Question

Use standard reduction potentials to calculate \(\mathscr{E}^{\circ}, \Delta G^{\circ}\), and \(K\) (at \(298 \mathrm{~K}\) ) 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 $$\begin{aligned}\mathrm{Au}(\mathrm{CN})_{2}^{-}+\mathrm{e}^{-} \longrightarrow \mathrm{Au}+2 \mathrm{CN}^{-} & \mathscr{C}^{\circ} &=-0.60 \mathrm{~V} \\ \mathrm{Zn}(\mathrm{CN})_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn}+4 \mathrm{CN}^{-} & \mathscr{E}^{\circ} &=-1.26 \mathrm{~V} \end{aligned}$$

Short answer

In conclusion, we have found that the standard cell potential for the gold production reaction is 0.66 V, the standard Gibbs free energy change is -127.6 kJ/mol, and the equilibrium constant is \(1.49\times10^{17}\).

Step-by-step solution

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

Since we are given the standard reduction potentials of both half-reactions, we can calculate the standard cell potential by subtracting the reduction potential of the half-reaction that needs to be reversed (oxidation) from the reduction potential of the reaction that will remain as it is (reduction). $$\mathscr{E}^{\circ}_{cell} = \mathscr{E}^{\circ}_{red} - \mathscr{E}^{\circ}_{ox}$$ We need to reverse the second half-reaction because Zn needs to be oxidized, not reduced. Therefore, we will keep the first half-reaction's standard reduction potential as is and reverse the second half-reaction's standard reduction potential. $$\mathscr{E}^{\circ}_{cell} = (-0.60\,\text{V}) - (-1.26\,\text{V}) = 0.66\,\text{V}$$

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

To calculate the standard Gibbs free energy change, we can use the following formula: $$\Delta G^{\circ}=-nF\mathscr{E}^{\circ}$$ where \(n\) is the total number of electrons transferred in the balanced redox reaction \(F\) is the Faraday constant, equal to \(96485\frac{\text{C}}{\text{mol}}\) \(\mathscr{E}^{\circ}\) is the standard cell potential For our reaction, \(n = 2\) for transferring two electrons. $$\Delta G^{\circ} = - (2\,\text{mol} \cdot 96485\,\frac{\text{C}}{\text{mol}} \cdot 0.66\,\text{V}) = -127599\,\text{J/mol}$$ Converting to kJ/mol for convenience: $$\Delta G^{\circ} = -127.6\,\text{kJ/mol}$$

Calculate the equilibrium constant, \(K\)

We can relate the equilibrium constant to the standard Gibbs free energy change using the following equation: $$\Delta G^{\circ} = -RT\ln{K}$$ where \(R\) is the universal gas constant, equal to \(8.314\frac{\text{J}}{\text{mol}\cdot\text{K}}\) \(T\) is the absolute temperature in Kelvin \(K\) is the equilibrium constant We are given the temperature as 298 K. Rearranging the equation, we have: $$K = e^{\frac{-\Delta G^{\circ}}{RT}}$$ Plugging in the known values: $$K = e^{\frac{-(-127599\,\text{J/mol})}{(8.314\,\frac{\text{J}}{\text{mol}\cdot\text{K}})(298\,\text{K})}}$$ Calculating for \(K\): $$K \approx 1.49\times10^{17}$$ In conclusion, we have found that the standard cell potential for the gold production reaction is 0.66 V, the standard Gibbs free energy change is -127.6 kJ/mol, and the equilibrium constant is \(1.49\times10^{17}\).

Key concepts

Cell Potential Calculation

The concept of cell potential is fundamental in electrochemistry. When we calculate the standard cell potential (\(\mathscr{E}^{\circ}_{\text{cell}}\)), we are essentially determining the voltage difference between two half-cells in a galvanic cell. This voltage is derived from the standard reduction potentials of each half-reaction.
To calculate \(\mathscr{E}^{\circ}_{\text{cell}}\), follow these simple steps:

• Identify the two half-reactions involved in the redox process.
• Determine which half-reaction will undergo reduction and which one will undergo oxidation. This might require reversing one of the potentials given.
• Subtraction is key: The cell potential is calculated by subtracting the standard potential of the oxidation reaction from the reduction reaction.

In our example, the reaction is for producing gold, using the half-reactions: - Gold complex reduction: \(\mathrm{Au}( ext{CN})^{-}_{2} + \mathrm{e}^{-} \rightarrow \mathrm{Au} + 2\text{CN}^{-}\), with \(\mathscr{E}^{\circ} = -0.60 \text{V}\). - Zinc complex oxidation (reversed): \(\mathrm{Zn} + 4\text{CN}^{-} \rightarrow \mathrm{Zn}( ext{CN})_{4}^{2-} + 2\mathrm{e}^{-}\), with \(\mathscr{E}^{\circ} = -1.26 \text{V}\), which we reverse. By substituting these values into the formula and accounting for the direction of electron flow, we calculate the standard cell potential as 0.66 V, indicating a spontaneous reaction.

Gibbs Free Energy Change

Gibbs free energy change (\(\Delta G^{\circ}\)) is a measure of the work potential of a reaction under standard conditions. It informs us on whether a reaction can occur spontaneously. A negative \(\Delta G^{\circ}\) suggests a spontaneous process.
Once we have the cell potential, we can easily calculate \(\Delta G^{\circ}\) using the formula: \[ \Delta G^{\circ} = -nF\mathscr{E}^{\circ} \] - \(n\) represents the number of electrons transferred during the redox reaction. - \(F\) is Faraday's constant, approximately \(96485 \text{ C/mol}\). - \(\mathscr{E}^{\circ}\) is the cell potential determined previously.
For the reaction of gold production from cyanide complexes, 2 moles of electrons are transferred, leading us to a calculated \(\Delta G^{\circ}\) of \(-127.6 \text{ kJ/mol}\). This value reflects the energy change when 1 mole of the reaction occurs and confirms that the process is favorably spontaneous under standard conditions.

Equilibrium Constant Calculation

The equilibrium constant \(K\) provides valuable insight into the extent of a reaction at equilibrium. It is directly related to the Gibbs free energy change through the equation:
\[ \Delta G^{\circ} = -RT\ln{K} \] Rearranging this gives us the expression for \(K\): \[ K = e^{\frac{-\Delta G^{\circ}}{RT}} \] - \(R\) is the universal gas constant, valued at \(8.314 \frac{\text{J}}{\text{mol}\cdot\text{K}}\). - \(T\) is the temperature in Kelvin (298 K for standard conditions).
Using the determined \(\Delta G^{\circ}\) of \(-127599 \text{ J/mol}\), and substituting these values into our formula, we find \(K \approx 1.49\times10^{17}\).
This large value of \(K\) indicates a reaction that favors the formation of products almost entirely, signifying a highly successful conversion of reactants to products at equilibrium.