Question

Use standard reduction potentials to calculate \(\mathscr{C}^{\circ}, \Delta G^{\circ},\) and \(K\) (at 298 K) for the reaction that is used in production of gold: $$2 \mathrm{Au}(\mathrm{CN})_{2}-(a q)+\mathrm{Zn}(s) \longrightarrow 2 \mathrm{Au}(s)+\mathrm{Zn}(\mathrm{CN})_{4}^{2-}(a q)$$ The relevant half-reactions are $$\begin{aligned} \operatorname{Au}(\mathrm{CN})_{2}^{-}+\mathrm{e}^{-} \longrightarrow \mathrm{Au}+2 \mathrm{CN}^{-} & \mathscr{C}^{\circ}=-0.60 \mathrm{V} \\ \mathrm{Zn}(\mathrm{CN})_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn}+4 \mathrm{CN}^{-} & \mathscr{C}^{\circ}=-1.26 \mathrm{V} \end{aligned}$$

Short answer

In summary, for the redox reaction $$2\;\mathrm{Au}(\mathrm{CN})_{2}^{-}+\mathrm{Zn}(s)\longrightarrow 2\;\mathrm{Au}(s)+\mathrm{Zn}(\mathrm{CN})_{4}^{2-},$$ the standard cell potential \(\mathscr{C}^{\circ}\) is 0.66 V, the standard free energy change \(\Delta G^{\circ}\) is -127,640 J/mol, and the equilibrium constant \(K\) at 298 K is approximately \(5.90\times10^{20}\).

Step-by-step solution

Determine the balanced equation for the redox reaction

We are provided with the balanced redox reaction already: $$2\;\mathrm{Au}(\mathrm{CN})_{2}^{-}+\mathrm{Zn}(s)\longrightarrow 2\;\mathrm{Au}(s)+\mathrm{Zn}(\mathrm{CN})_{4}^{2-}$$

Calculate the standard cell potential (\(\mathscr{C}^{\circ}\))

First, let's identify the two half-reactions involved. The given reduction half-reactions are: $$\mathrm{Au}(\mathrm{CN})_{2}^{-}+\mathrm{e}^{-}\longrightarrow \mathrm{Au}+2\;\mathrm{CN}^{-}\;\;\;\;\mathscr{C}^{\circ}=-0.60\;\mathrm{V}$$ $$\mathrm{Zn}(\mathrm{CN})_{4}^{2-}+2\;\mathrm{e}^{-}\longrightarrow \mathrm{Zn}+4\;\mathrm{CN}^{-}\;\;\;\;\mathscr{C}^{\circ}=-1.26\;\mathrm{V}$$ For the second equation to match the overall reaction, it must be reversed and turned into an oxidation half-reaction: $$\mathrm{Zn}+4\;\mathrm{CN}^{-} \longrightarrow\mathrm{Zn}(\mathrm{CN})_{4}^{2-}+2\;\mathrm{e}^{-}\;\;\;\;\mathscr{C}^{\circ}=1.26\;\mathrm{V}$$ Now, we can find the standard cell potential by adding the standard reduction potentials of the two half-reactions: $$\mathscr{C}^{\circ}_{\text{cell}}= \mathscr{C}^{\circ}_{\text{red}}+\mathscr{C}^{\circ}_{\text{ox}}= -0.60\;\mathrm{V}+1.26\;\mathrm{V}=0.66\;\mathrm{V}$$

Calculate the standard free energy change (\(\Delta G^{\circ}\))

We can calculate the standard free energy change using the standard cell potential: $$\Delta G^{\circ}=-nFE_{\text{cell}}^{\circ}$$ F is the Faraday constant \((96,\!485\;\mathrm{J}\cdot\mathrm{mol}^{-1}\cdot\mathrm{V}^{-1})\), n is the number of transferred electrons (2 in this case), and \(E_{\text{cell}}^{\circ}\, (= \mathscr{C}_{\text{cell}}^{\circ})\) is the standard cell potential. $$\Delta G^{\circ}=-(2\;\text{mol}\;\mathrm{e}^{-})(96,\!485\;\mathrm{J}\cdot\mathrm{mol}^{-1}\cdot\mathrm{V}^{-1})(0.66\;\mathrm{V})=-127,\!640\;\mathrm{J}\cdot\mathrm{mol}^{-1}$$

Calculate the equilibrium constant (\(K\))

Finally, we can calculate the equilibrium constant using the following relationship: $$\Delta G^{\circ}=-RT\ln K$$ R is the gas constant \((8.314\;\mathrm{J}\cdot\mathrm{mol}^{-1}\cdot\mathrm{K}^{-1})\), and T is the temperature (298 K). $$K = \exp{\left(-\frac{\Delta G^{\circ}}{RT}\right)}= \exp{\left(-\frac{-127,\!640\;\mathrm{J}\cdot\mathrm{mol}^{-1}}{(8.314\;\mathrm{J}\cdot\mathrm{mol}^{-1}\cdot\mathrm{K}^{-1})(298\;\mathrm{K})}\right)}\approx 5.90\times10^{20}$$ In conclusion, for this redox reaction, the standard cell potential is 0.66 V, the standard free energy change is -127,640 J/mol, and the equilibrium constant is approximately \(5.90\times10^{20}\).

Key concepts

Redox Reactions

Redox reactions are fundamental processes in chemistry where one species gains electrons while another loses them. They stand for reduction-oxidation reactions. In our example of gold production:

• The reduction half-reaction: \(\mathrm{Au}(\mathrm{CN})_{2}^{-}+\mathrm{e}^{-}\longrightarrow \mathrm{Au}+2\mathrm{CN}^{-}\) depicts gold ions gaining electrons to form gold metal.
• The oxidation half-reaction: \(\mathrm{Zn}+4\mathrm{CN}^{-} \longrightarrow\mathrm{Zn}(\mathrm{CN})_{4}^{2-}+2\mathrm{e}^- \) illustrates zinc metal losing electrons to form zinc cyanide complex ions.

These reactions are crucial in refining and extracting metals, acting as elegant "electron bookkeeping" procedures. The total reaction combines these half-reactions and ensures that the number of electrons lost equals those gained, keeping the charge balanced.
Remember, the driving force of a redox reaction is the movement of electrons from the donor (oxidized element) to the acceptor (reduced element). Always balance electrons gained and lost to ensure the reaction is properly accounted for.

Standard Cell Potential

The standard cell potential \(\mathscr{C}^{\circ}\) measures the voltage produced by a redox reaction under standard conditions (all solutes at 1 M concentration, gases at 1 atm, and a temperature of 298 K). This potential tells us how energetically favorable a reaction is.

Calculating Standard Cell Potential:

In our gold extraction reaction, we start with two half-reactions:

• Reduction of gold: \(\mathscr{C}^{\circ}_{\text{red}}=-0.60\,\mathrm{V}\)
• Oxidation of zinc: \(\mathscr{C}^{\circ}_{\text{ox}}=1.26\,\mathrm{V}\)

The standard cell potential is obtained by adding these two potentials:\[\mathscr{C}^{\circ}_{\text{cell}}=\mathscr{C}^{\circ}_{\text{red}}+\mathscr{C}^{\circ}_{\text{ox}}= -0.60\,\mathrm{V} + 1.26\,\mathrm{V} = 0.66\,\mathrm{V}\]This positive value indicates a spontaneous reaction, meaning it occurs naturally without any extra energy supplied.

Free Energy Change

The change in free energy \(\Delta G^{\circ}\) gives insight into the spontaneity and energy dynamics of a chemical reaction. When \(\Delta G^{\circ}\) is negative, a reaction is spontaneous, releasing energy. Calculations involve the following relationship:\[\Delta G^{\circ} = -nFE_{\text{cell}}^{\circ}\]

Components of Calculation:

• n: Number of moles of electrons transferred (2 in this exercise).
• F: Faraday constant \(96,485\,\mathrm{J\cdot mol^{-1}\cdot V^{-1}}\).
• E: Standard cell potential, already calculated as \(0.66\,\mathrm{V}\).

By placing these values into the formula, \(\Delta G^{\circ}= -(2)(96,485)(0.66)\approx-127,640\,\mathrm{J\cdot mol^{-1}}\).
This negative value confirms spontaneity, where energy is released as the reaction proceeds.

Equilibrium Constant

The equilibrium constant \(K\) quantifies the ratio of product concentrations to reactant concentrations at equilibrium. A very large \(K\) suggests the formation of products is highly favored, indicating a virtually complete reaction.

Relationship to Free Energy:

The free energy change and equilibrium constant are connected through:\[\Delta G^{\circ} = -RT \ln K\]where:

• R: The gas constant \(8.314\,\mathrm{J\cdot mol^{-1}\cdot K^{-1}}\).
• T: Absolute temperature in Kelvin (298 K in this example).

Rearranging for \(K\), we find:\[K = \exp\left(-\frac{\Delta G^{\circ}}{RT}\right)\]Substitute the determined values: \(K = \expÂ\left(-\frac{-127,640}{(8.314)(298)}\right) \approx 5.90\times10^{20}\).
This extremely high value underscores the reaction's shift towards forming products under given conditions.