Nerst Equation (Electrochemistry)

[cuttooth]

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!

 

[Barfolumu]

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?

 

[Blueshift5]

The molarity of H2O is not required for the calculation of the equilibrium constant K. The concentration of H2O does not change significantly in the reaction and can be considered constant. Therefore, it does not affect the value of K and can be omitted in the calculation.

In this case, the concentration of H+ and HSO4- is given as 8.1 M, which means that both ions are present at a concentration of 8.1 M in the solution. This can be interpreted as a combined solution with a total concentration of 8.1 M for both H+ and HSO4-.

To calculate E, we can first calculate the equilibrium constant K using the given concentrations:

K = [H2O]^2 / [H+]^2[H2SO4-]^2 = [1]^2 / [8.1]^2[8.1]^2 = 1/ 8.1^4

Then, we can plug in this value of K into the Nernst equation to calculate E:

E = 2.04 V - 0.0257/2 * ln K = 2.04 V - 0.0257/2 * ln (1/8.1^4) = 2.04 V + 0.0257/2 * 4 * ln 8.1

E = 2.04 V + 0.0257 * ln 8.1 = 2.04 V + 0.0257 * 2.091 = 2.095 V

Therefore, the value of E for this battery at a concentration of 8.1 M for both H+ and HSO4- is 2.095 V.