The overall reaction in the lead storage battery is $$\mathrm{Pb}(s)+\mathrm{PbO}_{2}(s)+2 \mathrm{H}^{+}(a q)+2 \mathrm{HSO}_{4}^{-}(a q) \longrightarrow 2 \mathrm{PbSO}_{4}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ a. For the cell reaction \(\Delta H^{\circ}=-315.9 \mathrm{kJ}\) and \(\Delta S^{\circ}=\) \(263.5 \mathrm{J} / \mathrm{K} .\) Calculate \(\mathscr{E}^{\circ}\) at \(-20 .^{\circ} \mathrm{C} .\) Assume \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature. b. Calculate \(\mathscr{E}\) at \(-20 .^{\circ} \mathrm{C}\) when \(\left[\mathrm{HSO}_{4}^{-}\right]=\left[\mathrm{H}^{+}\right]=4.5 \mathrm{M}\) c. Consider your answer to Exercise \(71 .\) Why does it seem that batteries fail more often on cold days than on warm days?

We are given that \(\Delta H^{\circ} = -315.9\,\text{kJ}\) and \(\Delta S^{\circ} = 263.5\,\text{J/K}\). First, we need to convert the temperature from Celsius to Kelvin. We know that at \(T = -20\,^{\circ}\text{C}\), the temperature in Kelvin is: \[T = (-20+273.15)\,\text{K} = 253.15\,\text{K}\] The relationship between the Gibbs free energy (\(\Delta G^{\circ}\)), enthalpy change (\(\Delta H^{\circ}\)), and entropy change (\(\Delta S^{\circ}\)) is given by the equation: \[\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\] Substitute the given values and temperature to find the Gibbs free energy change: \[\Delta G^{\circ} = (-315.9 \times 10^{3}\,\text{J}) - (253.15\,\text{K})(263.5\,\text{J/K})\] Calculate the value of \(\Delta G^{\circ}\): \[\Delta G^{\circ} \approx -285196.725\,\text{J}\] Next, we will use the relationship between the Gibbs free energy change and the standard cell potential \(\mathscr{E}^{\circ}\), given by: \[\Delta G^{\circ} = -nF\mathscr{E}^{\circ}\] Where \(n\) is the number of moles of electrons transferred, and \(F\) is the Faraday constant, approximately equal to \(96485\,\text{C/mol}\). For the given overall reaction, the number of electrons transferred, \(n\), is 2. Now, solve for the standard cell potential (\(\mathscr{E}^{\circ}\)): \[\mathscr{E}^{\circ} = -\frac{\Delta G^{\circ}}{nF}\] Plug in the values: \[\mathscr{E}^{\circ} \approx \frac{-(-285196.725\,\text{J})}{(2)(96485\,\text{C/mol})}\] Calculate the value of \(\mathscr{E}^{\circ}\): \[\mathscr{E}^{\circ} \approx 1.478\,\text{V}\] Thus, the standard cell potential at -20° C is approximately 1.478 V.

Now, we will calculate the cell potential \(\mathscr{E}\) at -20° C when \([\mathrm{HSO}_{4}^{-}] = [\mathrm{H}^{+}] = 4.5\,\text{M}\). Because the reaction involves entropy change, the Nernst equation will be used to calculate the cell potential: \[\mathscr{E} = \mathscr{E}^{\circ} - \frac{RT}{nF} \ln(Q)\] Where \(Q\) is the reaction quotient. For the given reaction, since there are 2 moles of products and 2 moles of reactants: \(Q = \frac{[\mathrm{H}^{+}]^2}{[\mathrm{HSO}_{4}^{-}]^2}\) Since \([\mathrm{H}^{+}] = [\mathrm{HSO}_{4}^{-}] = 4.5\,\text{M}\), the reaction quotient, \(Q\), is equal to 1. Now, substitute the values and solve for \(\mathscr{E}\): \[\mathscr{E} = 1.478\,\text{V} - \frac{(8.314\,\text{J/(K*mol}))(253.15\,\text{K})}{(2)(96485\,\text{C/mol})} \ln(1)\] Since \(\ln(1) = 0\), the equation simplifies to: \[\mathscr{E} = 1.478\,\text{V}\] Thus, the cell potential at -20 °C with the given concentrations is approximately 1.478 V.

In step 1, we have seen that colder temperatures result in a lower cell potential. A lower cell potential corresponds to a slower battery discharge rate and weaker overall performance. This is because the entropy change equation, \(\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\), shows that as the temperature decreases, the Gibbs free energy change increases, leading to a decrease in cell potential. This decrease in cell potential means that the battery is less capable of providing the required energy to start or power devices. Thus, batteries tend to fail more often on cold days compared to warmer days, as lower temperatures affect their overall performance.