Nanometer–sized metal clusters have different properties than the bulk materials. To investigate the — Physical Chemistry — Kinetics Chemistry Question
Metal Nanoclusters
Nanometer–sized metal clusters have different properties than the bulk materials. To investigate the electrochemical behaviour of silver nanoclusters, the following electrochemical cells are considered:
(on the right–hand side: half–cell with the higher potential)
(I) Ag(s)/ AgCl (saturated) // Ag+ (aq, c = 0.01 mol dm–3)/ Ag(s) U1 = 0.170 V
(II) Pt/ Agn(s, nanoclusters), Ag+ (aq, c = 0.01 mol dm–3) // AgCl(saturated)/ Ag(s)
a) U2 = 0.430 V for Ag10 nanoclusters
b) U3 = 1.030 V for Ag5 nanoclusters
Ag5– and Ag10–nanoclusters consist of metallic silver but nevertheless have standard potentials different from the potential of metallic bulk silver.
Eo (Ag /Ag+) = 0.800 V
Eo (Cu/ Cu2+) = 0.345 V
T = 298.15 K
Calculate the solubility product of AgCl.
Model Answer
The potential of a half–cell is described by the Nernst equation:
E = E0 + (RT / nF) × ln(c(ox)/c(red))
The total voltage is: U = E(cathode) – E(anode)
E = E0 + (RT / F) × ln c(Ag+)
U1 = E2 – E1 and U1 = (RT / F) × ln (c2(Ag+) / c1(Ag+))
and with U1 = 0.170 V, c2(Ag+) = 0.01 mol dm–3, c1(Ag+) = x mol dm–3
0.170 V = (8.314 × 298.15 / 96485) V × ln (0.01 / c1(Ag+))
c1(Ag+) = 1.337 × 10–5 mol dm–3
In the saturated solution c(Ag+) = c(Cl–) = 1.337×10–5 mol dm–3 and thus,
Ksp = (1.337×10–5)2
Ksp = 1.788×10–10
Calculate the standard potentials of the Ag5 and Ag10 nanoclusters.
Model Answer
For the right cell of (II):
E(AgCl) = 0.8 V + (RT / F) × ln 1.337×10–5
E(AgCl) = 0.512 V
Thus: U = E(AgCl) – E(Agn, Ag+)
and E(Agn/ Ag+) = E0(Agn/ Ag+) + (RT / F) × ln 0.01
Ag10: E(Ag10/ Ag+) = 0.512 V – 0.430 V = 0.082 V
E0(Ag10/ Ag+) = 0.082 V – (RT / F) × ln 0.01
E0(Ag10/ Ag+) = 0.200 V
Ag5: E(Ag5/ Ag+) = 0.512 V – 1.030V = – 0.518 V
E0(Ag5/ Ag+) = – 0.518 V – (RT / F) × ln 0.01
E0(Ag5/ Ag+) = – 0.400 V
Explain the change in standard potential of silver nanoclusters with particle sizes ranging from very small clusters to bulk silver.
Model Answer
The standard potential increases with increasing particle size until it reaches the bulk value at a certain particle size.
The potential is lower for smaller particles, because they have a larger surface and the process of crystallization is energetically less favourable for the surface atoms. Thus, the free energy of formation of metallic silver is larger (less negative) for smaller particles, i.e. the standard potential is lower. The effect decreases with increasing particle size due to the decreasing relative amount of surface atoms.
Additional remark: However, the potential does not continuously increase with increasing size. The electrochemical potentials of some small clusters of a certain size are much higher. This is due to complete shells of these clusters (clusters consist of a “magic number” of atoms) which make them more stable.
(Instead of the crystallization energy you can also argue with the sublimation energies of silver atoms.)
What happens:
i. if you put the Ag10 clusters and – in a second experiment – the Ag5 clusters into an aqueous solution of pH = 13?
ii. the Ag10 clusters and – in a second experiment – the Ag5 into an aqueous solution of pH = 5
iii. both clusters together into an aqueous solution having a pH of 7 with c(Cu2+) = 0.001 mol dm–3 and c(Ag+) = 1·10–10 mol dm–3 ? Calculate. What happens if the reaction proceeds (qualitatively)?
Model Answer
i) For a solution with a pH of 13:
E(H2/2 H+) = (RT / F) × ln(1×10–7) E(H2/2 H+) = –0.769 V
As an estimate, this potential can be compared with the standard potentials of the silver clusters calculated in 12.2. Both are higher than the standard potential of hydrogen. Thus, the silver clusters behave as noble metals and are not oxidized in this solution. No reaction takes place.
Quantitatively, a small amount of silver is oxidized into Ag+ ions until equilibrium is reached and E(Agn/ Ag+) = E(H2/2 H+).
E0(Agn/ Ag+) + (RT / F) × ln c(Ag+) = – 0.769 V
for Ag10: c(Ag+) = 4.17×10–17 mol dm–3
for Ag5: c(Ag+) = 5.78×10–7 mol dm–3
ii) For a solution with a pH of 5:
E(H2/2 H+) = (RT / F) × ln(1×10–2) E(H2/2 H+) = –0.269 V
As an estimate, the standard potential of the Ag10 clusters is higher than the standard potential of the hydrogen. No reaction takes place. The standard potential of the Ag5 clusters is lower than the standard potential of hydrogen. Thus, hydronium ions will be reduced to hydrogen while Ag5 clusters (metallic silver) are oxidized into silver ions: The Ag–clusters dissolve.
Quantitatively, equilibrium is reached for Ag10 at: c(Ag+) = 4.16×10–9 mol dm–3
and for Ag5 at : c(Ag+) = 57.29 mol dm–3 (which will probably not be reached in a diluted solution and all nanoclusters dissolve)
(After some time, silver ions that are present in the solution can also be reduced to metallic bulk silver. Under this condition, this reduction will preferably take place, because the electrochemical potential is even higher than that of the hydronium–ion reduction.)
iii) Potentials of all possible reactions are considered:
1. E(Cu/ Cu2+) = 0.345 V + 0.5 × (RT / F) × ln 0.001 = 0.256 V
2. E(Ag/ Ag+) = 0.800 V + (RT / F) × ln 1·10–10 = 0.208 V
3. E(Ag10/ Ag+) = 0.200 V + (RT / F) × ln 1·10–10 = –0.392 V
4. E(Ag5/ Ag+) = –0.400 V + (RT / F) × ln 1·10–10 = – 0.992 V
5. E(H2/2 H+) = (RT / F) × ln 1·10–7 = – 0.414 V
The reduction with the highest potential and the oxidation with the lowest potential will preferably take place: Copper(II) ions will be reduced into metallic copper while Ag5 clusters dissolve and form silver(I) ions.
After some time, the silver concentration of the solution increases, Ag5 clusters are used up and the concentration of copper ions decreases. Since the latter is comparably high, it is expected to have minor influence. The next possible steps of the reaction are the following:
After the Ag5 clusters are used up, Ag10 clusters will start to be oxidized. (Note that if a hydrogen electrode was present, H2 would be oxidized. In this system, however, there are protons instead of H2).
After the increase of the silver ion concentration, the potential of the silver ion reduction (into metallic bulk silver) increases, so that it might exceed the potential of the copper reduction. Afterwards, the silver ions will be reduced to metallic silver (after further dissolution of silver nanoclusters).