Chemistry and Development of a Battery

Battery Chemistry

Table of Contents

1. Experimental

1.1 Materials and Apparatus

1.2 Schematic

1.3 Experimental Procedure

Part A: Discharging a 12V lead-acid battery:

Part B: Making an alkaline battery:

2. Results and Discussion

2.1 Part A: Discharge of a Lead-acid battery

2.11 Relationship between the overall potential and the standard  potential of a cell

2.12 Using the Nernst Equation to predict a loss in measured potential

2.13 Plots of battery voltage and power over time for lead-acid battery

2.14 Energy expended, Specific Energy Density, and Power Density of the lead-acid battery during discharge processes

2.2 Part B: Making a battery

2.21 Labelled sketch of the Zinc-MnO2 battery cell

2.22 Comparison of Standard Potential and relation to the Nernst equation

2.23 Energy density and capacity of the cell

2.24 Factors that limit cell performance of constructed Zn-MnO2 batteries

2.25 Combination of Anode/Cathode material that will give the maximum theoretical specific energy density

3. Conclusions and Recommendations

4. Nomenclature

5. References

6. Appendices

6.1 Appendix A [Sample Calculations]

Part A Sample Calculations:

Part B Sample Calculations:

6.2 Appendix B [Report Questions]

6.3       Appendix C [Raw Data]

1.    Experimental

1.1 Materials and Apparatus

• Galvanized wire
• Zinc and Aluminum foil
• Carbon black
• MnO₂ powder
• KOH and NaCl pellets
• SoilMoist® granules
• Deionized water
• 2-mm Pencil lead
• Filter paper
• 5-mL Plastic syringe
• 400-mL plastic container/beaker
• 20-mL glass vial with cap
• Pasteur pipette and bulb
• DC Voltmeter
• LED test lights
• Wires and clips
• Leas acid battery
• Potentiostat
• 30K booster

1.2 Schematic

1.3 Experimental Procedure

Part A: Discharging a 12V lead-acid battery:

First, the Ref3000 Potentiostat and the 30K Booster is turned on and the Gamry Framework software was initiated. Then, the blue plug from the Ref3000 Potentiostat and the green lead from the 30K Booster was attached to the negative terminal. Similarly, the orange plug from the Ref3000 Potentiostat and the red lead from the 30K Booster was attached to the positive terminal. After connecting everything that is required, a new folder called “Group P” and this file will be the file destination for the data collected during the lab. Subsequently, test #5 Discharge was selected, and the parameters were updated according to the lab manual. After the set-up is done, the test was started. After completing the test, the data file was imported into Excel with the instructions stated in the lab manual. At the end, the battery was charged fully, and the mass of the battery and the discharge capacity was recorded.

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Part B: Making an alkaline battery:

First, the plunger from a 5 mL plastic syringe was removed and the barrel was cut around the 5.5 mL mark. Then, for the anode, a piece of 4 cm x 5 cm zinc foil was cut and then weighed. Similarly, for the cathode, 25 wt% activated carbon and 75 wt% MnO₂ was weighted and mixed in a capped vial by shaking vigorously to mix thoroughly. Wet SoilMoist gels were added to the syringe barrel to seal the tip. Subsequently, a piece of 5 cm x 10 cm of filter paper was cut, and the cathode mixture and a 6 cm long piece of pencil lead were placed on the filter paper and rolled up with the bottom edge being folded. The filter paper is wrapped around the anode and then inserted into the trimmed syringe body such that the pencil lead protrudes from the syringe tip and the anode protrudes from the top. The entire assembly is then submerged into a cooled electrolyte solution of KOH for a few minutes. Finally, the syringe was taken out of the electrolyte solution using tweezers and weighted using a tared blue weighing tray on the microbalance. The cell potential was measured using a DC voltmeter and tested on LED lights. This process is repeated 3 times, with varying parameters for each battery made, such as changing the anode, cathode, or the electrolyte solution. 

For a more detailed procedure, please refer to Lab Manual ^[1]

2.    Results and Discussion

2.1  Part A: Discharge of a Lead-acid battery

2.11 Relationship between the overall potential and the standard  potential of a cell

From the pre-lab calculations, it is known that the standard cell potential of each lead-acid battery cell is +2.04 V. To produce a 12 V lead-acid battery that was used for part (a) of the lab, multiple cells can be aligned in series to produce a potential of 12 V. The individual voltages of each cell are summed up to create a higher total voltage. From a simple calculation, it is known that 6 individual lead-acid cells will produce a voltage of 12 V. This theory is valid because the cells are placed in series, and the total voltage for a series connection is simply the addition of all the components’ voltages. 

2.12 Using the Nernst Equation to predict a loss in measured potential

Assuming that the electrolyte inside the battery is approximately 1.0M H₂SO₄, the reaction that occurs will be as follows:

$\mathrm{Overall\; Reaction}:\mathrm{ Pb }\left(\mathrm{s}\right)+\mathrm{Pb}{\mathrm{O}}_2\mathrm{ }\left(\mathrm{s}\right)+4\mathrm{ }{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+\mathrm{ }2\mathrm{ S}{\mathrm{O}}_4^{2–}\mathrm{ }\left(\mathrm{aq}\right)\rightleftharpoons 2\mathrm{ PbS}{\mathrm{O}}_4\mathrm{ }\left(\mathrm{s}\right)+4\mathrm{ }{\mathrm{H}}_2\mathrm{O }\left(\mathrm{l}\right)$

$\mathrm{Cathode }\left(\mathrm{Reduction}\right):{\mathrm{PbO}}_2\mathrm{ }\left(\mathrm{s}\right)+\mathrm{S}{\mathrm{O}}_4^{2–}\mathrm{ }\left(\mathrm{aq}\right)+4\mathrm{ }{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+2\mathrm{ }{\mathrm{e}}^{–}\rightleftharpoons 2\mathrm{ PbS}{\mathrm{O}}_4\mathrm{ }\left(\mathrm{s}\right)+2\mathrm{ }{\mathrm{H}}_2\mathrm{O }\left(\mathrm{l}\right)$

$\mathrm{Anode }\left(\mathrm{Oxidation}\right):\mathrm{Pb }\left(\mathrm{s}\right)+\mathrm{S}{\mathrm{O}}_4^{2–}\mathrm{ }\left(\mathrm{aq}\right)\rightleftharpoons \mathrm{PbS}{\mathrm{O}}_4\mathrm{ }\left(\mathrm{s}\right)+2\mathrm{ }{\mathrm{e}}^{–}$

For the reaction above, the standard cell potential is +2.04 V

The reaction quotient. Q_C, for the reaction shown above is $\frac{1}{\left[H_{2}S{O}_4\right]}$

. However, since we do not have the concentrations for both H⁺ and SO₄^2-, a value for ln (Q_C) in the Nernst equation cannot be computed. Hence, it is impossible to predict whether there will or will not be a loss in the measured potential based only on the experimental data collected. However, if the value for ln (Q_C) > 1 with the restriction ln (Q_C) ≠ 0, then an increase in the measured potential will be observed. Likewise, if the value for ln (Q_C) < 1 with the restriction ln (Q_C) ≠ 0, then an increase in the measured potential will be observed.

 2.13 Plots of battery voltage and power over time for lead-acid battery

From Figures 2.13a and 2.13b, we observe that both instances experience the same trend, which is a decrease in voltage as time progresses. However, there is a noticeable difference in the rate of discharge, where the voltage drops with respect to time. In Figure 2.13a, it is observed that it takes approximately 719s for the potential to drop from 12 V to 10 V, whereas in Figure 2.13b, it is observed that it takes approximately 87.5s for the potential to drop from 12 V to 10 V.

Similarly, Figures 2.13c and 2.13d demonstrate similar trends with a different rate of power consumption. Although the start and end point of the slow discharge and fast discharge are at different values, the general trend that occurs for both cases is the same. It is also observed that the fast discharge has a steeper slope as compared to the slow discharge, which is evident from the change in power in a considerably short amount of time in Figure 2.13d. From both Figures 2.13c and 2.13d, it is also seen that the amount of power drawn from the battery as time progresses increases, which corresponds to the large voltage drops seen in Figure 2.13a and 2.13b above.

With reference to Peukert’s Law, the Peukert coefficient can be determined for both the slow and fast discharge of the lead-acid battery. Peukert’s law defines the capacity of a lead-acid battery and how the capacity changes during the slow and fast discharge process. ^[2] Peukert’s law allows us to understand the longevity of a battery under a given load. Generally, the greater the discharge rate of the battery, the lower the storage capacity will be.

$\mathrm{Peuker}{\mathrm{t}}^{‘}\mathrm{s\; Law\; is\; given\; by}:\mathrm{ }\mathit{ t}=H⸱{\left[\frac{C}{\mathit{IH}}\right]}^k\mathrm{ }$

Where t is the time taken to discharge in hours, H is the rated discharge time in hours, C is the capacity rating of the battery in amp-hours, and k is the Peukert’s coefficient.

For the slow discharge, the Peukert’s coefficient is calculated to be k = 1.58

For the fast discharge, the Peukert’s coefficient is calculated to be k = 1.68

Since the value of the Peukert’s coefficient for the slow discharge is smaller than the value of the Peukert’s coefficient for the fast discharge, our calculations are correct as the values are coherent with existing theories. The sample calculations for the Peukert’s coefficient will be shown in Appendix A.

2.14 Energy expended, Specific Energy Density, and Power Density of the lead-acid battery during discharge processes

To determine the energy expended by the battery during discharge, the area under the power curve for the slow and fast discharge curves were found using MATLAB. Using the trapezoidal rule in MATLAB, the energy expended during the slow discharge is determined to be 11.3194 W⸱h, while the energy expended during the fast discharge is determined to be 4.5981 W⸱h. Using this, the specific energy density and the average power density can also be subsequently determined since the weight of the lead-acid battery is known to be 2080.2g. The average power was determined using Excel.

The energy expended, the specific energy density and the power density for both discharge processes are as follows:

From Table 2.14, several observations can be made. It is observed that the energy expended by the slow discharge is higher than of the fast discharge. This is because the slower discharge takes a longer time, hence more energy is being used for the slower discharge. Energy density is the capacity of the battery to store energy, and it is observed the specific energy density of the slow discharge is higher than of the fast discharge. This agrees with existing theories, as a slower discharge can store energy longer than a fast discharge, which is why a slower discharge takes a longer time. Power density is the ability of the battery to take on and deliver power, and it is observed that the power density of the fast discharge is higher than of the slow discharge. This makes sense since the fast discharge is capable of taking and delivering more power as compared to the slow discharge. The sample calculation for the energy expended, specific energy density and power density will be shown in Appendix A.

2.2  Part B: Making a battery

Using zinc as the anode, MnO₂ and carbon powder as the cathode, and KOH solution as an electrolyte, a Zinc-MnO₂ battery cell was constructed by hand. To investigate how different factors affect the battery’s performance, one parameter was varied for 4 batteries that were constructed in the lab.

For the first battery, zinc plate was used as an anode, a 1:3 ratio of carbon powder to MnO₂ was used as the cathode, and a solution of KOH was used as the electrolyte.

Consequently, a similar set-up of the first battery was done for the second battery, but instead of zinc plate, a zinc wire was used as the anode. This allows us to observe the effect of changing the anode material.

Subsequently, a similar set-up of the first battery was done for the third battery, but instead of using only KOH as the electrolyte, a solution mixture of KOH and NaCl was used as the electrolyte. In this case, this will allow us to inspect and observe the effect of changing the electrolyte.

Lastly, a similar set-up of the first battery was done for the fourth battery, but instead of using carbon powder, carbon granules were used instead for the 1:3 ratio of carbon powder to MnO₂. This will allow us to observe the effects of changing the cathode material of the battery cell.

2.21 Labelled sketch of the Zinc-MnO₂ battery cell

2.22 Comparison of Standard Potential and relation to the Nernst equation

The following reaction takes place for the Zinc-MnO₂ battery:

$\mathrm{Zn }\left(\mathrm{s}\right)+2\mathrm{ Mn}{\mathrm{O}}_2\mathrm{ }\left(\mathrm{s}\right)\rightleftharpoons \mathrm{ZnO }\left(\mathrm{s}\right)+\mathrm{M}{\mathrm{n}}_2{\mathrm{O}}_3\mathrm{ }\left(\mathrm{s}\right)$

The theoretical standard potential for the reaction above is calculated to be 1.43 V.

The following table shows the potential measured for all 4 batteries constructed using a DC voltmeter:

From the table above, it is observed that there are slight deviations from the theoretical standard potential of the cell. However, there should be no deviation from the theoretical standard potential of the cell since the reactants and products from the reaction shown above are all in the solid phase, which equates to a reaction quotient (Q_c) of 1, and thus the Nernst equation will not predict any deviation in cell potential. This relation will be shown in the sample calculations in Appendix A.

2.23 Energy density and capacity of the cell

Using the cell potential, the equation that relates Gibbs free energy and voltage can be used.

$∆\mathrm{G}=–\mathrm{n}\mathcal{F}{\mathrm{E}}_{\mathrm{cell}}^{\mathrm{º}}$

In addition, the electric charge that passes through the battery cell and the energy density of the cell can be determined. Since the limiting reagent is MnO₂ in the Zn-MnO₂ battery cells, only the mass of MnO₂ and the stoichiometric quantities of other reactants are considered for our calculations.

The material that will limit the life of the cell would be the MnO₂ since it is the limiting reagent in the reaction. The sample calculations for the energy stored, specific energy density and the capacity will be shown in Appendix A.

2.24 Factors that limit cell performance of constructed Zn-MnO₂ batteries

Some possible factors that may limit cell performance of the constructed Zn-MnO₂ batteries include:

• Inaccurate weighing of reactants
• Poor contact between the parts of the constructed battery cell
• Direct contact between the electrodes
• Impurities in the electrolyte solution that prevent electron transfer

To improve the overall construction of the battery cell, more precise measurements would be essential to ensure that the battery cell is constructed in such a way that the cathode never makes contact with the anode. In addition, the battery cell should also be submerged in the electrolyte solution for a longer duration for more effective electron transfer at the anode and cathode.

2.25 Combination of Anode/Cathode material that will give the maximum theoretical specific energy density 

The anode/cathode combination that are available in this lab are Zn-MnO₂, Zn-Air, and Al-Air.

For the Zn-MnO₂ cell, the limiting reagent will be the MnO₂ powder (0.75 g), whereas in the Zn-Air and Al-Air cells, the limiting reagent would be the zinc (4.0 g) and aluminum (4.0 g) since air is in excess.

With the assumption that the mass of the battery is 8.0g, consisting of 4g of Zn/ Al anode and 0.25g of carbon with 0.75 of MnO₂,the maximum theoretical specific energy density are as follows:

The sample calculations for the maximum theoretical specific energy density will be shown in Appendix A.

3.    Conclusions and Recommendations

To conclude, the objective of this lab is to analyze the chemistry behind batteries and study the performance of a lead-acid battery, then create a battery using materials provided in the lab.

In Part (A) of the lab, it can be observed that there are similar trends that occur for both slow and fast discharges. This is evident from the plots of voltage against time, as well as the plot of power against time, where it is seen that voltage decreases over time and power increases over time for both discharges. Peukert’s law can be used in this instance to approximate the change in capacity of the rechargeable lead-acid battery at a different rate of discharge. Subsequently, the energy expended by the battery during the discharge can also be determined using the “Trapezoidal rule” function in MATLAB, and the specific energy density and the average power density of the cell can be determined using this. It is observed that as the current being drawn from the cell increases, the specific energy density of the cell decreases whereas the power density increases.

In Part (B) of the lab, four Zn-MnO₂ batteries were created using the materials provided in the lab. The voltage of the battery was measured using a DC voltmeter, and the voltage was compared with the theoretical standard potential. Using the actual voltage measured, the energy stored, specific energy density and capacity of the cell was also computed.

 To improve the results of the experiment, measurements should be taken more precisely to ensure no error and discrepancies being contributed from the minor inaccuracy. For example, the carbon powder to MnO₂ ratio should be strictly kept precise to 3:1 to ensure the most accurate results possible. Furthermore, it should be ensured that no carbon powder and MnO₂ powder is leaking out of the battery or no reactants are lost during the battery making process, as the loss of some reactants can potentially affect the experimental results. Furthermore, to have the most accurate results for discharging the lead-acid battery, sulfation should be avoided by recharging the battery immediately after the discharge is complete. This will help to prevent the lead sulfate build up, which may potentially affect the results of the discharge process.

4.    Nomenclature

$\mathrm{E}\mathrm{º}$

– Standard Reduction Potential at STP

${\mathrm{E}}_{\mathrm{anode}}^{\mathrm{º}}$

– Standard Reduction Potential of Anode at STP

${\mathrm{E}}_{\mathrm{cathode}}^{\mathrm{º}}$

– Standard Reduction Potential of Cathode at STP

$∆\mathrm{G}$

– Gibbs Free Energy

n – Number of electrons involved in the stoichiometry of the redox cell

$\mathcal{F}$

– Faraday’s constant ( $96485\frac{\mathrm{C}}{\mathrm{mol}}$

)

R – Ideal gas constant $(8.314\frac{\mathrm{J}}{\mathrm{mol}\mathrm{⸱}\mathrm{K}})$

T – Absolute Temperature (K)

K_c – Equilibrium constant for the overall cell reaction

Q_c – Reaction Quotient

V – Voltage

q – Charge (A⸱h)

r – Electrical Resistance (Ω)

5.    References

1. Moll, Jennifer. ChE 290 Chemical Engineering Laboratory 1 [Lab Manual]. University of Waterloo, Department of Chemical Engineering; 2018 [Accessed 17 Nov. 2018]
2. All About Lead Acid Batteries. (2018). Peukert’s Law and Exponent Explained. [Online] Available at: http://all-about-lead-acid-batteries.capnfatz.com/all-about-lead-acid-batteries/lead-acid-battery-fundamentals/peukerts-law-and-exponent-explained/  [Accessed 17 Nov. 2018].
3. Crown Batteries. (2017). Sulfation: What is it and how to avoid it? [Online]. Available at: http://www.crownbattery.com/news/sulfation-and-battery-maintenance [Accessed 17 Nov. 2018]

6.    Appendices

6.1  Appendix A [Sample Calculations]

Part A Sample Calculations:

$\mathrm{Predicting\; cell\; potential\; using\; the\; Nernst\; equation}:$

$\mathrm{The\; following\; redox\; reaction\; occurs\; in\; the\; Lead}–\mathrm{Acid\; cell}:$

$\mathit{Pb }\left(s\right)+\mathit{Pb}{O}_2 \left(s\right)+2\mathit{ S}{O}_4^{2–} \left(\mathit{aq}\right) \rightleftharpoons 2\mathit{ PbS}{O}_4 \left(s\right)+2 H_2\mathit{O }\left(l\right)\mathit{ }{\mathit{ E}}_{\mathit{cell}}^o=2.04\mathit{ V }$

$\mathrm{If }\left[{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4\right]=1.0\mathrm{M},$

$Q_c=\frac{1}{{\left[S{O}_4^{2–}\right]}^2}=\frac{1}{{\left[H_{2}S{O}_4\right]}^{2 }}=\frac{1}{1^2}=1=K_c$

$\mathrm{Using\; the\; Nernst\; equation},\mathit{ E}=E_{\mathit{cell}}^o–\frac{2.302\mathit{RT}}{\mathit{nF}}\mathit{ln}\left(K_c\right)\mathit{ }$

$E=2.04\mathit{ V}–\frac{2.302*8.314\frac{J}{\mathit{mol}⸱K}*298.15\mathit{K }}{2\mathit{ mol }{e}^{–}*96485\frac{C}{\mathit{mol }{e}^{–}}}\ln \left(1\right)=2.04\mathit{ V}$

$\mathrm{The\; Nernst\; equation\; predicts\; that\; the\; cell\; potential\; will\; be\; unchanged\; at }2.04\mathrm{ V}.$

$\mathrm{Calculation\; for }{\mathrm{Peukert}}^{‘}\mathrm{s\; Coefficient}:$

$\mathrm{Peuker}{\mathrm{t}}^{‘}\mathrm{s\; Law}:\mathrm{ }t=H⸱{\left[\frac{C}{\mathit{IH}}\right]}^k$

$\mathrm{For\; the\; slow\; discharge},$ $20\min *\frac{1 \mathit{hr}}{60\mathit{ min}}=20 \mathit{hr }{\left[\frac{7.5\mathit{ A}⸱h}{5.0\mathit{ A}*20 \mathit{hr}}\right]}^k\mathrm{ }$

$\frac{1}{60}={[0.075]}^k$

$\ln \left(\frac{1}{60}\right)=k*\ln (0.075)$

$k=1.580664966 \cong 1.58$

$\mathrm{For\; the\; fast\; discharge},$ $90\mathrm{ s}*\frac{1 \mathit{hr}}{3600\mathit{ s}}=20 \mathit{hr }{\left[\frac{7.5\mathit{ A}⸱h}{20\mathit{ A}*20 \mathit{hr}}\right]}^k\mathrm{ }$

$\frac{1}{800}={[0.01875]}^k$

$\ln \left(\frac{1}{800}\right)=k*\ln (0.01875)$

$k=1.681002968 \cong 1.68$

$\mathrm{Computing\; the\; Energy\; Expended\; by\; the\; battery\; during\; discharge\; using\; MATLAB}:$

$\mathrm{MATLAB\; Script\; for\; Slow\; Discharge}:$

$\mathrm{MATLAB\; Command\; Window\; for\; Slow\; Discharge}:$

$\mathrm{MATLAB\; Script\; for\; Fast\; Discharge}:$

$\mathrm{MATLAB\; Command\; Window\; for\; Fast\; Discharge}:$

$\mathrm{Calculations\; for\; the\; Specific\; Energy\; Density\; and\; the\; Power\; Density\; of\; the\; Lead}–\mathrm{Acid\; Battery}:$

$\mathrm{The\; mass\; of\; the\; battery\; is\; recorded\; to\; be }2080.2\mathrm{ g},\mathrm{ which\; is\; equivalent\; to }2.0802\mathrm{ kg}$

$\mathrm{For\; the\; slow\; discharge},\mathrm{ the\; Specific\; Energy\; Density\; is}$

: $11.3194\mathit{ W}⸱\mathit{h }÷2.0802\mathit{ kg}=5.44149601 \frac{W⸱h}{\mathit{kg}}=5.44\mathrm{ }\frac{W⸱h}{\mathit{kg}}$

$\mathrm{For\; the\; fast\; discharge},\mathrm{ the\; Specific\; Energy\; Density\; is}:$ $4.5981\mathit{ W}⸱\mathit{h }÷2.0802\mathit{ kg}=2.21041246 \frac{W⸱h}{\mathit{kg}}=2.21\mathrm{ }\frac{W⸱h}{\mathit{kg}}\mathrm{ }$

From our experimental data, it is observed that the slow discharge took 719.127s and the fast discharge took 87.5133s.

$\mathrm{For\; the\; slow\; discharge},\mathrm{ the\; Power\; Density\; is}$

: $5.44149601 \frac{W⸱h}{\mathit{kg}} ÷\frac{719.127\mathit{ s}}{\frac{3600\mathit{ s}}{\mathit{hr}}}=27.24050917 \frac{W}{\mathit{kg}}=27.24 \frac{W}{\mathit{kg}}$

$\mathrm{For\; the\; fast\; discharge},\mathrm{ the\; Power\; Density\; is}:$ $2.21041246 \frac{W⸱h}{\mathit{kg}} ÷\frac{87.5133\mathit{ s}}{\frac{3600\mathit{ s}}{\mathit{hr}}}=90.92886288 \frac{W}{\mathit{kg}}=90.93 \frac{W}{\mathit{kg}}\mathrm{ }$

Part B Sample Calculations:

Note: All Sample Calculations shown only applies to the first Zn-MnO₂ battery cell constructed

$\mathrm{Calculating\; Energy\; stored\; and\; Energy\; Density\; of\; Zn}–{\mathrm{MnO}}_2\mathrm{ cells}:\mathrm{ }$

$\mathrm{For\; the\; first\; Zn}–\mathrm{plate\; battery\; cell\; made\; in\; Part\; B}:\mathrm{ }$

$∆\mathrm{G}=–\mathrm{n}\mathcal{F}{\mathrm{E}}_{\mathrm{cell}}^{\mathrm{º}}$

$\mathit{\Delta G}=–2\mathit{ mol }{e}^{–}\times 96485 \frac{C}{\mathit{mol }⸱ e^{–}}\times 1.504\mathit{ V}=– 290226.88\mathit{ J}\cong – 290.227\mathit{ kJ}$

$\mathrm{Hence},\mathrm{ the\; Gibbs\; energy\; stored\; in\; the\; cell\; is\; approximately}–290.227\mathrm{ kJ}.$

$\mathrm{The\; theoretical\; Specific\; Energy\; Density\; of\; the\; cell\; is }\frac{\left|\mathit{\Delta G}\right|}{\mathit{Weight\; of\; Reactants }}$

$=\frac{290.22688\mathit{ kJ}}{\left(0.7504+3.6735\right)\mathit{g }}= 65.60430389 \frac{\mathit{kJ}}{g}= 65.604\frac{\mathit{kJ}}{g}$

$\mathrm{The\; practical\; Specific\; Energy\; Density\; of\; the\; cell\; is }\frac{\left|\mathit{\Delta G}\right|}{\mathit{Mass\; of\; Battery }}$

$=\frac{290.22688\mathit{ kJ}}{9.466\mathit{ g }}= 30.65992816 \frac{\mathit{kJ}}{g} \cong 30.660\frac{\mathit{kJ}}{g}$

$\mathrm{Determining\; the\; limiting\; reagent\; and\; the\; \#\; of\; moles\; of }{\mathrm{e}}^{–} \mathrm{transferred}:$

$\mathrm{The\; following \; chemical\; reaction\; occurs\; in\; the\; Zn}–\mathrm{Mn}{\mathrm{O}}_2\mathrm{ cell}:$

$\mathit{Zn }\left(s\right)+2\mathit{ Mn}{O}_2 \left(s\right)\rightleftharpoons \mathit{ZnO }\left(s\right)+M{n}_2{O}_3 \left(s\right)$

$\mathrm{For\; the\; first\; Zn}–{\mathrm{MnO}}_2\mathrm{ battery\; cell\; constructed}:\mathrm{ }$

$m_{\mathit{Zn}}= 3.6735\mathit{ g }$

$m_{\mathit{Mn}{O}_2}= 0.7504\mathit{ g}$

$n_{\mathit{Zn}}=\frac{3.6735\mathit{ g}}{65.38\frac{g}{\mathit{mol}}}= 0.0561869\mathit{ mol}$

$n_{\mathit{Mn}{O}_2}=\frac{0.7504\mathit{ g}}{86.94\frac{g}{\mathit{mol}}}= 0.00863123\mathit{ mol}$

$\mathrm{From\; the\; number\; of\; moles\; present},\mathrm{ it\; is\; evident\; that\; Mn}{\mathrm{O}}_2\mathrm{ is\; the\; limiting\; reagent}.$

$\mathrm{Since\; we\; know\; limiting\; reagent\; is\; Mn}{\mathrm{O}}_2,\mathrm{ the\; number\; of\; mol\; of }{\mathrm{e}}^{–} \mathrm{transferred\; can\; be\; determined\; by}:\mathrm{ }$

$\mathrm{\#\; of\; mol\; of }{\mathrm{e}}^{–}\mathrm{ transferred}=\mathrm{ }\frac{\mathit{Mass\; of\; Limiting\; Reagent}}{\mathit{Molar\; Mass\; of\; Limiting\; Reagent}}$

$\mathrm{Theoretical\; \#\; of\; mol\; of }{\mathrm{e}}^{–}\mathrm{ transferred}=\mathrm{ }\frac{0.75\mathit{ g}}{86.9368\frac{g}{\mathit{mol}}}= 8.626956594*{10}^{–3}\mathit{ mol }{e}^{–}$

$\mathrm{Actual\; \#\; of\; mol\; of }{\mathrm{e}}^{–}\mathrm{ transferred}=\mathrm{ }\frac{0.7504\mathit{ g}}{86.9368\frac{g}{\mathit{mol}}}= 8.631557637*{10}^{–3}\mathit{ mol }{e}^{–}$

$\mathrm{Estimating\; the\; Capacity\; of\; Zn}–{\mathrm{MnO}}_2\mathrm{ battery\; cells}:$

$\mathrm{With\; the\; \#\; of }{\mathrm{mol\; of\; e}}^{–} \mathrm{transferred\; known},\mathrm{ the\; capacity\; can\; be\; determined }$

$q=n\mathcal{F}$

$\mathrm{Theoretical\; Amount\; of\; Charge}:\mathit{ q}=8.626956594*{10}^{–3}\mathit{ mol }{e}^{–}* 96485\frac{C}{\mathit{mol }{e}^{–}}=832.3719069\mathit{ C}$

$\mathrm{Since} 3600\mathit{ C}=1\mathit{ A}⸱h$

$\mathrm{Theoretical\; Capacity}: 832.3719069\mathit{ C}*\frac{A⸱h}{3600\mathit{ C}}=0.2312144186\mathit{ A}⸱\mathit{h }\cong 0.23121\mathit{ A}⸱\mathit{h }$

$\mathrm{Actual\; Amount\; of\; Charge}:\mathit{ q}=8.631557637*{10}^{–3}\mathit{ mol }{e}^{–}* 96485\frac{C}{\mathit{mol }{e}^{–}}=832.8158386\mathit{ C}$

$\mathrm{Since} 3600\mathit{ C}=1\mathit{ A}⸱h$

$\mathrm{Actual\; Capacity}: 832.8158386\mathit{ C}*\frac{A⸱h}{3600\mathit{ C}}=0.2313377329\mathit{ A}⸱\mathit{h }\cong 0.23134\mathit{ A}⸱\mathit{h }$

$\mathrm{Finding\; the\; maximum\; theoretical\; Specific\; Energy\; Density\; for\; all\; Anode}–\mathrm{Cathode\; Combinations}:$

$\mathrm{The\; Anode}–\mathrm{Cathode\; combinations\; and\; the\; reactants\; are\; as\; follows}:$

$\mathit{\Delta G}= –\mathrm{n}\mathcal{F}{\mathrm{E}}_{\mathrm{cell}}^{\mathrm{º}}$

$\mathrm{Maximum\; Theoretical\; Specific\; Energy\; Density}=\frac{\left|\mathit{\Delta G}\right|}{\mathit{Weight\; of\; Reactants}}$

$\mathrm{For\; the\; Zn}–{\mathrm{MnO}}_2 \mathrm{cell},\mathrm{ n}=2\mathrm{ and }{\mathrm{E}}_{\mathrm{cell}}^{\mathrm{º}}=+1.43 \mathrm{V\; from\; literature}$

$\mathit{\Delta G}=–2\mathit{ mol }{e}^{–}\times 96485 \frac{C}{\mathit{mol }⸱ e^{–}}\times 1.43\mathit{ V}=– 275947.1\mathit{ J}\cong – 275.9471\mathit{ kJ}$

$\mathrm{Maximum\; Theoretical\; Specific\; Energy\; Density}=\frac{275.9471\mathit{ kJ}}{0.75\mathit{ g\; Mn}{O}_2+4.0\mathit{ g\; Zn }}=58.0941 \frac{\mathit{kJ}}{g}$

$\mathrm{Hence},\mathrm{ the\; Maximum\; Theoretical\; Specific\; Energy\; Density\; is }58.1\mathrm{ }\frac{\mathrm{kJ}}{\mathrm{g}} \mathrm{for\; the\; Zn}–{\mathrm{MnO}}_2 \mathrm{cell}.$

$\mathrm{For\; the\; Zn}–\mathrm{Air} \mathrm{cell},\mathrm{ n}=2\mathrm{ and }{\mathrm{E}}_{\mathrm{cell}}^{\mathrm{º}}=+1.59 \mathrm{V\; from\; literature}$

$\mathit{\Delta G}=–2\mathit{ mol }{e}^{–}\times 96485 \frac{C}{\mathit{mol }⸱ e^{–}}\times 1.59\mathit{ V}=– 306822\mathit{ J}\cong – 306.822\mathit{ kJ}$

$\mathrm{Maximum\; Theoretical\; Specific\; Energy\; Density}=\frac{ 306.822 \mathit{ kJ}}{4.0\mathit{ g\; Zn }}=76.7056 \frac{\mathit{kJ}}{g}$

$\mathrm{Hence},\mathrm{ the\; Maximum\; Theoretical\; Specific\; Energy\; Density\; is }70.7\mathrm{ }\frac{\mathrm{kJ}}{\mathrm{g}} \mathrm{for\; the\; Zn}–\mathrm{Air} \mathrm{cell}.$

$\mathrm{For\; the\; Al}–\mathrm{Air\; cell},\mathrm{ n}=2\mathrm{ and }{\mathrm{E}}_{\mathrm{cell}}^{\mathrm{º}}=+1.59\mathrm{ V\; from\; literature}$

$\mathit{\Delta G}=–2\mathit{ mol }{e}^{–}\times 96485 \frac{C}{\mathit{mol }⸱ e^{–}}\times 2.71\mathit{ V}=– 784423\mathit{ J}\cong – 784.423\mathit{ kJ}$

$\mathrm{Maximum\; Theoretical\; Specific\; Energy\; Density}=\frac{ 784.423 \mathit{ kJ}}{4.0\mathit{ g\; Al }}=196.106 \frac{\mathit{kJ}}{g}$

$\mathrm{Hence},\mathrm{ the\; Maximum\; Theoretical\; Specific\; Energy\; Density\; is }196.1\mathrm{ }\frac{\mathrm{kJ}}{\mathrm{g}} \mathrm{for\; the\; Al}–\mathrm{Air} \mathrm{cell}$

6.2  Appendix B [Report Questions]

1. The output voltage decreases upon discharge over time as a result of lead sulfate being formed on the surface and in the body of the plates, which is known as sulfation. ^[3] The sulfate has a higher resistance as compared to the lead, causing the internal resistance of the cell to increase, and thus, contributing to a drop in voltage. In addition, PbO₂ also contributes to the overall resistance, resulting in a greater voltage drop in the given time frame. This drop in voltage causes the power output to decrease as well, since voltage and power are related by the equation P = V⸱I. In contrast, the alkaline batteries and the Al-Air batteries have a slower rate of voltage drop due to having a lower resistance. Hence, the alkaline batteries and the Al-Air batteries can last longer than the lead-acid battery.

2. The reaction that occurs is as follows:

$\mathit{Pb }\left(s\right)+\mathit{Pb}{O}_2 \left(s\right)+2\mathit{ S}{O}_4^{2–} \left(\mathit{aq}\right) \rightleftharpoons 2\mathit{ PbS}{O}_4 \left(s\right)+2 H_2\mathit{O }\left(l\right)$

 In the reaction shown above, 2 electrons are involved in the process.

It is known that 1 Amp hour is equivalent to 3600 C, and the number of moles of electrons can be determined using this information.

$q=n_e\mathcal{F}$

$3600\mathit{ C}=n_e*96485 \frac{C}{\mathit{mol }{e}^{–}}$

$n_e=0.0373114992\mathit{ mol }{e}^{–}$

Based on the stoichiometric ratios shown in the chemical reaction above, one mole of lead is consumed for every 2 moles of electrons. Consequently, two moles of sulfuric acid are consumed for every two moles of electrons.

  $n_{\mathit{Pb}}=n_e/2$

$n_{\mathit{Pb}}=0.0373114992\mathit{ mol }{e}^{–}*\frac{1\mathit{ mol\; Pb}}{2\mathit{ mol }{e}^{–}} =0.0186557496\mathit{ mol\; Pb}$

$m_{\mathit{Pb}}=M_{\mathit{Pb}}{n}_{\mathit{Pb}}$

$m_{\mathit{Pb}}= 0.0186557496\mathit{ mol\; Pb}* 207.2 \frac{g}{\mathit{mol\; Pb}} = 3.865471317\mathit{ g\; Pb}$

$n_{H_{2}S{O}_4}=n_e$

$n_{H_{2}S{O}_4}=0.0373114992\mathit{ mol} H_{2}S{O}_4$

$m_{H_{2}S{O}_4}=M_{H_{2}S{O}_4}*n_{H_{2}S{O}_4}$

$m_{H_{2}S{O}_4}=0.0373114992\mathit{ mol }{H}_{2}S{O}_4* 98.08 \frac{g}{\mathit{mol }{H}_{2}S{O}_4} = 3.659511841\mathit{ g} H_{2}S{O}_4$

From the calculations, it is determined that the theoretical amount of lead and sulfuric acid consumed by 1 amp-hour is approximately 3.865g and 3.660g respectively. In batteries, lead is supplied in excess with reference to the theoretical amount since the efficiency is not 100% due to the possibility of side reactions occurring. This is commonly known as lead shedding and is contributed by the lead sulfate, which causes an irreversible process of producing crystals that cannot be turned back into lead. As a result, the battery cannot be charged back to its full capacity. Hence, extra lead being supplied allows us to considerably slow down the capacity fade.

3. Two possible factors that can contribute to a permanent reduction in lead-acid battery performance include having high internal resistance, and the occurrence of plate corrosion. If the internal resistance of a battery is very high, the amount of current that would be able to flow will be restricted, decreasing the voltage. Hence, this will cause the battery to heat up and cause a loss of electrolytes. In addition, plate corrosion is also an important factor that can contribute towards a permanent reduction in lead-acid battery performance. As the lead-acid discharges, a layer of lead sulfate is formed on the surface and in the body of the plates, leading to corrosion and thus, permanently reduces the performance of the battery.

4. The 9 V in an alkaline battery is produced by connecting 6 alkaline batteries in series. This is because voltage is additive in a series connection, and since each alkaline battery is 1.43 V, approximately 6 alkaline batteries can be connected in series to produce a total voltage of 9 V.

5. Based on the given uncertainty in voltage measurements across the battery and the current through the leads, the range of possible values for the measured voltage are 5.7 V ± 0.0285 V. Similarly, the range of possible values for the current through the leads are 0.7 A ± 0.0035 A.

The equation that relates voltage with energy is:

$E=V⸱I⸱\mathit{t }$

Since P = V⸱I and E = P⸱t

The time elapsed was given to be 30 minutes, which is equivalent to 1800 seconds. The maximum error range possible for the energy dissipated for the battery over the observed time can be determined by substituting in the maximum and minimum values into the equation.

$\mathrm{E}=5.7\mathit{ V}\times 0.7\mathit{ A}\times 1800\mathit{ s}=7182\mathit{ J }$

$E_{\mathit{min}}=\left(5.7–0.0285\right)\mathit{ V}\times \left(0.7–0.0035\right)\mathit{ A}\times 1800\mathit{ s}=7110.35955\mathit{ J }$

$E_{\mathit{max}}=\left(5.7+0.0285\right)\mathit{ V}\times \left(0.7+0.0035\right)\mathit{ A}\times 1800\mathit{ s}=7253.99955\mathit{ J }$

$\mathrm{Hence},\mathrm{ the\; maximum\; error\; range\; for\; the\; energy\; dissipated\; is }7110.35955\mathrm{ J\; to }7253.99955\mathrm{ J}.$

To determine the uncertainty in energy, the partial derivatives are taken.

The error for V was found to be ± 0.0285 V, and the error for I was found to be ± 0.0035 A.

$\partial E=\pm \mathit{ E}\sqrt{{\left(\frac{\partial V}{V}\right)}^2+{\left(\frac{\partial I}{I}\right)}^2}$

For the given data, substituting the appropriate values produces the following uncertainty in energy.

$\partial E=\pm 7182\sqrt{{\left(\frac{0.0285}{5.7}\right)}^2+{\left(\frac{0.0035}{0.7}\right)}^2}= \pm 50.78440902\mathit{ J}$

$\mathrm{Hence},\mathrm{ the\; estimated\; uncertainty\; in\; energy\; dissipated\; is\; approximately }50.78\mathrm{ J}.$

$\mathrm{The\; range\; for\; the\; energy\; dissipated\; is }7131.215591\mathrm{ J}<\mathrm{E}<7232.784409\mathrm{ J}.$

6.3 Appendix C [Raw Data]

Experimental Data collected for Part B: