution. Connect the Cu and Ag half-cells with the string. 5. Turn on your LabPro with a voltage sensor attached to measure DC voltage. Place the Cu strip into the well with Cu(NO 3 ) 2 and the Ag strip into the well with AgNO 3. Record the voltage. Pay attention to the sign (+ or ) of the voltage reading. 6. Repeat this procedure for the remaining metal electrodes and 0.10 M solutions listed in Table I. Always using the copper half-cell with the black lead of the voltage probe as reference. Use a fresh salt bridge for each new half-cell constructed. Record all voltages with signs in Data Table I. 7. When finished, wash the metal pieces, catching the rinse in a waste beaker, and return each metal piece to its container. Also, rinse the test-plate into the waste beaker. Procedure Part B: Nonstandard Cell Potentials Complex Ion Formation 1. Set up the Cu-Ag cell as shown in Figure 1. Use the porous crucibles provided by the stockroom, not your small crucible. Use 10 ml of 1 M CuSO 4 solution in the crucible and 10 ml of 0.10 M AgNO 3 in the beaker. Use a piece of copper for the copper electrode and a piece of silver for your silver electrode. When making cell measurements, the voltmeter probes must only touch the copper and silver electrodes, do not contact them with the electrolyte solutions. 2. Record the temperature of one of the solutions. 3. Record the cell voltage, (black probe on Cu, red probe on Ag) in the space provided. This will be used for E cell. 4. Add 10 ml of 6 M NH 3 solution to the CuSO 4 solution, stirring carefully with a stirring rod. Measure the cell potential again after the potential has stabilized. 5. Pour the solutions into your waste beaker. Determination of the K sp for AgCl 1. Remove the crucible from the beaker. Discard the solutions in the appropriate waste containers. Clean the crucible by using a small amount of nitric acid and rinsing with distilled water several times. Dispose of the acid and rinse in the Cu 2+ ion waste container. Reassemble the cell this time placing the CuSO 4 in the beaker and 10 ml of 1 M KCl in the crucible. Place the siver and copper electrodes in their respective ions solutions. 2. Add two drops of 0.10 M AgNO 3 to the KCl solution and stir. An equilibrium involving Ag + and Cl will be established as AgCl precipitates. 3. Record the temperature of one of the solutions. 4. Record the cell voltage, (black probe on Cu, red probe on Ag) in the space provided. In this case the [Ag + ] may be very low, which will decrease the cell potential to such an extent that the polarity may change. When making cell measurements, the voltmeter probes must only touch the copper and silver electrodes, do not contact them with the electrolyte solutions. 5. When finished, wash the Cu and Ag pieces, catching the rinse in your waste beaker, and return each metal piece to its container. 6. Pour the solutions into your waste beaker. Clean-up Dispose of the your waste solution in the waste container provided for this experiment. Voltaic Cells.doc 4 April 18, 2008

5 Name Part A: Voltaic Cells Results Data Table I. Complete the following table based on your results with Cu as the reference cell. Half-cell electrode Ag Pb Zn Al Mg E reduction relative to Cu (V) Anode half-cell Cathode half-cell E cell (V) Overall redox rxn between the half-cells Use your E reduction values relative to Cu to make a list of relative reduction potentials for the measured half-cells. Include Cu in the list. List the most positive half-cell reduction potential at the top of Table II, the least positive on the bottom. Write the corresponding REDUCTION half-rxn for each metal. Convert your half-cell voltages to voltages relative to the standard hydrogen electrode. (The conversion is simple) Finally, report the literature standard reduction potential for each metal. Table II. Experimental half-reaction potentials. Reduction half-reaction E half-rxn (V) (measured in lab) E half-rxn (V) (relative to the SHE) Literature E red (V) 1. Based on your table of measured relative E half-rxn : a) What metal is the strongest reducing agent? b) What ion is the strongest oxidizing agent? 2. Refering to your measured E half-rxn results, if you were to place the black probe of the voltmeter on the zinc electrode and the red probe on the silver electrode: a) What voltage would the voltmeter read? b) Which electrode would act as the anode? c) Write the overall redox reaction that would occur. 3. Why are accurate measurements for Mg and Al difficult to obtain? (Hint: Carefully read the electrochemistry chapter of the text book. The answer is there) Voltaic Cells.doc 5 April 18, 2008

6 Part B: Nonstandard Cell Results Name Copper(II) Complex Ion Formation Experimental Data: Voltage before adding ammonia Voltage after adding ammonia = E cell = E cell Temperature 1. Write the net ionic equation, including phase labels, for the redox reaction that takes place between the half-cells before adding ammonia. (This will define Q) 2. Write the net ionic equation, including phase labels, for the chemical reaction that takes place when aqueous ammonia is added to a solution containing copper(ii) ions. 3. Use the Nernst equation to calculate the concentration of copper(ii) ions in solution after adding the ammonia. 4. After addition of the ammonia, does the cell redox reaction become more or less favorable? Explain why this occurs. Voltaic Cells.doc 6 April 18, 2008

7 Determination of K sp for AgCl Experimental Data: Voltage after adding AgNO 3 (aq) to the KCl(aq) Temperature Name = E cell 1. Write the net-ionic equation, including phase labels, for the chemical reaction that takes place when AgNO 3 (aq) is added to a KCl(aq) solution. 2. Use the Nernst equation to calculate the concentration of Ag + ions in solution based on your cell measurements. You will need E cell from the complex ion formation data. Use the same definition of Q that was used for the complex ion formation questions. 3. Using the concentrations of Ag + and Cl ions in solution, calculate K sp for silver chloride. a) What is the literature value of K sp for silver chloride? b) How well does your result agree with the literature value? Voltaic Cells.doc 7 April 18, 2008

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