A cell is set up with copper and lead electrodes in contact with CuSO4(aq) and Pb(NO3)2(aq), respectively, at 25°C. The standard reduction potentials are Pb2+ + 2e–? Pb E° = –0.13 V Cu2+ + 2e–? Cu E° = +0.34 V

If sulfuric acid is added to the Pb(NO3)2 solution, forming a precipitate of PbSO4, the cell potential

a. It is impossible to tell what will happen.

b. decreases.

c. is unchanged.

d. increases.

Now look at the matter closely. Given the values of electrode potential stated in the question, one can see that lead will function as the anode and copper as the cathode since lead has a more negative electrode potential.

This implies that at the anode, the half reaction going on is this;

Pb(s) -------> Pb^2+(aq) + 2e.

There will be a build up of Pb^2+ in the anode compartment. The addition of H2SO4 and formation of PbSO4 favours the removal of the Pb^2+ ions;

Pb^2+(aq) + SO4^2-(aq) -------> PbSO4(s)

As this continues, Pb^2+ concentration begins to decrease, in order to maintain equilibrium, more Pb^2+ is formed thereby increasing the current and voltage flowing in the cell as more electrons are transferred from anode to cathode(more current flows).

andromache andromache · Answer 2 · 2022-03-23T07:44:29+01:00

In the case when the sulfuric acid is added to the Pb(NO3)2 solution increases so the cell potential is increased.

Impact on cell potential:

Here the anode and copper treated as the cathode because it contains more negative electrode potential. Due to this, it should be half reaction

Pb(s) -------> Pb^2+(aq) + 2e.

Also, there should be the addition of H2SO4 and creation of the PbSO4 that eliminates the Pb^2+ ions.

So,

Pb^2+(aq) + SO4^2-(aq) -------> PbSO4(s)

Here the concentration of Pb^2+ should reduced for maintaining the equilibrium.

Learn more about cell here: https://brainly.com/question/21038516