IChO 2011 · Theory problem 6

Thiosulfate leaching of gold

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Extraction of gold using sodium cyanide, a very poisonous chemical, causes environmental problems and gives rise to serious public concern about the use of this so called “cyanide process”. Thiosulfate leaching of gold has been considered as an alternative. In this process, the main reagent is ammonium thiosulfate, \(\ce{(NH4)2S2O3}\), which is relatively non-toxic. Although this process appears to be environmentally benign, the chemistry involved is very complex and needs to be studied thoroughly. The solution used for leaching gold contains \(\ce{S2O3^2-}\), \(\ce{Cu^2+}\), \(\ce{NH3}\), and dissolved \(\ce{O2}\). The solution must have a pH greater than 8.5 to allow free ammonia to be present.

According to the proposed mechanism, a local voltaic micro-cell is formed on the surface of gold particles during the leaching process and operates as follows:

Figure 1

6.1 Write the overall cell reaction for this voltaic cell.

6.2 In the presence of ammonia \(\ce{[Cu(S2O3)3]^5-}\) is oxidized by \(\ce{O2}\) back to \(\ce{[Cu(NH3)4]^2+}\). Write a balanced equation for this oxidation-reduction reaction in basic solution.

6.3 In this leaching process, the \(\ce{[Cu(NH3)4]^2+}\) complex ion functions as catalyst and speeds up the dissolution of gold. Write the net overall oxidation-reduction reaction for dissolution of the gold metal, which is catalyzed by \(\ce{[Cu(NH3)4]^2+}\) complex ion.

6.4 Draw the coordination geometries of the metal in \(\ce{[Au(NH3)2]+}\) and \(\ce{[Au(S2O3)2]^3-}\) complex ions, indicating the coordinating atoms.

The formation constants, \(K_f\), of \(\ce{[Au(NH3)2]+}\) and \(\ce{[Au(S2O3)2]^3-}\) complexes are \(1.00\cdot10^{26}\) and \(1.00\cdot10^{28}\), respectively. Consider a leaching solution in which the equilibrium concentrations of the species are as follows:

\([\ce{S2O3^2-}]=0.100;\quad[\ce{NH3}]=0.100\) and the total concentration of gold(I) species = \(5.50\cdot10^{-5}\ \mathrm{mol\,dm^{-3}}\).

6.5 Calculate the percentage of gold(I) ion that exists in the form of thiosulfate complex.

When the concentration of \(\ce{O2}\) is not high enough and pH > 10, \(\ce{S2O3^2-}\) reduces \(\ce{[Cu(NH3)4]^2+}\) to \(\ce{[Cu(S2O3)3]^5-}\) with the formation of tetrathionate ion \(\ce{S4O6^2-}\):

Figure 2

In basic solution tetrathionate disproportionates to trithionate, \(\ce{S3O6^2-}\), and thiosulfate.

6.6 Write a balanced equation for this disproportionation reaction.

6.7 When the \(\ce{O2}\) concentration is too high, \(\ce{S2O3^2-}\) is oxidized by oxygen to yield trithionate and sulfate ions. Write a balanced equation for this reaction.

Official English text of the problem.

Answers

Names and SMILES for the species used here are collected at the end of the Solution.

  • 6.1 \[ \begin{aligned} &\ce{Au(s) + [Cu(NH3)4]^2+ + 5S2O3^2-}\\ &\qquad\ce{-> [Au(S2O3)2]^3- + [Cu(S2O3)3]^5- + 4NH3} \end{aligned} \]

  • 6.2 \[ \begin{aligned} &\ce{4[Cu(S2O3)3]^5- + 16NH3 + O2 + 2H2O}\\ &\qquad\ce{-> 4[Cu(NH3)4]^2+ + 12S2O3^2- + 4OH-} \end{aligned} \]

  • 6.3 \[ \boxed{\ce{4Au(s) + 8S2O3^2- + O2 + 2H2O -> 4[Au(S2O3)2]^3- + 4OH-}} \]

  • 6.4 Both gold(I) complexes are linear, with coordination number 2 and a donor–gold–donor angle of \(180^\circ\). In \(\ce{[Au(NH3)2]+}\), the donors are nitrogen atoms; in \(\ce{[Au(S2O3)2]^3-}\), they are the terminal sulfur atoms of the thiosulfate ligands.

  • 6.5 \[ \boxed{99.0\%} \]

  • 6.6 \[ \boxed{\ce{4S4O6^2- + 6OH- -> 2S3O6^2- + 5S2O3^2- + 3H2O}} \]

  • 6.7 \[ \boxed{\ce{2S2O3^2- + 2O2 -> S3O6^2- + SO4^2-}} \]

Solution

All dissolved ions and ammonia in the following equations are aqueous; the oxygen is dissolved oxygen.

6.1 — Overall cell reaction

Adding the two given anode steps cancels the diamminegold(I) intermediate and ammonia:

\[ \ce{Au(s) + 2S2O3^2- -> [Au(S2O3)2]^3- + e-}. \]

Gold is oxidized from oxidation state 0 to \(+1\).

Adding the two cathode steps cancels the diamminecopper(I) intermediate:

\[ \ce{[Cu(NH3)4]^2+ + 3S2O3^2- + e- -> [Cu(S2O3)3]^5- + 4NH3}. \]

Copper is reduced from \(+2\) to \(+1\). Adding these net half-reactions cancels one electron:

\[ \begin{aligned} &\ce{Au(s) + [Cu(NH3)4]^2+ + 5S2O3^2-}\\ &\qquad\ce{-> [Au(S2O3)2]^3- + [Cu(S2O3)3]^5- + 4NH3}. \end{aligned} \]

The total charge is \(-8\) on each side.

6.2 — Regeneration of the copper(II) complex

In \(\ce{[Cu(S2O3)3]^5-}\), the three thiosulfate ligands contribute charge \(-6\), so copper has oxidation state \(+1\). In \(\ce{[Cu(NH3)4]^2+}\), ammonia is neutral, so copper has oxidation state \(+2\).

The oxidation half-reaction, including ligand exchange, is therefore

\[ \ce{[Cu(S2O3)3]^5- + 4NH3 -> [Cu(NH3)4]^2+ + 3S2O3^2- + e-}. \]

The oxygen reduction half-reaction in basic solution is

\[ \ce{O2 + 2H2O + 4e- -> 4OH-}. \]

Multiplying the copper half-reaction by four and adding gives

\[ \begin{aligned} &\ce{4[Cu(S2O3)3]^5- + 16NH3 + O2 + 2H2O}\\ &\qquad\ce{-> 4[Cu(NH3)4]^2+ + 12S2O3^2- + 4OH-}. \end{aligned} \]

Both sides have total charge \(-20\).

6.3 — Net dissolution of gold

Multiply the reaction in 6.1 by four and add the regeneration reaction in 6.2. The following cancel:

  • Four tetraamminecopper(II) ions and four tris(thiosulfato)cuprate(I) ions.
  • Sixteen ammonia molecules.
  • Twelve of the twenty thiosulfate ions initially appearing on the reactant side.

The result is

\[ \ce{4Au(s) + 8S2O3^2- + O2 + 2H2O -> 4[Au(S2O3)2]^3- + 4OH-}. \]

Thus, four gold atoms each lose one electron, and one oxygen molecule accepts the four electrons. The copper complex is regenerated, explaining its catalytic role.

6.4 — Coordination geometries

Both complexes contain gold(I), which has a \(5d^{10}\) configuration and characteristically forms linear complexes with two donor atoms.

For diamminegold(I), \(\ce{[Au(NH3)2]+}\), each ammonia ligand donates its nitrogen lone pair.

  • Draw gold at the center of a straight N–Au–N arrangement.
  • Attach three hydrogen atoms to each nitrogen.
  • Mark the N–Au–N angle as \(180^\circ\), and enclose the complex in brackets with overall charge \(+1\).

For bis(thiosulfato)aurate(I), \(\ce{[Au(S2O3)2]^3-}\), each thiosulfate ligand binds through its terminal sulfur atom: the sulfur atom that is not bonded to oxygen. This sulfur donor is favored by the soft gold(I) ion.

  • Draw a straight S–Au–S arrangement with angle \(180^\circ\).
  • Each coordinating sulfur is bonded to a second sulfur bearing three oxygen atoms.
  • Enclose the complete complex in brackets with overall charge \(3-\).

The linear geometry concerns the two donor atoms and gold; it does not require all atoms of the ligands to lie on that line.

6.5 — Percentage present as the thiosulfate complex

The formation equilibria are

\[ \ce{Au+ + 2NH3 <=> [Au(NH3)2]+} \]

and

\[ \ce{Au+ + 2S2O3^2- <=> [Au(S2O3)2]^3-}. \]

Their formation constants give

\[ K_{f,N} =\frac{[\ce{[Au(NH3)2]+}]}{[\ce{Au+}][\ce{NH3}]^2}, \qquad K_{f,T} =\frac{[\ce{[Au(S2O3)2]^3-}]}{[\ce{Au+}][\ce{S2O3^2-}]^2}. \]

Dividing eliminates the free gold(I) concentration:

\[ \frac{[\ce{[Au(S2O3)2]^3-}]}{[\ce{[Au(NH3)2]+}]} = \frac{1.00\times10^{28}}{1.00\times10^{26}} \left(\frac{0.100}{0.100}\right)^2 =100. \]

To check the contribution of uncomplexed gold(I), use the gold mass balance:

\[ C_{\mathrm{Au}} = [\ce{Au+}] +[\ce{[Au(NH3)2]+}] +[\ce{[Au(S2O3)2]^3-}]. \]

Using numerical concentrations in \(\mathrm{mol\,dm^{-3}}\),

\[ C_{\mathrm{Au}} = [\ce{Au+}] \left(1+10^{26}(0.100)^2+10^{28}(0.100)^2\right) = [\ce{Au+}]\left(1+10^{24}+10^{26}\right). \]

Therefore,

\[ [\ce{Au+}] = \frac{5.50\times10^{-5}}{1+10^{24}+10^{26}} = 5.45\times10^{-31}\ \mathrm{mol\,dm^{-3}}. \]

The free-ion contribution is negligible. The complex concentrations are

\[ [\ce{[Au(NH3)2]+}] = 5.45\times10^{-7}\ \mathrm{mol\,dm^{-3}}, \]

\[ [\ce{[Au(S2O3)2]^3-}] = 5.45\times10^{-5}\ \mathrm{mol\,dm^{-3}}. \]

Using the unrounded values, the required percentage is

\[ \frac{10^{26}}{1+10^{24}+10^{26}}\times100\% = 99.0099\ldots\% = \boxed{99.0\%}. \]

6.6 — Disproportionation of tetrathionate

The sums of the sulfur oxidation numbers in tetrathionate, trithionate, and thiosulfate are respectively

\[ \begin{array}{c|c|c} \text{Ion} & \text{Sum for sulfur} & \text{Average for sulfur}\\ \hline \ce{S4O6^2-} & +10 & +2.5\\ \ce{S3O6^2-} & +10 & +10/3\\ \ce{S2O3^2-} & +4 & +2 \end{array} \]

Formation of trithionate increases the average sulfur oxidation state, while formation of thiosulfate decreases it, consistent with disproportionation.

Let \(a\) tetrathionate ions produce \(b\) trithionate ions and \(c\) thiosulfate ions. Sulfur conservation and electron balance require

\[ 4a=3b+2c, \qquad 10a=10b+4c. \]

These give the smallest integer ratio

\[ a:b:c=4:2:5. \]

Balancing oxygen, hydrogen, and charge with hydroxide and water gives

\[ \boxed{\ce{4S4O6^2- + 6OH- -> 2S3O6^2- + 5S2O3^2- + 3H2O}}. \]

There are 16 sulfur atoms, 30 oxygen atoms, and 6 hydrogen atoms on each side; both sides have charge \(-14\).

6.7 — Oxidation of thiosulfate by excess oxygen

Balancing sulfur and charge gives two thiosulfate ions for one trithionate ion and one sulfate ion.

The sulfur oxidation-number sum increases from

\[ 2(+4)=+8 \]

to

\[ (+10)+(+6)=+16. \]

Thus, eight electrons are released. The corresponding oxidation half-reaction in basic solution is

\[ \ce{2S2O3^2- + 8OH- -> S3O6^2- + SO4^2- + 4H2O + 8e-}. \]

Two oxygen molecules accept these electrons:

\[ \ce{2O2 + 4H2O + 8e- -> 8OH-}. \]

Adding and cancelling electrons, hydroxide, and water gives

\[ \boxed{\ce{2S2O3^2- + 2O2 -> S3O6^2- + SO4^2-}}. \]

Both sides contain four sulfur atoms and ten oxygen atoms and have charge \(-4\).

Species names and SMILES

The coordination complexes below use dative-bond SMILES: arrows point from donor atoms toward the metal. These encode connectivity, but the linear gold coordination geometries must be specified separately as in 6.4. The sulfur oxyanions are represented by one localized bonding and charge description.

Species and name SMILES
\(\ce{Au}\), gold atom [Au]
\(\ce{Au+}\), gold(I) ion [Au+]
\(\ce{NH3}\), ammonia N
\(\ce{O2}\), dioxygen O=O
\(\ce{H2O}\), water O
\(\ce{OH-}\), hydroxide ion [OH-]
\(\ce{S2O3^2-}\), thiosulfate ion [O-]S(=O)(=O)[S-]
\(\ce{[Au(NH3)2]+}\), diamminegold(I) ion [NH3]->[Au+]<-[NH3]
\(\ce{[Au(S2O3)2]^3-}\), bis(thiosulfato)aurate(I) ion [Au+](<-[S-]S(=O)(=O)[O-])<-[S-]S(=O)(=O)[O-]
\(\ce{[Cu(NH3)4]^2+}\), tetraamminecopper(II) ion [Cu+2](<-[NH3])(<-[NH3])(<-[NH3])<-[NH3]
\(\ce{[Cu(NH3)2]+}\), diamminecopper(I) ion [NH3]->[Cu+]<-[NH3]
\(\ce{[Cu(S2O3)3]^5-}\), tris(thiosulfato)cuprate(I) ion [Cu+](<-[S-]S(=O)(=O)[O-])(<-[S-]S(=O)(=O)[O-])<-[S-]S(=O)(=O)[O-]
\(\ce{S4O6^2-}\), tetrathionate ion [O-]S(=O)(=O)SSS(=O)(=O)[O-]
\(\ce{S3O6^2-}\), trithionate ion [O-]S(=O)(=O)SS(=O)(=O)[O-]
\(\ce{SO4^2-}\), sulfate ion [O-]S(=O)(=O)[O-]

The gold-atom SMILES does not represent the extended metallic lattice of solid gold.

The solution was checked against the official solution.