Galvanic Cells

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JHU INTRO. CHEM. LAB SUMMER 2016 Electrochemistry is considered to be a difficult  Practice problems that will help you with the post- topic in introductory chemistry. Even students who lab assignment. are able to successfully do quantitative problems on Similar information is available for all experiments. this topic often show misconceptions when asked to In addition, there is information on Blackboard that qualitatively describe the processes that occur in a will be useful for all experiments. This includes galvanic or electrolytic cell. Yet it is important. Development of electrochemical technology for  A schedule of all experiments and tests rechargeable batteries has led to the success of  Laboratory notebook instructions wireless technology. Electrochemistry will be  Information on how to use Chem21labs instrumental in reducing our dependence on limited  Information on spreadsheets and graphing oil supplies through the development of fuel cells for  Information on accuracy of equipment and cars and storage cells for solar power. Understanding significant figures. electrochemistry also enables the prediction of the spontaneous direction of oxidation-reduction (redox) Objectives reactions. These reactions are among the most  Use a volumetric flask to make solutions with important in both chemistry and biochemistry. accurate concentrations from a solid. (You will A brief review of this topic is presented here, but need to do this on the lab skills exercise later this you will also need to refer to your textbook (Atkins, semester.) Jones, and Laverman, Chemical Principles, 6th ed.,  Construct a galvanic cell and identify its parts. Ch. 14) or any other freshman chemistry textbook. In This includes recognizing addition, there are good animations of galvanic cells o direction of electron flow on the internet. o chemical half-reactions at the electrodes the identity of the anode and cathode In the prelab homework for this experiment, you o the contents of the beakers will calculate the amounts of the salts you will need o to make the solutions for the two half-cells in Part A.  Use the specialized notation for describing Record this calculation in your notebook to make it galvanic cells. quicker for you to get started in the lab. Check the  Predict the voltage for galvanic cells using results with your lab partner and with your TA before standard reduction potentials. actually making the solutions. You will also need to  Use the Nernst equation to calculate the calculate theoretical voltages for cells composed of concentration dependence of the potential. one half-cell containing copper sulfate and a copper  Construct a galvanic cell that produces a potential electrode and the other half-cell composed of due to concentration differences in the half-cells unknowns chosen from 5 possibilities. You should and use this to review solubility product concepts. also do these calculations before you come to lab and record them in your notebook. GALVANIC CELLS In a galvanic cell, electrical potential is Blackboard resources generated by the difference in electrical reduction  Experiment → Exp 1 potential of two half-cells so that electrons flow through the external wire from the anode where they o Laboratory instructions. You are required to print these and bring them with you to lab. are produced by the oxidation to the cathode, where they are used in the reduction. Batteries are galvanic o Background information with link to an animation cells. Galvanic and electrolytic cells have the same fundamental definition of anode and cathode and the o Online video on concepts and techniques same processes occur at the electrodes. In galvanic o Answers to the inlab worksheet are posted after lab cells the reactions are spontaneous, caused by the different reduction potentials of the ions in the o Link to a techniques video on using a volumetric flask to make solution with different half-cells. In electrolytic cells, as you will accurate concentrations from a solid. (You see in the next experiment, the reactions require the will need to do this on the lab skills exercise work supplied by an external voltage. later this semester.) In this experiment, you will make several galvanic cells and measure the voltage produced. In

Part A, you will make one galvanic cell from copper different solution containing Nn+ cations and Y- and lead. In Part B, you will be given several anions. The lower case n and m denote the oxidation electrodes and unknown salt solutions and you will be number of the metal. A salt bridge allows ions to pass asked to identify the unknowns from the voltage into the cells to maintain electrical neutrality but produced by the half-cells when combined with the prevents gross mixing of the two solutions. A copper half-cell from Part A. In Part C you will make voltmeter connected between the electrodes will show a concentration cell and use the voltage produced to that a voltage difference exists between the two sides. determine the concentration of copper ions in a The voltmeter reading will be positive when the saturated solution of copper oxalate. This will be used positive terminal is attached to the electrode with the to determine the Ksp of copper oxalate. higher reduction potential, labeled M in this diagram. All oxidation-reduction reactions can be written The other electrode is attached to the COM terminal as the sum of two half-reactions. One half-reaction on the voltmeter. Electrons flow through the involves oxidation, the loss of electrons. The other voltmeter from the N electrode to the M electrode. half-reaction involves reduction, the gain of electrons. The reaction at the N electrode is N(s)  Nn+(aq) + In order to compare half-reactions, it is conventional ne-. The N electrode is the anode, because oxidation to write them all as reduction reactions. They are then occurs at this site (the N loses electrons, becoming tabulated in lists that indicate the relative tendency for Nn+). The reaction at the M electrode is Mm+(aq) + each half-reaction to occur. The higher the reduction me-M(s). The M electrode is the cathode since the potential, the more likely the reduction. The values Mm+ ions are reduced at this site and gain electrons. that are assigned are only relative values since all that In your experiment, you will be able to observe which can be observed is the voltage of a complete reaction. electrode is the anode and which is the cathode by The values are all referenced to the reduction of H+ observing the connections to the terminals on the ions, which is arbitrarily given the potential of zero. voltmeter that produce a positive reading. The following table of standard reductions Reactions at the anode and cathode take place potentials used in this experiment is taken from the simultaneously in the apparatus. At the same time, Chem21labs reference pages. ions from the salt bridge travel to the solutions, keeping the solutions neutral. The electrons produced o Standard Reduction Potentials at 25 C by the oxidation of N generate a current in the wire Half-Reaction Eo(volts) that travels through the voltmeter to the M electrode, where electrons are used in the reduction of the Mm+ Ag+ + e- Ag(s) 0.799 in the solution at the cathode. The voltmeter measures Cu2+ + 2e- Cu(s) 0.337 the difference in the potentials between the two half- Pb2+ + 2e- Pb(s) -0.126 reactions. A small lamp could be substituted for the voltmeter and the movement of electrons would cause Sn2+ + 2e- Sn(s) -0.136 the lamp to glow. Zn2+ + 2e- Zn(s) -0.763 A major source of confusion is about what Al3+ + 3e- Al(s) -1.660 processes occur in galvanic (and electrolytic) cells. Mg2+ + 2e- Mg(s) -2.370 Here it helps to look at a picture and you should look at the pictures in sections 14.4 and 14.5 in your Consider the apparatus shown here: textbook or at the animation on the internet referenced on Blackboard. The important things to observe are com + voltmeter that electrons flow in the external wire while ions, not electrons, travel in the solutions and in the salt bridge. M N The salt bridge is used to complete the circuit and electrode electrode maintain electrical neutrality in the solutions. When positive ions are created at the anode, negative ions will travel from the salt bridge into the solution on the m+ n+ M N anode side so that the solution remains electrically X- Y- neutral overall. At the same time, the positively charged ions near the cathode are being reduced so positive ions from the salt bridge must travel into that salt bridge solution so that it remains electrically neutral and the Here M is the metal electrode in contact with a salt bridge remains electrically neutral as well. solution containing Mm+ cations and X- anions and N The voltmeter cannot be used to measure the is a different metal electrode in contact with a potential of a half-reaction by itself; it measures the

difference in potentials between two half-reactions. where E is the voltage of the cell, Eo is the voltage Consequently, you can only determine relative values of the cell under standard conditions, n is the number for the half-reaction potentials. They are given of electrons transferred in the reaction, R is the absolute numerical values by reference to the universal gas constant 8.314472 J mol-1 m-1, T is the hydrogen half-reaction: temperature in Kelvin, F is the Faraday constant + - 96,485.3399 and Q is the reaction quotient for the 2H (aq) + 2e  H2(g) reaction. If you substitute these values including the for which the half-cell potential is arbitrarily set at conversion factor for natural log to base ten log and zero. The standard cell in electrochemistry is one in determine ∆E at 25oC, you get the more familiar form which the half-cell is combined with a hydrogen of this equation electrode under standard conditions (concentrations = . 1M). For example, if a standard copper half-cell is ∆ ∆ . (4) connected to a hydrogen half-cell, a potential However, you should use the more general form difference of 0.337V is observed at 25oC. If a standard shown in equation 3 since you will be able to use an lead cell is connected to a hydrogen half-cell, a experimental value for the temperature. potential difference of -0.126 V is observed at 25oC. For half-reactions, the Nernst equation can be The standard voltage of a galvanic cell written as composed of two different half-cells is calculated from the difference between the standard potentials of (5) the two half cells. For example, for the cell shown in where Q is the reaction quotient for the half-reaction. the diagram on the previous page: h m+ n+ For example, for the lead-copper cell, the E = Ecathode - Eanode = E(M M) - E(N N) overall net ionic reaction is (1) Cu2+(aq) + Pb(s)  Cu(s) + Pb2+(aq) (6) Electrochemistry uses specialized notation. Half- For this reaction, Q = [Pb2+]/[Cu2+] and n=2, so the cells are denoted by a single vertical line denoting a Nernst equation can be written as metal-solution interface. A galvanic cell (comprised of two half-cells) is written as: ∆ ∆ (7) NNn+(conc m+(conc)M (2) )‖M Eo for the lead-copper cell is +0.463 V from the The double vertical lines between the half-cells standard reduction potentials of copper and lead. As denote the salt bridge or ion-permeable membrane. long as the concentration of Cu2+ and Pb2+ are equal, The anode is on the left and the cathode is on the right. the log term is zero. However, if the concentrations Because the reduction happens at the cathode, in are not equal, the concentrations must be included to galvanic cells it is always the half-cell with the determine the theoretical voltage. The concentration highest reduction potential. For an actual cell, the ratio enters in the argument of the log term. This concentrations of the solutions are included since that means that a large concentration differential produces affects the voltage, as will be discussed later. only a small voltage change. For example, if you were The voltage of a battery is determined by the to measure the voltage of the cell with [Pb2+] = 2.00 difference in the half-cell potentials. The voltage lasts M and [Cu2+] = 0.200 M at 25oC, the voltage would until one of the reactions producing the difference in be 0.463 – (0.05917/2)log(10) = 0.433 V, only a 6.5% half-cell potentials has exhausted the reactants, at difference in the voltage for a factor of 10 difference which point we say the battery is "dead". in the concentrations in the two cells. The situation with respect to the concentrations CONCENTRATION DEPENDENCE is more complicated when the cations have different In this experiment, the concentrations of the charges. For example, if the cation being oxidized has solutions are not 1M, which makes the experiment an oxidation number of +2, the half reaction is performed under nonstandard conditions. The N(s) N2+(aq) + 2e- (8) dependence of the voltage on concentration is given If the anion has an oxidation number of +1, then the by the Nernst equation. A good discussion of half reaction for the cation being reduced is concentration effects is given in section 14.9 of your + - text. This equation is based on thermodynamic M (aq) + e M(s) (9) considerations. When these are combined into a full cell, the reaction The Nernst equation for a full galvanic cell is equation is written as N(s) + 2M+(aq) N2+(aq) + 2M(s) (10) ∆ ∆ (3) 20

The reaction quotient Q = [N2+]/[M+]2 (11) salt bridges in Parts A and B and a strip of filter paper with the denominator term squared due to the to use as the salt bridge in Part C. stoichiometry. The Nernst equation is written as Part A: Construct a lead-copper galvanic cell. ∆ ∆ (12) You will need to prepare two solutions: CuSO4 ~0.05M 100 ml quantity again with the denominator in the log term squared. If Pb(NO ) ~0.05M 20 ml quantity non-standard concentrations are used in either of the 3 2 half-cells, the log term must be considered in the Determine the amounts of chemical you will need to calculation for the theoretical value of the voltage. If make these solutions. Your final concentration should the half-cell chemical equations involve different be within 10% of these concentrations. The copper numbers of electrons, then you must pay attention to (II) sulfate salt is hydrated, with the molecular the stoichiometry when you write the Nernst equation formula CuSO4∙5H2O. The lead nitrate salt is not since it affects the powers in the reaction quotient. hydrated. Check your calculations with your lab Voltages also develop between two half-cells partner and answer the first question on the in-lab having the same chemical species but different worksheet before making the solutions. concentrations. This will be the case in Part C of this Note: To avoid long lines at the balances, you should experiment where you will make a galvanic cell only get one salt at a time. Make up the first solution consisting of one half-cell with the concentration of before getting the second salt. By then, the lines will cupric ion of 0.05M and one half-cell with the be gone. Students who monopolize the balances concentration of cupric ion that is several orders of before everyone has had a chance to get at least one magnitude lower. In the lower concentration half-cell, of the compounds will lose points. the concentration will be limited by the solubility Use the following procedure to fill volumetric product of an insoluble copper salt. The half-cell flasks when you need to dissolve a compound and potential for each half-cell will be accurately know the concentration. A video showing 0.337 (13) this technique is on Blackboard. You will be tested individually on making solutions as part of the lab The half-cell with the higher concentration of Cu2+ skills exercise later this semester. will have the larger reduction potential, so it will be 1) Dissolve the measured amount of compound in a the cathode. As a result, for this special case, beaker, using about 50% of the volume of the volumetric flask. So, for example, you should ∆ (14) dissolve the copper(II) sulfate in about 50 ml of since ∆Eo = 0 because the standard reduction water in a beaker. Also rinse any solid remaining potentials cancel. in the weighing pan into the beaker using your wash bottle. Then pour that into the volumetric SAFETY flask. Using safe laboratory practices is important when 2) Rinse any remains in the beaker into the using unknowns. Some of the metal salts are toxic. volumetric flask. Use about half the remaining Lead nitrate, for example, is a poison. Others, such as volume. So, for example, you should rinse the aluminum chloride and silver nitrate, are corrosive. copper sulfate beaker with about 20-25 ml of Silver nitrate will also turn your skin black after it is water and pour it into the volumetric flask. exposed to light. Always wear gloves and goggles. 3) Repeat step 2, using less water. Keep the chemicals under the hood and work with 4) Use your wash bottle to carefully fill the your face as far from the chemicals as possible. volumetric flask to a few mm below the line. Be careful – it fills very quickly once you reach the EXPERIMENTAL PROCEDURE narrow neck of the flask. You will have one lab period to complete all three 5) Use your eyedropper to fill the flask so that the parts of this experiment. Obtain a reaction plate with bottom of the meniscus is on the line. 6 wells, a multimeter, 2 alligator clips, and a glass 6) Then, cover the flask with parafilm®. Hold your thermometer from the stockroom along with two finger over the parafilm® and invert the flask copper metal strips and one lead metal strip. You will several times to fully mix the solution. also need a 20-ml volumetric flask, a 100-ml volumetric flask, and a 1-ml pipette and pipettor. Make the lead(II) nitrate solution in a similar Your TA will give you short pieces of string to use as manner, except you will only be making 20 ml using the 20 ml volumetric flask. 21

Make the lead-copper cell by filling one of the Turn on the multimeter to the voltage position, two center wells with copper(II) sulfate solution and being sure that your multimeter is set to measure the other center cell with lead nitrate solution. direct current. You will be able to record your voltage Polish the copper and lead electrode strips with to the most number of significant digits if you set the emery paper until they are shiny. Rinse with water. multimeter dial to the lowest voltage range at which Bend them so that they will be able to stay in the the voltage is on scale. If your multimeter reading is appropriate well in the reaction plate. Then connect negative, switch the leads on the electrodes. The the red and black multimeter leads to the copper and voltage should settle quickly if your connections are lead strips using the alligator clips. (Depending on clean and tight. If not, remove the electrodes, clean, your voltmeter, either turn the dial to V==, or make and repeat. Record the voltage and record the sure the button is in the “out” position. You will be temperature in one of the wells. Make a sketch of the measuring a direct current, DC, voltage, not an apparatus that produces a positive voltage in your alternating current, AC, voltage.) It is conventional, notebook showing the voltage and which connection but not essential, for the red cable to be connected to on the voltmeter is attached to which electrode. This the + connection on the multimeter and the black will help you identify which electrode is the anode cable to the COM (or negative) connection. The and which is the cathode. In addition, you should electrons that will be produced at the anode will flow complete the sketch for part A on your in-lab through the cable to the multimeter, through the other worksheet. You also need a sketch in your notebook, cable and then to the cathode, where they will be used but the one in your notebook only needs labels for the to reduce the metal ions to solid metal. Make sure all + and com connections from the voltmeter, the your connections are tight. Do not just insert the contents of the half-cells, the metals in the electrodes, voltmeter leads into the end of the alligator clip, you the temperature, and the voltage. need to clip the lead directly to the electrode to obtain After you have recorded the temperature and a tight connection, as shown below. voltage, pour the solutions from the wells into a beaker for disposal in the large drum and rinse the

wells with tap water. You can also dispose of your remaining lead nitrate solution in the drum. You will need your remaining copper sulfate solution for parts

NOT B and C.

Part B: Unknowns You will be given a sheet with your unknown Your TA will have a beaker containing ~0.1M assignments at the start of the lab. Bring this sheet KNO3 and will have presoaked some small lengths of with you to obtain your assigned 2 unknown string for use as salt bridges. Use your forceps to pick electrodes and 2 unknown salt solutions from your one and carry it over a beaker or watch glass. Then, TA. Label the beakers or flasks and also the wells you put one end in the Cu2+ solution and the other end in use for the unknown solutions since all are clear and the Pb2+ solution. colorless and the experiment will not be successful if Place the lead electrode in the lead solution and you mix them up. Fill one of the center wells with the copper electrode in the copper solution. The copper (II) sulfate solution. Construct galvanic cells electrodes should not be touching the salt bridge. The using each of your unknowns in wells adjacent to the apparatus should look like this: copper cell. Use new salt bridges for each full galvanic cell to avoid contaminating the solutions in salt bridge the wells. Measure the temperature in one of the cells. Connect the voltmeter so that you obtain a positive voltage reading for each unknown when combined with your copper half-cell. After you have recorded the temperature and voltage, make a drawing in your notebook showing the connections and the voltage for each of the cells. Include the unknown number with

each of your sketches in your notebook since you will voltmeter need that for your post-lab assignment. In addition to the sketches in your notebook, you also need to make and label the sketches on your in- O O COM V lab worksheet.

The unknowns will be taken from the following one of the beakers. Just as in part A, you should list: silver electrode with AgNO3 solution, aluminum connect the leads to the electrodes so as to obtain a electrode with AlCl3 solution, magnesium electrode positive reading on the multimeter. This will help you with MgSO4 solution, tin electrode with SnCl2 determine which half-cell contains the anode and solution, or zinc electrode with ZnSO4 solution. The which has the cathode. Make a sketch in your concentrations of the solutions are such that the log notebook. Answer question 3 on the in-lab term in the Nernst equation can be ignored when worksheet and then complete the sketch on the identifying the unknowns based on calculating their worksheet. theoretical voltages when combined with your copper half-cell. DATA ANALYSIS FOR KSP CALCULATION Before you discard the solutions in the wells, The solubility product of copper oxalate is 2+ 2- identify the 2 unknowns you were assigned and turn Ksp = [Cu ] [C2O4 ] (15) the completed form in to your TA. You and your lab As part of the Chem21labs assignment, you will partner can figure this out working together, or each determine the solubility product by determining the 2+ 2- may do it separately. It’s up to you. Do not discard concentrations of the Cu and C2O4 ions. your solutions until you both have made the To determine [Cu2+] in the copper oxalate cell, identifications. You will lose points for every you will use the measured voltage of the cell and the unknown for which you need to get a repeat solution. known concentration of the Cu2+from the solution you Return the electrodes to your TA when you are made in Part A in the Nernst equation as shown in finished. The solutions of the unknowns should all go equation 13. in the large drums for disposal. Save your remaining You determine [C O 2-] from the ionization copper sulfate solution for part C. Answer question 2 4 constants of oxalic acid. Oxalic acid (H2C2O4) is a 2 on your in-lab worksheet before continuing. weak diprotic acid whose first ionization constant at 25oC is 5.9 x 10-2 and whose second ionization Part C: Concentration cell o -5 2+ constant at 25 C is 6.4 x 10 . To calculate the Construct a Cu concentration cell with a large 2- concentration of C2O4 in the solution, you make use concentration ratio. You will do this in two small of the fact that the second ionization constant is much beakers, rather than in the reaction well plate. smaller than its first. The dissociation of oxalic acid In order to get a very small cupric ion can be written as follows: concentration on one side of this cell so that a + - -2 H2C2O4 = H + HC2O4 Ka1= 5.9 x 10 (16) measurable voltage is developed, you will use a - + 2- -5 saturated solution of solid copper oxalate in water. HC2O4 = H + C2O4 Ka2= 6.4 x 10 (17) + - Copper oxalate is only slightly soluble; the solubility To a good approximation, all the H and HC2O4 product determines the concentration of the copper in come from the first step and therefore have the same the solution. The voltage measurement will allow the concentration. Since + 2- - solubility product of the salt to be calculated. Ka2 = [H ][C2O4 ]/[HC2O4 ] (18) + - Get 50 ml of the 0.1M solution of oxalic acid and [H ] = [HC2O4 ], (19) from a carboy. Prepare the copper oxalate solution by it follows that adding 2 ml of your ~0.05M CuSO solution to the 50 4 2- ml of 0.1M oxalic acid solution. This solution has a Ka2 = [C2O4 ] (20) large excess of oxalate ion compared to cupric ion. This is a surprisingly simple relationship that sets Stir the solution gently for at least 5 minutes to allow the concentration of the fully dissociated conjugate the reaction to reach equilibrium. The copper oxalate base equal to the second ionization constant. It is true precipitate (CuC2O4) will probably appear only as a for all diprotic acids as long as the second ionization slight cloudiness in the solution or may not be visible constant is much smaller than the first ionization at all. Even if you can’t see the precipitate, this constant. solution now contains Cu2+ ions at a concentration determined by the solubility product of copper EXCESS CHEMICAL DISPOSAL oxalate. This will be the solution used in one half-cell Heavy metal ions can cause environmental and the ~0.05M CuSO4 solution you made in part A damage. They are toxic to many forms of marine and will be the solution for the other half-cell. Prepare a plant life. All the metal ion solutions from Parts A salt bridge using a piece of filter paper dipped in and B should be neutral so you do not need to add KNO3 to connect the two half-cells. Polish two copper baking soda before disposing of them in the excess electrodes. Put one in each beaker and measure the chemical drums. Your solution containing copper voltage developed by the cell and the temperature in oxalate must be neutralized since the oxalate ion 23

hydrolyzes to oxalic acid. Dispose of all these UNKNOWN IDENTIFICATION solutions in the excess chemical drums. Any leftover Before you leave the lab, turn in your completed oxalic acid (not containing metal ions) should also be unknown assignments to your TA along with the neutralized and poured into the drums. Rinse your electrodes. beakers and volumetric flasks out with a large amount of tap water. Rinse your electrodes before returning ASSIGNMENT them. Give the unknown electrodes back to your TA Complete the Chem21labs assignment. Doing the and the copper and lead electrodes back to the practice problems on Blackboard will help with some stockroom with the rest of the equipment. Your used of the calculations. salt bridges can be thrown in the trash.