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 $$ \begin{aligned} \mathrm{Au}(\mathrm{CN})_{2}^{-}+\mathrm{e}^{-} \longrightarrow \mathrm{Au}+2 \mathrm{CN}^{-} & \mathscr{E}^{\circ}=-0.60 \mathrm{~V} \\ \mathrm{Zn}(\mathrm{CN})_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn}+4 \mathrm{CN}^{-} & \mathscr{E}^{\circ}=-1.26 \mathrm{~V} \end{aligned} $$

Short answer

The standard cell potential for the gold production reaction is \(\mathscr{E}^{\circ} = 0.66 \mathrm{~V}\). Using this value, the standard Gibbs free energy change is \(\Delta G^{\circ} = -127.68 \mathrm{~kJ/mol}\), and the equilibrium constant at \(298 \mathrm{~K}\) is \(K \approx 4.39\times10^{21}\).

Step-by-step solution

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

To calculate the standard cell potential for the given redox reaction, we will use the standard reduction potentials of the half-reactions. Remember that the standard cell potential is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode. Here, we should identify which half-reaction is the reduction and which half-reaction is the oxidation. The given reaction is: $$ 2 \mathrm{Au} (\mathrm{CN})_{2}^{-}(a q)+\mathrm{Zn}(s) \longrightarrow 2 \mathrm{Au}(s)+\mathrm{Zn}(\mathrm{CN})_{4}^{2-}(a q) $$ The first half-reaction is: $$ \mathrm{Au}(\mathrm{CN})_{2}^{-}+\mathrm{e}^{-} \longrightarrow \mathrm{Au}+2 \mathrm{CN}^{-} \qquad \mathscr{E}^{\circ}=-0.60 \mathrm{~V} $$ The second half-reaction is: $$ \mathrm{Zn}(\mathrm{CN})_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn}+4 \mathrm{CN}^{-} \quad \mathscr{E}^{\circ}=-1.26 \mathrm{~V} $$ In the overall redox reaction, Zn is oxidized to Zn(CN)\(_4^{2-}\) and Au(CN)\(_2^-\) is reduced to Au. Therefore, the first half-reaction is the reduction half-reaction (cathode) and the second half-reaction is the oxidation half-reaction (anode). Now, let's calculate the standard cell potential (\(\mathscr{E}^{\circ}\)): $$ \mathscr{E}^{\circ} = \mathscr{E}_{cathode}^{\circ} - \mathscr{E}_{anode}^{\circ} = (-0.60) -(-1.26) = 0.66 \mathrm{~V} $$

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

To calculate the standard Gibbs free energy change, we will use the following formula: $$ \Delta G^{\circ} = -nFE^{\circ} $$ Here, \(n\) is the number of electrons transferred, \(F\) is Faraday's constant (\(96485 \mathrm{~C/mol}\)), and \(\mathscr{E}^{\circ}\) is the standard cell potential. In this case, 2 electrons are transferred in the overall redox reaction (both half-reactions show the transfer of 2 electrons and are already balanced). Now, plug in the values and calculate \(\Delta G^{\circ}\): $$ \Delta G^{\circ} = -(2)(96485 \mathrm{~C/mol})(0.66 \mathrm{~V}) = -127680.6 \mathrm{~J/mol} = -127.68 \mathrm{~kJ/mol} $$

Calculate the equilibrium constant (\(K\))

To calculate the equilibrium constant, we will use the following formula: $$ \Delta G^{\circ} = -RT\ln {K} $$ Where \(R\) is the gas constant (\(8.314 \mathrm{~J/(mol \cdot K)}\)), \(T\) is the temperature in Kelvin (given as \(298 \mathrm{~K}\)), and \(K\) is the equilibrium constant. To find \(K\), we can rearrange this formula: $$ K = e^{\frac{-\Delta G^{\circ}}{RT}} $$ Now, plug in the values and calculate \(K\): $$ K = e^{\frac{-( -127680.6 \mathrm{~J/mol})}{(8.314 \mathrm{~J/(mol \cdot K)})(298 \mathrm{~K})}} = 4.39\times10^{21} $$ The equilibrium constant (\(K\)) for the reaction is approximately \(4.39\times10^{21}\).

Key concepts

Standard Reduction Potential

The concept of standard reduction potential (\(\mathscr{E}^{\circ}\)) is crucial when analyzing redox reactions since it tells us how readily a substance gains electrons during a chemical reaction. Each half-reaction in a redox system has its own standard reduction potential which is measured in volts. The values are determined under standard conditions, which typically involve 1 M concentration and 298 K temperature.
When comparing two half-reactions, the one with a higher (more positive) reduction potential is the more favorable reaction for gaining electrons, thus it acts as the cathode in a galvanic cell.

• For example, in the given problem, the reduction of Au(CN) \((\mathrm{Au} (\mathrm{CN})_{2}^{-} + \mathrm{e}^{-} \rightarrow \mathrm{Au} + 2 \mathrm{CN}^{-}\)) has a standard reduction potential of \(-0.60 \mathrm{~V}\).
• Meanwhile, Zn(CN) \((\mathrm{Zn}(\mathrm{CN})_{4}^{2-} + 2 \mathrm{e}^{-} \rightarrow \mathrm{Zn} + 4 \mathrm{CN}^{-}\)) has a potential of \(-1.26 \mathrm{~V}\).

To find the standard cell potential \(\mathscr{E}^{\circ}\) for the entire reaction, the standard reduction potential of the anode is subtracted from that of the cathode. This difference tells us the driving force of the reaction and in this case, it is \(0.66 \mathrm{~V}\) indicating that the reaction is spontaneous at standard conditions.

Gibbs Free Energy

Gibbs free energy change (\(\Delta G^{\circ}\)) is a vital concept in determining whether a chemical reaction occurs spontaneously. A negative value of \(\Delta G^{\circ}\) signifies a spontaneous process, meaning it releases energy.
The relationship between the standard cell potential and Gibbs free energy is given by the equation:\[\Delta G^{\circ} = -nF\mathscr{E}^{\circ}\]Where:

• \(n\) = number of moles of electrons transferred in the reaction.
• \(F\) = Faraday's constant, with a value of \(96485\ \mathrm{C/mol}\).

In our specific example, the transfer of two moles of electrons and a standard cell potential of \(0.66\ \mathrm{V}\) result in \(\Delta G^{\circ} = -127.68 \mathrm{~kJ/mol}\). This negative value confirms the reaction's spontaneity under standard conditions.

Equilibrium Constant

The equilibrium constant (\(K\)) quantifies the ratio of concentrations of products to reactants at equilibrium for a reversible reaction. It tells us the extent to which reactants turn into products. The greater the value of \(K\), the more the reaction favors the formation of products.
The relationship between Gibbs free energy and the equilibrium constant is represented by:\[\Delta G^{\circ} = -RT\ln K\]Where:

• \(R\) is the universal gas constant (\(8.314 \mathrm{~J/(mol\cdot K)}\)).
• \(T\) is the temperature in Kelvin.

Rearranging to solve for \(K\) gives:\[K = e^{\frac{-\Delta G^{\circ}}{RT}}\]Plugging in the values provided, we calculate that \(K \approx 4.39\times10^{21}\), indicating a strong preference for products at equilibrium. This large value affirms that the reaction is product-favored under the given conditions.

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, entail the exchange of electrons between substances. One reactant gets reduced (gains electrons), while another is oxidized (loses electrons). These reactions are central to numerous biological and industrial processes, including this case of gold production.
Redox reactions typically involve two half-reactions:

• Reduction half-reaction, where a substance gains electrons.
• Oxidation half-reaction, where a substance loses electrons.

In the original exercise, the reduction half-reaction is \(\mathrm{Au} (\mathrm{CN})_{2}^{-} + \mathrm{e}^{-} \longrightarrow \mathrm{Au} + 2 \mathrm{CN}^{-}\), and the oxidation half-reaction is \(\mathrm{Zn} + 4 \mathrm{CN}^{-} \longrightarrow \mathrm{Zn}(\mathrm{CN})_{4}^{2-} + 2 \mathrm{e}^{-}\). By balancing these two half-reactions, we form the full redox reaction, highlighting both the movement of electrons and the fundamental interplay of energy changes, which gives rise to the concepts of potential, spontaneity, and equilibrium.