Voltaic Cells using a Series of Different Metals SECTION #1 10-Oct-2014 The following data was generated from a simple "lemon battery", using the following metals: (1) galvanized finishing nail (Zn-plated iron nail), 2 inch (2) common iron finishing nail, 1 inch (3) Ni - 1964 Canadian Nickel (99.9% Ni) (4) Cu - 1994 Canadian Penny (98% Cu, 0.5% Tin, 1.5% Zn) (5) Ag - Canadian 1967 Dime, 50% Ag, 50% Cu? (or 80% Silver, 20% Cu - both types were minted in 1967) Table A Anode (negative {-} electrode) Voltaic Cell Data for Five Metals [voltage (V) & current (I); T = Theoretical ] Cathode (positive {+} electrode) Silver (Ag) Copper (Cu) Nickel (Ni) Iron (Fe) V I V I V I V I Zn (zinc) 1.045 V (T=1.561) 268 μa 0.973 V (T=1.10V) 223 μa 0.918 V (T=0.526) 235 μa 0.438 V (T=0.322) 59 μa Fe (iron) 0.643 V (T=1.239) 66 μa 0.513 V (T=0.779) 55 μa 0.407 V (T=0.204) 51 μa Ni (nickel) 0.180 V (T=1.035) 7 μa 0.053 V (T=0.103) 8 μa Cu (copper) 0.094 V (T=0.460) 4 μa Ag (silver) Data obtained from 3 different runs on two different days and data was averaged. Anode & Cathode Notes: cannot be the same metal Negative voltage Negative voltage Table B Standard Reduction Potentials (E 0, Volts) Zn (zinc) -0.762 V Electrons flow towards the more positive electrode potential; thus, the higher metal in the table will be the Anode (-) and the lower metal in the table will be the Cathode (+), no matter which pair is chosen. Fe (iron) Ni (nickel) Cu (copper) Ag (silver) -0.440 V -0.236 V 0.339 V 0.799 V
SECTION #1 Diagram A Electrons move towards the more positive electrode potential. ANODE (-) CATHODE (+) -2-1 0 1 2 Zn (-0.76 V) Fe (-0.44 V) Ni (-0.236 V) Cu ( 0.339 V) Ag (0.799 V) NOTES: 1. Voltaic cells: use the acronym VANO, which stands for Voltaic/Anode/Negative/Oxidation; thus, in a voltaic cell, the Anode is Negative and OXIDATION occurs at the Anode. {NOTE: Oxidation ALWAYS occurs at the ANODE, in both Voltaic Cells & Electrolytic Cells} {Electrolytic cells - EAPO: electrolytic cell, the Anode is Positive & OXIDATION occurs at Anode} 2. The metal on the left of a pair of metals will be the ANODE, e.g., - for the Zn/Fe pair, Zn is the Anode (-), thus, Fe will be positive (+), the cathode - for the Fe/Cu pair, Fe is the Anode (-), thus, Cu will be positive (+), the cathode 3. The voltage for a cell can be determined from the following equation: E cell = E cathode - E anode E cell = E (+) - E (-) e.g. for Zn/Fe pair: E cell = E Fe - E Zn = -0.44 - (-0.762) = 0.322 V Analysis of Experimental Data: 1. The voltages were not exactly the same as the theoretical, but at least trends were noticed, which were consistent with the theory. Also note that the theoretical voltages are based on Standard conditions (i.e., 1 Molar concentration & 1 atmosphere) and the lemon battery is NOT at standard conditions. 2. Going from left to right, the voltages decrease, as expected from the theoretical voltages. 3. Going from top to bottom, the voltages decrease, as expected from the theoretical voltages. 4. The current was more difficult to measure than the voltage and additionally, the lower currents were more difficult to measure; although, a general trend was observed i.e., the lower the voltage, the lower the current. 5. The voltages for Silver (Ag) as the cathode are considerably smaller than the theoretical, but it should be noted that I was using a Canadian Dime, of which I do not know the composition of the coin (could be as low as only 50% Ag) and, as well, it does have a significant amount of corrosion.
SECTION #2 10-Oct-2014 Batteries in SERIES: 1. Theoretically, when batteries are connected in series, the voltage will be a multiple of the number of batteries and the current will remain constant. 2. I have connected two "lemon batteries" or cells in series as follows: galvanized nail (-) (+) wire link (-) (+) lemon copper penny, 1994 1.885 V Table C # of cells Expected Experimental % of V for 1 cell % of 1 cell I 1 0.973 V 0.973 V 100% 223 μa 100% 2 1.946 V 1.885 V 193% 143 μa 64% 4 3.892 V 3.71 V 381% 175 μa 78% 3. The experimental voltage across this combination of two cells = 1.885 V. 4. The theoretical value should be 2 x 1.10V = 2.20 V 5. Thus, the experimental voltage of 1.885 V = 86% of the 2.20 V. 6. If one were to consider the experimental voltage obtained for one cell, i.e., 0.973V, the combined cell voltage for two cells should be 2 x 0.973 V = 1.946 V. 7. Thus, the experimental voltage of 1.885 V is 97% of the expected 1.946 V. 8. The experimental voltage for four cells should be 4 x 0.973 V = 3.892 V. 9. Experimentally, I obtained 3.71 V for 4 cells or 95% of the expected voltage. 10. The current for one cell as experimentally obtained is 223 μa, Table A. 11. The current for two cells should be unchanged and thus be 223 μa. 12. Experimentally, I found the current of two cells to be 143 μa. 13. Experimentally, I found the current of four cells to be 175 μa. 14. The average of the currents for the two cell and the four cell batteries is approx 71% (avg of 64 & 78). 15. Thus, the voltages are doubling, but the currents are remaining approx constant at 71% of the expected current (i.e. the currents for the 2 cell & 4 cell are NOT 200% & 400% as would be expected from a doubling). Thus, in line with the theoretical expectations. Note: The "lemon batteries" wear down very quickly (due to the internal resistance of the battery), so it is not surprising to see the variation in the data. Using V=IR, I have calculated an average internal resistance for a one cell lemon battery to be 6300 Ω, based on an average. As internal resistance of the battery increases, the voltage drops slightly, but the current drops off rapidly.
SECTION #3 12-Oct-2014 Batteries in PARALLEL 1. Theoretically, when batteries are connected in parallel, the current will be a multiple of the number of batteries and the voltage will remain constant. Note: that is just the opposite from the series example. 2. I have connected two "lemon batteries", or cells, in parallel as in the diagram using Zn (-) & Cu (+). 3. The following data was obtained: Table D # of cells Expected Experimental % of I for 1 cell % of V for1 cell 1 223 μa 223 μa 100% 0.973 V 100% 2 446 μa 366 μa 164% 0.981 V 101% 4 892μA 645 μa 289% 0.976 V 100 % (-) lemon galvanized nail copper penny, 1994 (+) lemon 0.981 V 366 μa
SECTION #3 12-Oct-2014 Batteries in PARALLEL continued: 4. Note that when the batteries are connected in parallel, the current is approximately the multiple of the number of cells, i.e., - when two batteries are connected in parallel, the current is 164% of the expected current (with the expected being 200%) and - when 4 batteries are connected in parallel, the current is 289% of the expected current (with the expected being 400%). 5. Again, as previously indicated, currents are more difficult to measure than the voltages, so it is not surprising to see the variation. 6. The batteries connected in parallel approximate what one would expect from theory. 7. Batteries hooked up as indicated can generate approximately 1 V (0.976V) and a current of 645 μa.
SECTION #4a 14-Oct-2014 Battery Efficiency: 1. As indicated previously, the "lemon batteries" are not very efficient due to the rapid increase in "internal resistance" of the battery. 2. I have connected a single "lemon battery", as in section #1, with Zn (galvanized nail) as the anode {negative (-) terminal} and Cu (penny) as the cathode {positive (+) terminal} and monitored the current and voltage (separate experiments) under continuous load conditions (thus, a closed circuit with current continuously flowing). 3. The following table gives the results of the experiment: Table E Time, min % of starting V % of starting I 0 0.970 100% 280 μa 100% 0.5 0.950 98% 212 μa 76% 2 0.938 97% 194 μa 69% 5 ----- ----- 192 μa 69% 7 ----- ----- 192 μa 69% 15 0.860 89% 184 μa 66% 20 ----- ----- 187 μa 67% 30 0.860 89% 194 μa 69% 60 0.878 90% 196 μa 70% 4. One can see from the 2 minute time point, the voltage has dropped only 3% (97% of initial conditions); whereas, the current has dropped by approximately 30% (69% of initial conditions). 5. From the 60 minute time point, the voltage has dropped by 10% (90% of initial conditions); whereas, the current has dropped by 30% (70% of initial conditions). 6. Thus, the current of the battery drops off at a faster rate than the voltage produced by the battery. 7. Resistance (Ohms, Ω) for 0 minute time point: V = IR 0.970 V = (0.000280 Amps) x R R = 3464 Ω 8. Resistance for 60 minute time point: V = IR 0.878 V = (0.000916 A) x R R = 4479 Ω 9. Thus, the resistance of the battery is 4478/3464 x 100 = 130% of the starting conditions of the battery; thus, the internal resistance has increased by 30%.
SECTION #4b 15-Oct-2014 Just How Much? 1. In the preceding experiment, #4a (1 cell lemon, Zn/Cu), how much electrical charge (measured in coulombs, Coul), energy (in Joules) or power (measured in Watts) is being generated? is measured in Amps (A). A single cell "lemon battery", with Zn (galvanized nail) as the anode {negative (-) terminal} and Cu (penny) as the cathode {positive (+) terminal}. Time frame = 1 hour. 2. From table E, the one cell "lemon battery" was able to sustain about 191 μa (0.000191 Amp) for the 1 hour (3600 Sec) time frame of the experiment (average current from 2 min to 60 min time points). 3. Coulombs (Coul) is a measure of charge and can be calculated from the following: 1 Amp = 1 Coul / 1 Sec, thus, #Coul = Amps x Sec For our experiment: #Coul = 0.000191 A x 3600 Sec = 0.688 0.69 Coul of charge. - a 1.5 V, AA battery can produce 7200 Coul (based on 2 Amp Hours) - thus, our one cell "lemon battery" during the 1 hour time frame was able to generate about 1/100 of 1% (or 0.01%) of a AA battery! 4. Joules (J) is a measure of energy and can be calculated from: Volt = Joule/Coul For our experiment: #Joules = V x Coul = (0.878 V) x (0.69 Coul) = 0.606 J of energy 5. Watts (W) is a measure of power and can be calculated from: W = V x A = (0.878 V) x (0.000191 A) = 0.000168W = 168μW. Solar cell powered calculators operate on a μw scale; thus, our one cell "lemon battery" can power such a calculator. Faraday Constant 1. The Faraday (F) constant is the number of Coulombs of charge equivalent to 1 mole of electrons. 1F = 96,485 Coul/1 mole electrons thus, 1 mole of electrons = 96,485 Coul of charge 2. From table E, we can see that at 2 minutes, the battery had 0.938V and a current of 194 μa (or 0.000194A); thus, at this point in the experiment, the Coul & # of moles of electrons would be: #Coulombs = (A)(Sec) = (0.000194 A)(120 sec) = 0.0233 Coul of charge # moles electrons = (0.0233 Coul)/(96485 Coul/mole electrons) = 2.4 x 10-7 moles of electrons or, the number can be represented as 0.24 x 10-6 moles of electrons 3. From table E, we can see that at 60 minutes, the battery had 0.878V and a current of 196 μa (or 0.000196A); thus, at this point in the experiment, the Coul & # of moles of electrons would be: #Coulombs = (A)(Sec) = (0.000196 A)(3600 sec) = 0.706 Coul of charge # moles electrons = (0.706 Coul)/(96485 Coul/mole electrons) = 7.3 x 10-6 moles of electrons
SECTION #4b 19-Oct-2014 Faraday Constant continued: 4. The Faraday Constant can also be calculated from the following equation: Faraday Constant = (J)/(V x moles of electrons) We know experimentally that: J = V x Coul = (0.878 V) x (0.69 Coul) = 0.606 J Volts = 0.878 V moles of electrons = 7.3 x 10-6 moles of electrons {which in our case can only be determined from knowing the Faraday constant as in #3} Thus, from the current, Volts and time of the experiment, the Faraday Constant would be: F = (0.606 J)/(0.878 V x 7.3 x 10-6 ) = 94,549 Coul/mole of electrons This value is 94549/96485 x 100 = 98% of the true Faraday Constant! SECTION #5a 16-Oct-2014 LIME versus LEMON: 1. The acidity, (thus ph) of solutions or fruits can vary and the Handbook of Chemistry & Physics provides the following data: Table F Fruit / Solution ph Vinegar 2.4-3.4 Lemon 2.2-2.4 Lime 1.8-2.0 2. Thus, a one cell "lime battery" was set up using the Canadian 1967 Dime as the cathode (+) and the indicated metals as the Anode (-). Table G Anode (-) Silver, Cathode (+) Zn 0.950 V 184 μa Fe 0.494 V 44 μa Ni 0.065 V 9 μa 3. The voltage and current characteristics for the lime battery have very similar trends to the Lemon battery. Thus, there is no advantage to using the "Lime" over the "Lemon" as the acid source. 4. The one cell "lime battery" was shown to have similar internal resistance characteristics as seen for the lemon battery: - Time zero = 0.950 V & 0.189 μa - Time 30 min = 0.938 V (99%) & 0.084 μa (44%) i.e., the current decreases more rapidly than the voltage.
SECTION #6a 16-Oct-2014 Cathode with more Surface area: 1. In the experiments with the lemon, a Canadian Dime was used, thus, for the following experiments, a Canadian 1961 Dollar was used (80% Ag & 20% Cu), to see if it would change the characteristics of the battery. The $1 Canadian is the Cathode (positive {+} electrode). A lime was used for convenience. It should be noted that I do not know the Ag content of the Canadain dime used in the original experiments, because in 1967, both types of coins were minted (50% Ag or 80% Ag). 2. As well, the Anode (-) was set up as 3 galvanized nails (Zn), as opposed to the single Zn plated nail. Table H Time (min) Silver, Cathode (+), $1, placed about 1/2 of the way into the lime 0 not measured 538 μa 1 not measured 265 μa 5 not measured 170 μa 10 0.681 V 145 μa 3. The initial current is significantly higher with the $1 coin (538 μa) than with the Dime (268 μa). Note: the area of the dime is π r 2 = π (18mm) 2 = 1021 mm 2, versus π (36mm) 2 = 4071 for the $1 coin. Also note that with the dime, I used approx 90% of the area (920mm 2 ) versus 50% for the $1, (2036mm 2 ), which about a factor of 2X, which is similar to the initial current difference, 538 vs 268 μa. 4. #Coul = Amps x Sec = 0.000145 A x 600 sec = 0.087 C, for the 10 minute time period. 5. For an unknown reason, the voltage was lower with the $1 coin (0.681V), versus the Dime (approx 1V). 5. W = V x A = (0.671 V) x (0.000145 A) = 0.000097 W = 97 μw, for the 10 minute time period. SECTION #6b 16-Oct-2014 1. For the following experiment, the Canadian 1961 Dollar was used (80% Ag & 20% Cu) and inserted most of the way into the Lime. The $1 Canadian is the Cathode (positive {+} electrode). A lime used. 2. As well, the Anode (-) was set up as 5 galvanized nails (Zn). I used approx 90% of the $1 coin. 3. 90% of the $1 coin = 3664 mm 2, which is a factor of 4X the dime, consistent with a 4 fold current difference, i.e., 268 μa (Dime) and 838 μa ($1), i.e., 268 x 4 =1072, thus, 78% of expected. Table I Time (min) Silver, Cathode (+), most of the way into the lime; 5 galvanized nails as the Anode (-) 0 0.695 V 838 μa 1 not measured 421 μa 5 not measured 273 μa 10 not measured 216 μa 30 not measured 164 μa 60 not measured 158 μa 1. The larger anode and/or cathode produced significantly larger initial currents than previously obtained. 2. The resistance at 60 minutes was measured directly and found to be 4000 Ω (4K Ω).