Question

A student is given a lead-acid battery, which has a standard reduction potential of -0.356 V. The battery is undergoing a reaction where Pb(s) + SO4^2-(aq) -> PbSO4(s) + 2e^-. If the concentration of PbSO4 in the anode compartment is 1.15 M and the concentration of PbSO4 in the cathode compartment is 0.55 M, what is the power output of the battery, assuming it has a current of 1.25 A?

Answer 1

To determine the power output of the battery, we first need to find the cell potential  E_cell  under non-standard conditions using the Nernst equation:E_cell = E -  RT/nF  * ln Q Where:E = standard reduction potential  -0.356 V R = gas constant  8.314 J/molK T = temperature  assuming room temperature, 298 K n = number of electrons transferred  2 for this reaction F = Faraday's constant  96485 C/mol Q = reaction quotientFor the reaction quotient  Q , we have:Q = [PbSO4]_cathode / [PbSO4]_anodeSubstituting the given concentrations:Q = 0.55 / 1.15  0.478Now, we can plug the values into the Nernst equation:E_cell = -0.356 -   8.314 * 298  /  2 * 96485   * ln 0.478 E_cell  -0.356 + 0.0129E_cell  -0.3431 VNow that we have the cell potential, we can calculate the power output  P  using the formula:P = I * E_cellWhere I is the current  1.25 A . Substituting the values:P = 1.25 *  -0.3431 P  -0.4289 WSince the power output is negative, it means that the battery is actually consuming power rather than producing it. This could be due to the concentrations of the reactants and products, which are causing the reaction to be non-spontaneous under these conditions.