A cell is set up with copper and lead electrodes in contact with CuSO 4 (aq, 1.0

Question

A cell is set up with copper and lead electrodes in contact with CuSO4(aq, 1.0 M) and Pb(NO3)2(aq, 1.0 M), respectively, at 25C. The standard cell potential is 0.47 V for the reaction:

Cu2+ + Pb --> Cu + Pb2+

If solid sodium sulfite is added to the CuSO4 solution, a precipitate of CuSO3will form.

A. Describe how the concentrations of copper (II) ion and lead (II) ion would change. Assume no change in the volume of solution after adding sodium sulfate.

B. Using Nernst equation, predict how the cell potential will change. Consider your answer for the previous question.

If possible, please provide a detailed solution to accompany your answer.

Explanation / Answer

A. Describe how the concentrations of copper (II) ion and lead (II) ion would change. Assume no change in the volume of solution after adding sodium sulfate.

Solution :- When the sodium sulfate Na2SO4 is added it forms the precipitate of the CuSO3

Means the concentration of the free Cu^2+ ions will decrease therefore the concentration of the Pb^2+ will also decrease because Pb^2+ is formed when the Pb reduces the Cu^2+

So both Cu^2+ and Pb^2+ ion concentration will decrease on the addition of the sodium sulfate

B. Using Nernst equation, predict how the cell potential will change. Consider your answer for the previous question.

Solution :-

Cell potential changes with the change in the concentration of the electrolyte solutions , so when the concentrations of the Cu^2+ and Pb^2+ decreases that means the cell potential will also decrease.

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