[12%] The standard reduction potentials of some compounds of platinum and silver at 298 K are given — Electrochemistry Chemistry Question
Problem Context
[12%] The standard reduction potentials of some compounds of platinum and silver at 298 K are given below.
Half-reaction E°, V
Pt2+(aq) + 2e– → Pt(s) +1.188
PtCl4 2-(aq) + 2e– → Pt(s) + 4 Cl–(aq) +0.758
PtCl6 2-(aq) + 2e– → PtCl4 2-(aq) + 2 Cl–(aq) +0.726
Ag+(aq) + e– → Ag(s) +0.799
What is Kf for the PtCl4 2- ion?
Model Answer
Pt2+(aq) + 2e– ® Pt(s) ∆G° = –2F(1.188 V)
Pt(s) + 4 Cl–(aq) ® PtCl4 2-(aq) + 2e– ∆G° = –2F(–0.758 V)
Pt2+(aq) + 4 Cl–(aq) ® PtCl4 2-(aq) ∆G° = –2F(1.188 – 0.758 V)
= –83.0 kJ mol-1
Kf = e–∆G°/RT = e–(–83000 J mol-1)/(8.314 J mol-1 K-1)(298 K)
Kf = 3.5 ´ 1014
What is E° for the reduction of hexachloroplatinate(IV) to give platinum metal?
PtCl6 2-(aq) + 4e– → Pt(s) + 6 Cl–(aq)
Model Answer
PtCl6 2-(aq) + 2e– ® PtCl4 2-(aq) + 2 Cl–(aq) ∆G° = –2F(0.726 V)
PtCl4 2-(aq) + 2e– ® Pt(s) + 4 Cl–(aq) ∆G° = –2F(0.758 V)
PtCl6 2-(aq) + 4e– ® Pt(s) + 6 Cl–(aq) ∆G° = –2F(0.726 + 0.758 V)
For the summed half-reaction, ∆G° = –4FE°. So E° = (0.726 + 0.758 V)/2 = 0.742 V.
An electrolytic cell is set up as shown below.
What is the minimum potential that would need to be applied in order to cause the platinum electrode to dissolve to form PtCl4 2-(aq) and the Ag+(aq) to deposit on the Ag electrode?
Model Answer
The net reaction of the electrolysis is:
2 Ag+(aq) + Pt(s) + 4 Cl–(aq) ® 2 Ag(s) + PtCl4 2-(aq)
For this reaction, E° = 0.799 V – 0.758 V = 0.041 V
Under the nonstandard conditions of the electrolysis setup:
E = E° – (RT/2F) ln ([PtCl4 2-] / ([Ag+]^2[Cl-]^4))
E = 0.041 V – 0.01284 V•ln([0.001] / [0.1]^2[0.1]^4)
E = –0.048 V
Thus, at least 0.048 V of potential must be applied for electrons to flow spontaneously to oxidize the Pt and reduce the Ag+.
As electrolysis is carried out, is it thermodynamically possible for a significant amount of hexachloroplatinate(IV) to be produced at the anode? Justify your answer.
Model Answer
There are several ways to analyze this situation. One is to consider the electrolytic oxidation to form PtCl6 2-:
4 Ag+(aq) + Pt(s) + 6 Cl–(aq) ® 4 Ag(s) + PtCl6 2-(aq)
E° = 0.799 V – 0.742 V = 0.057 V
This reaction is thus even more favorable under standard conditions than formation of PtCl4 2-!
Under these nonstandard conditions,
E = E° – (RT/4F) ln([PtCl6 2-]/([Ag+]^4[Cl-]^6))
E = 0.057 V – (0.00642 V) ln([PtCl6 2-]/([0.1]^4[0.1]^6))
E = –0.091 V – (0.00642 V)ln([PtCl6 2-])
Under the conditions of the electrolysis, this will be spontaneous if E > –0.048 V, which corresponds to [PtCl6 2-] < 1.3 ´ 10-3 M. So in fact PtCl6 2- will (thermodynamically) be produced exclusively at the beginning of the electrolysis and will continue to be produced until its concentration is quite similar to that of the initial concentration of PtCl4 2- (there will be some slight effect of the changing Cl– and Ag+ concentrations). This is definitely significant! (There may be kinetic factors at play that would favor one form over the other, but that is more difficult to predict, and the question specifically asked about the thermodynamic possibilities.)