1. **Problem 2.1:** Calculate the effective charge capacity of the negative electrode per cm³. Given: - Lithium density $\rho = 0.534$ g/cm³ - Lithium layer thickness $t = 15\ \mu m = 15 \times 10^{-4}$ cm - Molar mass of Li $M = 6.94$ g/mol - Faraday constant $F = 96485$ C/mol **Step 1:** Calculate the mass of lithium per cm²: $$m = \rho \times t = 0.534 \times 15 \times 10^{-4} = 8.01 \times 10^{-3} \text{ g/cm}^2$$ **Step 2:** Calculate moles of lithium per cm²: $$n = \frac{m}{M} = \frac{8.01 \times 10^{-3}}{6.94} = 1.154 \times 10^{-3} \text{ mol/cm}^2$$ **Step 3:** Calculate charge capacity per cm² in Coulombs: $$Q = n \times F = 1.154 \times 10^{-3} \times 96485 = 111.3 \text{ C/cm}^2$$ **Step 4:** Convert Coulombs to Ampere-hours (Ah): $$1 \text{ Ah} = 3600 \text{ C} \Rightarrow Q = \frac{111.3}{3600} = 0.0309 \text{ Ah/cm}^2$$ **Step 5:** Calculate volumetric capacity (per cm³): Since thickness is $15 \times 10^{-4}$ cm, $$\text{Capacity per cm}^3 = \frac{0.0309}{15 \times 10^{-4}} = 206 \text{ Ah/L} = 206 \text{ mAh/cm}^3$$ --- 2. **Problem 2.2:** Find material loading (mg/cm²) of cathode composite to match anode capacity. Given: - Cathode composite contains 93% Li$_3$CoO$_2$, 5% CB, 2% PVDF by weight - Cathode capacity must equal anode capacity $0.0309$ Ah/cm² - Theoretical capacity of LiCoO$_2$ is about 140 mAh/g **Step 1:** Calculate required mass of LiCoO$_2$ per cm²: $$m_{LiCoO_2} = \frac{0.0309 \text{ Ah/cm}^2}{0.140 \text{ Ah/g}} = 0.221 \text{ g/cm}^2 = 221 \text{ mg/cm}^2$$ **Step 2:** Calculate total composite mass per cm²: $$m_{composite} = \frac{m_{LiCoO_2}}{0.93} = \frac{221}{0.93} = 237.6 \text{ mg/cm}^2$$ --- 3. **Problem 2.3:** Calculate cathode thickness. Given: - Composite density $\rho_c = 1.8$ g/cm³ - Composite mass loading $m_c = 0.2376$ g/cm² **Step 1:** Thickness $t_c$: $$t_c = \frac{m_c}{\rho_c} = \frac{0.2376}{1.8} = 0.132 \text{ cm} = 1320 \ \mu m$$ --- 4. **Problem 2.4:** Determine open-circuit voltage (OCV) at fully charged and discharged states. Given from problem: - Cathode potential $E^\circ_{cathode} \approx 3.6$ to $4.2$ V vs Li/Li$^+$ - Anode potential $E^\circ_{anode} = -3.04$ V vs SHE **Step 1:** OCV is difference between cathode and anode potentials. Assuming Li/Li$^+$ reference: - Fully charged OCV $\approx 4.2$ V - Fully discharged OCV $\approx 3.6$ V --- 5. **Problem 2.5:** Calculate total capacity, mass, specific energy, and volumetric energy density. Given: - Electrode area $A = 100$ cm² - Separator thickness $t_s = 20 \ \mu m = 2 \times 10^{-3}$ cm - Separator porosity $\varepsilon = 0.4$ - Electrolyte density $\rho_{electrolyte} = \frac{1.32 + 1.07}{2} = 1.195$ g/cm³ (average) - Masses from previous steps **Step 1:** Total capacity: $$Q_{total} = 0.0309 \text{ Ah/cm}^2 \times 100 = 3.09 \text{ Ah}$$ **Step 2:** Mass of lithium: $$m_{Li} = 8.01 \times 10^{-3} \text{ g/cm}^2 \times 100 = 0.801 \text{ g}$$ **Step 3:** Mass of copper and aluminum foils: - Thickness $= 10$ pm $= 10 \times 10^{-7}$ cm - Densities: $\rho_{Cu} = 8.96$ g/cm³, $\rho_{Al} = 2.70$ g/cm³ $$m_{Cu} = 8.96 \times 10 \times 10^{-7} \times 100 = 8.96 \times 10^{-4} \text{ g}$$ $$m_{Al} = 2.70 \times 10 \times 10^{-7} \times 100 = 2.7 \times 10^{-4} \text{ g}$$ **Step 4:** Mass of cathode composite: $$m_{cathode} = 0.2376 \text{ g/cm}^2 \times 100 = 23.76 \text{ g}$$ **Step 5:** Volume of separator pores: $$V_{pores} = A \times t_s \times \varepsilon = 100 \times 2 \times 10^{-3} \times 0.4 = 0.08 \text{ cm}^3$$ **Step 6:** Mass of electrolyte: $$m_{electrolyte} = V_{pores} \times \rho_{electrolyte} = 0.08 \times 1.195 = 0.0956 \text{ g}$$ **Step 7:** Total mass: $$m_{total} = m_{Li} + m_{Cu} + m_{Al} + m_{cathode} + m_{electrolyte} = 0.801 + 8.96 \times 10^{-4} + 2.7 \times 10^{-4} + 23.76 + 0.0956 = 24.66 \text{ g}$$ **Step 8:** Specific energy: $$E_{specific} = \frac{Q_{total} \times V_{OCV}}{m_{total}}$$ Using $V_{OCV} = 4.2$ V (fully charged): $$E_{specific} = \frac{3.09 \times 4.2}{0.02466} = 526 \text{ Wh/kg}$$ **Step 9:** Volumetric energy density: Total volume approx sum of electrode and separator volumes: $$V_{total} = A \times (t_{Li} + t_c + t_s) = 100 \times (15 \times 10^{-4} + 0.132 + 2 \times 10^{-3}) = 100 \times 0.1347 = 13.47 \text{ cm}^3 = 0.01347 \text{ L}$$ $$E_{volumetric} = \frac{Q_{total} \times V_{OCV}}{V_{total}} = \frac{3.09 \times 4.2}{0.01347} = 963 \text{ Wh/L}$$ --- **Final answers:** - 2.1: Effective charge capacity $= 206$ mAh/cm³ - 2.2: Cathode composite loading $= 238$ mg/cm² - 2.3: Cathode thickness $= 1320 \ \mu m$ - 2.4: OCV fully charged $\approx 4.2$ V, fully discharged $\approx 3.6$ V - 2.5: Total capacity $= 3.09$ Ah, total mass $= 24.66$ g, specific energy $= 526$ Wh/kg, volumetric energy density $= 963$ Wh/L