The overall reaction in the lead storage battery is $\mathrm{Pb}(s)+\mathrm{PbO}_{2}(s)+2 \mathrm{H}^{+}(a q)+2 \mathrm{HSO}_{4}^{-}(a q) \longrightarrow$ $$\quad\quad\quad\quad\quad\quad\quad 2 \mathrm{PbSO}_{4}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ 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} \mathrm{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 69. Why does it seem that batteries fail more often on cold days than on warm days?

We will start by using the relation between the Gibbs free energy change ∆G°, the standard enthalpy change ∆H°, and the standard entropy change ∆S°: ∆G° = ∆H° - T∆S° Next, we need to relate ∆G° to the standard electromotive force (EMF) E° of the cell through the equation: ∆G° = -nFE° Where n is the number of moles of electrons transferred in the overall reaction, and F is Faraday's constant (96485 C/mol). Step 1: Determine the number of moles of electrons transferred (n) The balanced overall reaction is: Pb(s) + PbO₂(s) + 2 H⁺(aq) + 2 HSO₄⁻(aq) → 2 PbSO₄(s) + 2 H₂O(l) It can be seen that the redox reaction involves the transfer of 2 moles of electrons. Step 2: Calculate the standard Gibbs free energy change ∆G° Using the relationship ∆G° = ∆H° - T∆S°, we need to first convert -20°C to Kelvin: T = -20 + 273.15 = 253.15 K Now, plug in the given values of ∆H° (-315.9 kJ) and ∆S° (263.5 J/K) and calculate ∆G°: ∆G° = -(-315900 J) - (253.15 K)(263.5 J/K) = 17495.475 J Step 3: Calculate the standard cell potential E° Finally, we can calculate the standard cell potential E° using the equation: ∆G° = -nFE° E° = -∆G° / (nF) = -17495.475 J / (2 mol * 96485 C/mol) = 0.0905 V So, the standard cell potential E° at -20°C is 0.0905 V.

Now, we have to find the cell potential E at -20°C with the given concentrations of HSO₄⁻ and H⁺ ions. To do this, we'll use the Nernst equation: E = E° - (RT/nF) ln(Q) Where R is the gas constant (8.314 J/mol K), T is the temperature in Kelvin, n is the number of moles of electrons transferred, F is Faraday's constant, and Q is the reaction quotient. Step 1: Calculate the reaction quotient Q For the balanced overall reaction, the reaction quotient Q can be written as: Q = ([PbSO₄]²[H₂O]²) / ([Pb][PbO₂][H⁺]²[HSO₄⁻]²) Since the concentrations of solid and liquid species remain constant, we'll only include the concentrations of H⁺ and HSO₄⁻ ions: Q = ([H⁺]²[HSO₄⁻]²) = (4.5 M)²(4.5 M)² = 410.0625 Step 2: Calculate the cell potential E Plugging the known values into the Nernst equation: E = 0.0905 V - (8.314 J/mol K)(253.15 K) / (2 mol * 96485 C/mol) ln(410.0625) = -0.00582 V So, the cell potential E at -20°C with given ion concentrations is -0.00582 V.

Based on the results obtained in this exercise, it can be observed that the cell potential E decreases significantly at lower temperatures (-20°C). The cell potential E, which represents the driving force of the chemical reaction, directly affects the electrical output of the battery. The decrease in cell potential at low temperatures results in less electrical output, leading to a weaker battery performance. Batteries are more likely to fail on cold days because the decreased cell potential adversely affects the electrical output. Lower electrical output makes it difficult for the battery to provide enough power to start a vehicle or run electrical systems adequately. So, batteries tend to fail more often on colder days than on warmer ones.