Gold exists in two common positive oxidation states, \(+1\) and \(+3 .\) The standard reduction potentials for these oxidation states are $$ \begin{array}{ll}{\mathrm{Au}^{+}(a q)+\mathrm{e}^{-}} \quad {\longrightarrow \mathrm{Au}(s) \quad E_{\mathrm{red}}^{\circ}=+1.69 \mathrm{V}} \\\ {\mathrm{Au}^{3+}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au}(s)} \quad {E_{\mathrm{red}}^{\circ}=+1.50 \mathrm{V}}\end{array} $$ (a) Can you use these data to explain why gold does not tarnish in the air? ( b) Suggest several substances that should be strong enough oxidizing agents to oxidize gold metal. (c) Miners obtain gold by soaking gold-containing ores in an aqueous solution of sodium cyanide. A very soluble complex ion of gold forms in the aqueous solution because of the redox reaction $$ \begin{array}{rl}{4 \mathrm{Au}(s)+8 \mathrm{NaCN}(a q)+2} & {\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)} \\ {\longrightarrow} & {4 \mathrm{Na}\left[\mathrm{Au}(\mathrm{CN})_{2}\right](a q)+4 \mathrm{NaOH}(a q)}\end{array} $$ What is being oxidized, and what is being reduced in this reaction? (d) Gold miners then react the basic aqueous product solution from part (c) with Zn dust to get gold metal. Write a balanced redox reaction for this process. What is being oxidized, and what is being reduced?

Step by step solution

01

Part (a): Tarnishing of gold

Tarnishing happens when a metal reacts with a non-metal in the environment, forming a new compound on its surface. In the case of gold, the two common positive oxidation states are \(\text{Au}^{+}\) and \(\text{Au}^{3+}\). The standard reduction potentials of these oxidation states are +1.69 V and +1.50 V, respectively. The higher the reduction potential, the less likely the metal will oxidize, or react with other non-metal ions in the environment. Since gold has a high reduction potential, it is less likely to react with other non-metal ions in the air, such as oxygen or sulfur. Therefore, gold does not tarnish in the air.

02

Part (b): Substances that can oxidize gold

In order to oxidize gold, we need a strong oxidizing agent with a potential higher than that of gold. Some of the well-known oxidizing agents are F2 \((E^0_{\mathrm{red}} = +2.87\,\mathrm{V})\), Cl2 \((E^0_{\mathrm{red}} = +1.36\,\mathrm{V})\), \(Br_2 (E^0_{\mathrm{red}} = +1.07\,\mathrm{V})\), but note that only F2's standard reduction potential is higher than that of gold. Hence, F2 will oxidize gold metal.

03

Part (c): Identifying the components being oxidized and reduced

We are given the redox reaction: \[4 \,\text{Au}(s)+8 \,\text{NaCN}(aq)+2 \,\text{H}_{2} \text{O}(l)+\text{O}_{2}(g) \rightarrow 4 \,\text{Na}\left[\text{Au}(\text{CN})_{2}\right](aq) + 4 \,\text{NaOH}(aq)\] Here, gold (\(\text{Au}\)) is being oxidized (loses electrons) from \(\text{Au}(s)\) to \(\text{Au}^{+}\) in the complex ions, and oxygen (\(\text{O}_{2}\)) is being reduced (gains electrons) from \(\text{O}_{2}(g)\) to \(\text{OH}^-\) in \(\text{NaOH}\).

04

Part (d): Balanced redox reaction for obtaining gold metal

The product solution from part (c) reacts with Zn dust to obtain gold metal. The overall balanced redox reaction is: \[2 \,\text{Zn}(s) + 4 \,\text{Na}\left[\text{Au}(\text{CN})_{2}\right](aq) \rightarrow 2 \,\text{Zn}(\text{CN})_{2}(aq) + 4 \,\text{NaCN}(aq) + 4 \,\text{Au}(s)\] In this reaction, zinc (\(\text{Zn}\)) is being oxidized (loses electrons) from \(\text{Zn}(s)\) to \(\text{Zn}(\text{CN})^2_{2}\), and gold (\(\text{Au}\)) is being reduced (gains electrons) from \(\text{Au}^{+}\) in the complex ions to \(\text{Au}(s)\).

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