- Consider a concentration cell. Two Ag-electrodes are immersed in AgNO3 solutio

Question

- Consider a concentration cell. Two Ag-electrodes are immersed in AgNO3 solutions of different concentrations. When the two compartments have an AgNO3-concentration of 1 M and 0.1 M, respectively, the measured voltage is 0.065 V (note: T in not necessarily = 25°C ).
The electrochemical behavior of silver nanoclusters (Agn, with n the number of Ag atoms in the cluster) is investigated using the following electrochemical cells at 298 K:

I. Ag(s) | AgCl (saturated) || Ag+(aq, 0.01M) | Ag(s), E=0.170

II. (Pt electrode) Agn (s, nanocluster) | Ag+(aq, 0.01M) || AgCl (saturated) | Ag(s), with E = +1.030 V for Ag5 nanocluster and E = +0.430 V for Ag10 nanocluster
The standard reduction potential for Ag+ + e- Ag, is E0 = +0.800 V

Use this data to calculate the solubility product of AgCl..

Explanation / Answer

Using the Nernst equation,

Eo = Eocell - RT/nF logKsp

Given,

Ag+ + e- ---> Ag(s) ....0.80 V

I. Ag+(0.01M) + Cl- ----> AgCl(s) ....0.170 V

II. Agn + AgCl(s) -----> Ag+(0.01M) + Cl- + Ag(s) .....1.030 V for Ag5 and 0.430 V for Ag10

Now, Ksp AgCl = [Ag+][Cl-]

AgCl <===> Ag+ + Cl-

logKsp = -(1.030-0.430-0.170)/0.059

Ksp = 5.15 x 10^-8

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