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, 1.0M) and Pb(NO3)2(aq, 1.0M), respectively, at 25 degrees celsius. The standard cell potential is 0.47 for the reaction: Cu2+ + Pb --> Cu + Pb2+. If solid sodium sulfite is added to the CuSO4 solution, a precipitate of CuSO3 will 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.

Please provide a valid and detailed solution to each question.

Explanation / Answer

A)

Cu2+ + Pb --------------------------> Cu + Pb2+

SO3^-2 + Cu+2 ---------------------> CuSO3(s)

if Na2SO3 added Cu+2 concentration decreases . if Cu+2 concentration decreases in the reaction above the Pb reaction with Cu+2 also decreases that means production of Pb+2 concentration also decreases.

because there is no oxidation with out out reduction. if reduction decreases oxidation also decreases.

B)

E = Eocell - 0.0591 / 2 * log { [Pb+2] / [Cu+2] }

here n = 2   electrons transfered

E = 0.47 - 0.0591 / 2 * log { [Pb+2] / [Cu+2] }

Pb+2 and Cu+2 concentration decreases same amount . so cell potentail also decreases form the above equation.

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