[13%] A metallic alloy contains all of the group 11 metals (copper, silver, and gold). Half-reaction — Electrochemistry and Coordination Chemistry Chemistry Question
Problem Context
[13%] A metallic alloy contains all of the group 11 metals (copper, silver, and gold).
Half-reaction E°, V
Cu2+(aq) + 2 e– → Cu(s) +0.34
Ag+(aq) + e– → Ag(s) +0.80
Au+(aq) + e– → Au(s) +1.83
Au3+(aq) + 3 e– → Au(s) +1.52
AuCl4–(aq) + 3 e– → Au(s) + 4 Cl–(aq) +0.93
0.1000 g of this alloy is dissolved in 10 mL of 6 M nitric acid (a large excess), which leaves behind 0.0325 g of unreacted metallic gold. Explain why gold does not dissolve in nitric acid, but does dissolve in aqua regia, which is a mixture of nitric and hydrochloric acids.
Model Answer
The redox potential of Au to form either Au+(aq) or Au3+(aq) is so high that nitric acid is not a strong enough oxidant to allow this reaction. However, chloride ion binds strongly to Au3+, so that in aqua regia the relevant couple is the AuCl4–/Au couple, which has a lower potential.
What is the formation constant Kf for the AuCl4– complex ion?
Model Answer
Au3+(aq) + 3e– → Au(s) ∆G° = –3F(1.52 V)
Au(s) + 4 Cl–(aq) → AuCl4–(aq) + 3e– ∆G° = –3F(–0.93 V)
Au3+(aq) + 4 Cl–(aq) → AuCl4–(aq) ∆G° = –3F(0.59 V)
∆G° = –170.8 kJ mol–1 = –RTln(Kf) = –(8.314 J mol–1 K–1)(298 K)ln(Kf)
Kf = 8.6 × 10^29
To the filtrate from part a. is added aqueous ammonia to give a total volume of 50.0 mL, with the final pH of the solution equal to 4.00. What fraction of the Cu(II) ion in solution is in the form of Cu(NH3)4 2+? The Kf of Cu(NH3)4 2+ is 1.7 × 10^13 and the pKa of NH4+ is 9.25.
Model Answer
Since the nitric acid was employed in large excess over the dissolving metals, the moles of acid present will be close to those initially present, i.e. (6 mol L–1)(0.010 L) = 0.060 mol. Since the final pH = 4, essentially all the acid is neutralized (with little excess ammonia), so [NH4+] = 0.060 mol/0.050 L = 0.12 M. From the pH,
pH = pKa + log10([NH3]/[NH4+])
4.00 = 9.25 + log10([NH3]/[0.12])
[NH3] = 6.7 × 10–6 M
From the complex ion equilibrium,
[Cu(NH3)4 2+]/[Cu2+(aq)] = Kf[NH3]^4
[Cu(NH3)4 2+]/[Cu2+(aq)] = 3.4 × 10–8
Since this ratio is so tiny, it is essentially equal to the fraction of complexed copper.
The solution from part c. is subjected to constant-current electrolysis with a platinum cathode and its potential measured relative to a half-cell with a silver anode in a large volume of 1.000 M NaBr (with some silver bromide present). The electrolysis is carried out with a constant current of 12.0 mA and the potential measured as a function of time to give the graph shown below.
What is the percent by mass of copper and silver in the alloy?
Model Answer
The first equivalence point, where all the Ag+ has been reduced to Ag(s), occurs at 3400 s. This corresponds to (3400 s)(0.0120 C s–1)/(96500 C mol–1) = 4.23 × 10–4 mol Ag = 0.0456 g Ag = 45.6% of the 0.1000 g alloy sample. The percent Cu can be obtained by difference, given the 32.5% Au, as 21.9%. (Using the second equivalence point in the potentiometric titration at 8950 s gives 21.9% Cu as well.)
What is the Ksp of AgBr?
Model Answer
The initial potential, 0.61 V, is essentially a concentration cell between the Ag+ in the analyte (= 4.23 × 10–4 mol/0.0500 L = 8.46 × 10–3 M) and the Ag+ in the AgBr/Br– solution. So 0.61 V = (RT/F)ln([8.46 × 10–3]/[Ag+]), which gives [Ag+] = 4.1 × 10–13 in the AgBr reference cell. Since [Br–] = 1.00 M in this cell, then Ksp = 4.1 × 10–13.