Question

The cell reaction for the lead-acid battery is as shown below:

PbO₂ (s) + Pb (s) + 2 H₂SO₄ (aq) -> 2 PbSO₄ (s) + 2 H₂O (l)

            What will E_cell be for this reaction when [H₂SO₄] = 1.5 M?

Answer 1

The reaction taking place in this type of cell can be represented as:

At anode : Pb + SO₄^2- -----> PbSO₄ +2e- E⁰_oxidation = -0.36V

At cathode: PbO2 + SO₄^2- + 4H⁺ + 2e-   ----> PbSO₄ + 2H2O    E⁰_reducation = 1.69V

Net reaction : PbO2 (s) + Pb (s) + 2 H2SO4 (aq) ----> 2 PbSO4 (s) + 2 H2O (l)

E⁰_cell = E⁰_cathode - E⁰_anode = 1.69-(-0.36) = 2.05V

We can find the Ecell of the reaction using nernst equation

Ecell = E⁰_cell - 0.059/n(logKc)

Kc for the above reaction = 1/[H2SO4]² since all the other reactants and products are pure solid or liquids and their concentration will be taken as one and do not appear in Kc expression.

n = 2 (transfer of electron)

E⁰_cell = 2.05V

substitude all the values in equation Ecell = E⁰_cell - 0.059/n(logKc)

Ecell = 2.05V - 0.059/2log(1/[H₂SO₄]²)

[H₂SO₄​] = 1.5M (given in the question)

Ecell = 2.05V - 0.059/2log(1/[1.5]²)

on solving the ablove equation we get Ecell = 2.09V