How to Calculate the Nernst Potential for a Lead Storage Battery?

  • Thread starter Thread starter cuttooth
  • Start date Start date
  • Tags Tags
    Electrochemistry
Click For Summary
SUMMARY

The discussion focuses on calculating the Nernst potential for a lead storage battery using the Nernst Equation, E = E° - 0.0257/n * ln K. The standard potential (E°) for the reaction is established at 2.04 volts. The equilibrium constant (K) is derived from the concentrations of H+ and HSO4- ions, both at 8.1 M. Participants clarify that the concentration of water does not factor into the equilibrium constant for this calculation, and the confusion regarding the molarity of H2O is addressed.

PREREQUISITES
  • Understanding of the Nernst Equation and its components
  • Knowledge of equilibrium constants in chemical reactions
  • Familiarity with lead storage battery chemistry
  • Basic concepts of molarity and concentration in solutions
NEXT STEPS
  • Study the derivation and applications of the Nernst Equation in electrochemistry
  • Research the role of water in equilibrium constants for aqueous reactions
  • Explore the specific reactions and potentials involved in lead-acid batteries
  • Learn about the implications of concentration changes on cell potential in electrochemical cells
USEFUL FOR

Chemistry students, electrochemists, and professionals working with lead storage batteries will benefit from this discussion, particularly those focused on battery performance and electrochemical calculations.

cuttooth
Messages
3
Reaction score
0

Homework Statement


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.

Homework Equations



Nerst Equation:

E = E° - 0.0257/n * ln K

Equilibrium Constant:

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]^2First 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!
 
Physics news on Phys.org
My first point is a little nitpicky, but it's important:

The Nernst Equation is...

E = E° - 0.0257/n * ln Q

Where Q is written like an equilibrium constant, except that the reaction is not necessarily at equilibrium.

If a rxn takes place in an aqeous solution, and water is part of that equation, do you normally write the equilibrium constant with water? What would the concentration of water be, and in what way is that related to the concentration of solids in an aqeous rxn?
 

Similar threads

Replies
5
Views
4K
  • · Replies 1 ·
Replies
1
Views
10K
Replies
4
Views
2K
Replies
5
Views
20K