For a lead storage battery: (a) Sketch one cell that shows the anode, cathode, electrolyte, direction of electron and ion flow, and sign of the electrodes. (b) Write the anode, cathode, and overall cell reactions. (c) Calculate the equilibrium constant for the cell reaction (E° = 1.924 V). (d) What is the cell voltage when the cell reaction reaches equilibrium?

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1. Step 1: Understand the components of a lead storage battery. The anode is made of lead (Pb), the cathode is lead dioxide (PbO2), and the electrolyte is sulfuric acid (H2SO4). Electrons flow from the anode to the cathode through an external circuit, while ions move through the electrolyte.
2. Step 2: Write the half-reactions for the anode and cathode. At the anode, lead is oxidized: \( \text{Pb} \rightarrow \text{Pb}^{2+} + 2\text{e}^- \). At the cathode, lead dioxide is reduced: \( \text{PbO}_2 + 4\text{H}^+ + 2\text{e}^- \rightarrow \text{Pb}^{2+} + 2\text{H}_2\text{O} \).
3. Step 3: Combine the half-reactions to write the overall cell reaction. Ensure that the electrons cancel out: \( \text{Pb} + \text{PbO}_2 + 4\text{H}^+ \rightarrow 2\text{Pb}^{2+} + 2\text{H}_2\text{O} \).
4. Step 4: Use the Nernst equation to calculate the equilibrium constant \( K \) for the cell reaction. The equation is \( E^\circ = \frac{RT}{nF} \ln K \), where \( E^\circ \) is the standard cell potential, \( R \) is the gas constant, \( T \) is the temperature in Kelvin, \( n \) is the number of moles of electrons transferred, and \( F \) is Faraday's constant.
5. Step 5: Determine the cell voltage at equilibrium. At equilibrium, the cell voltage is zero because the forward and reverse reactions occur at the same rate, and there is no net electron flow.