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Calculating the [PB2+] in a voltaic Pb-PbI2 cell using the recorded cell potential

a concentration cell was made by putting 5 mL of 0.050 M Pb(NO3)2 solution into one half-cell in a test plate. for the neighboring well in the test plate, weplaced 5 mL of a PbI2 solution that was made by mixing 9 mL of 0.050 KI solution with 3 mL of 0.050 M Pb(NO3)2 solution inside. Pb Electrodes were placed in eachcell, and a KNO3-soaked string was used as the salt bridge.

data was collected and we recorded the average cell potential t be 0.02751 V

The question asks us to:

A .Use the cell potential for the Pb-PbI2 cell and the known [Pb2+] to calculate the [Pb2+] in
equilibrium with PbI2.

Can anyone help? thank you

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Answer #1

Pb - PbI2 cell

oxidation reaction : Pb(s) --->Pb+2 (sat. with PbI2) + 2e-

Reduction reaction : Pb+2 + 2e- ----> Pb(s)

Overall reaction :

Pb+2 -----> Pb+2 (sat)

Ecell = Eo - 0.0592/n log Q

Q = [Pb+2 (sat)] / [Pb+2]

Eo = 0

n = 2 electrons

0.238 = -0.0592/2 * log [Pb+2(sat)] / [0.13]

[Pb+2(sat)] = 1.184*10^-9 M

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