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.

The two nanoclusters Ag5 and Ag10-nanoclusters have standard potentials different from the potential of metallic bulk silver.

c. Calculate the standard potentials of Ag5 and of Ag10 nanoclusters. [for this part use Ksp(AgCl)=1.800·10-5;]

d. What happens, if you put the Ag10 nanoclusters and – in a second experiment – the Ag5 nanoclusters into an aqueous solution of pH=5? Estimate the consequences using the reduction potentials you calculated.

Answer 1

c)

To calculate the standard potentials of Ag5 and of Ag10 nanoclusters use the right cell of (II):

E(AgCl) = 0.8 V + R∙T∙F^-1 ∙ln((Ag+ *Cl-) ^1/2

E(AgCl) = 0.8 V + R∙T∙F^-1 ∙ln(1.80*10^-10)^1/2

E(AgCl) = 0.512 V

Es ist U = E(AgCl) - E(Agn. Ag+) and

E(Agn/ Ag+) = E0(Agn/Ag+) + R∙T∙F-1∙ln(0.01)

here, Ag+(aq, 0.01M)

For Ag10 nanoclusters:

here; E = +0.430 V for Ag10 nanocluster, then

Ag10: E(Ag10/Ag+) = 0.512 V - 0.430 V = 0.082 V

E0(Ag10/Ag+) = 0.082 V - R∙T∙F-1 ∙ln 0.01

E0(Ag10/Ag+)= 0.200 V

For Ag5 nanoclusters:

here; E = +1.030 for Ag5 nanocluster; then

Ag5: E(Ag5/Ag+) = 0.512 V - 1.030V = - 0.518 V

E0(Ag5/Ag+)) = - 0.518 V - R∙T∙F-1 ∙ln 0.01

E0(Ag5/Ag+)= -0.400 V

Hence the standard potentials of Ag5 and of Ag10 nanoclusters is E0(Ag5/Ag+)= -0.400 V and E0(Ag10/Ag+)= 0.200 V.

d)

first calculate the E (H2/2H+) with pH=5 as follows:

E (H2/2H+) = R∙T∙F-1∙ln (10-5)           

E (H2/2H+) = - 0.269 V

[E0(Ag10/Ag+)= 0.200 V > E(H2/2H+) = - 0.269 V]

[E(H2/2H+) = - 0.269 V] > E0(Ag5/Ag+))=-0.400 V]

With Ag10 clusters, no reaction takes place because

E0(Ag10/Ag+) of Ag10 clusters is greater than the standard potential of hydrogen.

[E(H2/2H+) = - 0.269 V] > E0(Ag5/Ag+))=-0.400 V]

while With Ag5 clusters, reaction takes place because

E0(Ag5/Ag+) of Ag5 clusters is less than the standard potential of hydrogen. here hydronium ions are reduced to hydrogen .