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 (\(\mathscr{E}^{\circ}\)) for the production of gold is 0.66 V. The standard Gibbs free energy change (\(\Delta G^{\circ}\)) is -127,360 J/mol, and the equilibrium constant (K) at 298 K is \(6.54 \times 10^{18}\).

Step-by-step solution

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

To obtain the overall reaction, add the given half-reactions: Au(CN)₂⁻ + e⁻ → Au + 2 CN⁻ \( \mathscr{E}^{\circ} = -0.60 V\) Zn + 4 CN⁻ → Zn(CN)₄²⁻ + 2 e⁻ \( \mathscr{E}^{\circ} = 1.26 V\) Note that the second half-reaction was reversed because Zn(s) is a reactant and a reducing agent. Also, note that when a half-reaction is reversed, the standard reduction potential changes its sign. Add the two half-reactions: 2 Au(CN)₂⁻ + Zn(s) → 2 Au(s) + Zn(CN)₄²⁻ Now, add the standard reduction potentials: \( \mathscr{E}^{\circ} = -0.60 V + 1.26 V\)

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

Calculate the sum of the standard reduction potentials from step 1: \( \mathscr{E}^{\circ} = -0.60 V + 1.26 V = 0.66 V\)

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

Use the relationship between the standard cell potential and the standard Gibbs free energy change: \(\Delta G^{\circ} = -nFE^{\circ}\) where n is the number of moles of electrons transferred, F is the Faraday constant (96,485 C/mol), and \( \mathscr{E}^{\circ}\) is the standard cell potential. In this case, n = 2 since 2 electrons are transferred in the overall reaction. Therefore, \(\Delta G^{\circ} = -(2)(96,485 C/mol)(0.66 V)\)

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

Calculate the standard Gibbs free energy change from step 3: \(\Delta G^{\circ} = -(2)(96,485 C/mol)(0.66 V) = -127,360 J/mol\)

Calculate the equilibrium constant, K

Use the relationship between the standard Gibbs free energy change and the equilibrium constant: \(\Delta G^{\circ} = -RT \ln K\) where R is the gas constant (8.314 J/(mol·K)), T is the temperature (298 K in this case), and K is the equilibrium constant. Rearrange the equation to solve for K: \(\ln K = -\frac{\Delta G^{\circ}}{RT}\) K = \(e^{-\frac{\Delta G^{\circ}}{RT}}\) Substitute the values of \(\Delta G^{\circ}\), R, and T: K = \(e^{-\frac{-127,360 J/mol}{(8.314 J/(mol·K))(298 K)}}\)

Find the equilibrium constant, K

Calculate the equilibrium constant from step 5: K = \(e^{-\frac{-127,360 J/mol}{(8.314 J/(mol·K))(298 K)}} = 6.54 \times 10^{18}\) Now we found the necessary values: the standard cell potential (\(\mathscr{E}^{\circ} = 0.66 V\)), the standard Gibbs free energy change (\(\Delta G^{\circ} = -127,360 J/mol\)), and the equilibrium constant (K = \(6.54 \times 10^{18}\)) at 298 K for the production of gold.

Key concepts

Cell Potential Calculation

Understanding the cell potential of an electrochemical cell is key for grasping the basics of electrochemistry. In essence, the standard cell potential, also represented as \( \mathscr{E}^{\circ} \), is the difference in potential between two half-cells in an electrochemical cell under standard conditions, which typically means solutions of 1 M concentration, at 1 atm of pressure, and a temperature of 298 K.

To calculate this value, one must first identify the relevant half-reactions involved in the overall cell reaction. Each half-reaction is associated with a standard reduction potential (SRP), which is a measure of the tendency of a chemical species to acquire electrons and be reduced. It's crucial to note whether each half-reaction functions as a reduction or an oxidation in the overall reaction, as it determines whether to use the SRP as is or to change its sign.

When combined, the standard cell potential can be calculated by summing the relevant SRPs while ensuring the correct signs based on the reaction direction.

Key Pointers for Cell Potential Calculation:

• Identify each half-reaction involved and their corresponding SRPs.
• Determine the direction of the reaction—whether it is going to be reversed or not.
• If a half-reaction is reversed (going from reduction to oxidation), invert the sign of its SRP.
• Sum the appropriate SRPs to find the overall cell potential.

Once you have the cell potential, you can make inferential leaps about the spontaneity of the reaction, potential energy it can provide in a galvanic cell, or the required energy input in an electrolytic cell.

Gibbs Free Energy Change

In thermodynamics, Gibbs free energy change (\( \Delta G^{\circ} \)) is a powerful predictor of the spontaneity of a chemical reaction under standard conditions. The relationship between the standard cell potential and Gibbs free energy is given by the equation \( \Delta G^{\circ} = -nFE^{\circ} \), where \( n \) represents the number of moles of electrons transferred in the electrochemical reaction, \( F \) is the Faraday constant (approximately 96,485 C/mol), and \( \mathscr{E}^{\circ} \) is the standard cell potential.

A negative value of \( \Delta G^{\circ} \) indicates a spontaneous process, whereas a positive value suggests non-spontaneity. The closer the number is to zero though, the more ‘reversible’ the reaction—or at least, the less inherently ‘driven’ it is in either direction.

Key Concepts for Understanding Gibbs Free Energy Change:

• Standard Gibbs free energy change is a measure of the 'driving force' behind a chemical reaction.
• The \( -nF \) term shows the direct proportionality of \( \Delta G^{\circ} \) to the cell potential and mol number of electrons transferred.
• If \( \Delta G^{\circ} \) is negative, the reaction is spontaneous; if positive, it is non-spontaneous.

Gibbs free energy change not only informs about spontaneity but can also be useful in calculating the equilibrium constant, providing a link between the electrochemical properties of a cell and the thermodynamic likelihood of a reaction.

Equilibrium Constant Calculation

The equilibrium constant (\( K \)) is a numerical value that expresses the ratio of concentrations of the products to the reactants at equilibrium for a given reaction under specific conditions, typically at a standard temperature of 298 K. Its relationship with the Gibbs free energy is captured by the equation \( \Delta G^{\circ} = -RT \ln K\), where \( R \) is the gas constant and \( T \) is the temperature in Kelvin.

By rearranging the equation, you can find \( K \) easily: \( K = e^{-\frac{\Delta G^{\circ}}{RT}} \). A large \( K \) value indicates a reaction with a greater extent of product formation under standard conditions.

Essential Points for Equilibrium Constant Calculation:

• The equilibrium constant gives an idea of the position of equilibrium under standard conditions.
• It is directly related to the Gibbs free energy change of the reaction.
• \( K \) can be calculated by exponentiating the negative ratio of \( \Delta G^{\circ} \) to the product of the gas constant and temperature.

The calculation of \( K \) is significant for predicting the concentrations of reactants and products at equilibrium. In practical terms, this informs chemists and engineers about how far a reaction will go, which is crucial when designing processes like the extraction of gold from its complexes in the exercise provided.