For the lead storage battery the overall reaction is:
Pb(solid) + PbO2(solid) + 2H+ + 2HSO4- --> 2PbSO4(solid) + 2H2O(liq.) for which standard potential, E°, is 2.04 volts.
Calculate E for this battery when the concentration of H2SO4 is 8.1 M, that is both H+ and HSO4- = 8.1 M.
E = E° - 0.0257/n * ln K
K= [H2O]^2 / [H+]^2[H2SO4-]^2
The Attempt at a Solution
E = 2.04 V - 0.0257/2 * ln [H2O]^2/[8.1]^2[8.1]^2
First off, I am confused as to whether the molarity of H2O is required for the calculation, or whether it is not used to calculate the equilibrium constant K. Secondly, in the wording of the problem where it says both H+ and HSO4- = 8.1M, does this mean that 8.1M solutions were prepared for each, or the two combined equals an 8.1M solution?
Assuming the molarity of H2O is not required, would the K be calculated as
K = [8.1]^2[8.1]^2 and then fit into the equation to calculate E? I'm a bit lost and some clarification would help. Thank you!