Question

Consider the following half-reactions: $$\begin{array}{ll} \operatorname{IrCl}_{6}^{3-}+3 \mathrm{e}^{-} \longrightarrow \operatorname{Ir}+6 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.77 \mathrm{V} \\ \mathrm{PtCl}_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt}+4 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.73 \mathrm{V} \\ \mathrm{PdCl}_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pd}+4 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.62 \mathrm{V} \end{array}$$ A hydrochloric acid solution contains platinum, palladium, and iridium as chloro-complex ions. The solution is a constant 1.0 \(M\) in chloride ion and \(0.020 \mathrm{M}\) in each complex ion. Is it feasible to separate the three metals from this solution by electrolysis? (Assume that \(99 \%\) of a metal must be plated out before another metal begins to plate out.)

Short answer

The reduction potentials for each half-reaction at the given concentrations are: 1. Iridium half-reaction: \[E_{\text{Ir}} = 0.77 - \frac{8.314 × 298}{3 × 96,485} \ln \frac{1}{0.020} \approx 0.69\text{ V}\] 2. Platinum half-reaction: \[E_{\text{Pt}} = 0.73 - \frac{8.314 × 298}{2 × 96,485} \ln \frac{1}{0.020} \approx 0.69\text{ V}\] 3. Palladium half-reaction: \[E_{\text{Pd}} = 0.62 - \frac{8.314 × 298}{2 × 96,485} \ln \frac{1}{0.020} \approx 0.58\text{ V}\] The required voltage to achieve 99% separation between Iridium and Platinum is: \[\Delta E_{\text{Ir-Pt}} = E_{\text{Ir}} - E_{\text{Pt}} \approx 0\text{ V}\] This suggests that the separation between Iridium and Platinum would be difficult. The voltage difference between Platinum and Palladium is: \[\Delta E_{\text{Pt-Pd}} = E_{\text{Pt}} - E_{\text{Pd}} \approx 0.11\text{ V}\] The required voltage for 99% separation between Iridium and Platinum (∆E) is less than the difference in the reduction potentials of Platinum and Palladium. Thus, it may not be feasible to separate all three metals from this solution by electrolysis.

Step-by-step solution

Write down the Nernst equation

The Nernst equation relates the reduction potential of a half-reaction to the standard reduction potential, temperature, and concentration of the reactants and products. The equation is given by: \[E = E^{°} - \frac{RT}{nF} \ln Q\] where: - E is the reduction potential of the half-reaction - \(E^{°}\) is the standard reduction potential of the half-reaction - R is the universal gas constant (R ≈ 8.314 J/(molK)) - T is the temperature in Kelvin - n is the number of electrons transferred in the half-reaction - F is Faraday's constant (F ≈ 96,485 C/mol) - Q is the reaction quotient, which is the ratio of the product of the activities of the species in the product over the product of the activities of the species in the reactants Since we will assume the temperature to be 25°C (298K) and the activities to be equal to the concentrations and the concentration of chloride ion is constant, the equation can be simplified to: \[E = E^{°} - \frac{RT}{nF} \ln \frac{1}{[\text{complex ion}]}\]

Calculate the reduction potentials at given concentrations

For each of the half-reactions given in the exercise, we will calculate the reduction potential by using the simplified Nernst equation: 1. Iridium half-reaction: \[E_{\text{Ir}} = 0.77 - \frac{8.314 × 298}{3 × 96,485} \ln \frac{1}{0.020}\] 2. Platinum half-reaction: \[E_{\text{Pt}} = 0.73 - \frac{8.314 × 298}{2 × 96,485} \ln \frac{1}{0.020}\] 3. Palladium half-reaction: \[E_{\text{Pd}} = 0.62 - \frac{8.314 × 298}{2 × 96,485} \ln \frac{1}{0.020}\] Once the reduction potentials for each half-reaction are calculated, we can compare them to determine if a successful separation is possible.

Analyze the results and check for separation feasibility

If there is a significant difference in the reduction potentials, it may be possible to separate the metals by applying a voltage between the values of reduction potentials of two adjacent metals. Apply a voltage between the reduction potentials of the first metal (highest reduction potential) and the second metal (next highest reduction potential). If it is possible to plate out 99% of the first metal before the second metal begins to plate out, the separation is considered feasible. Calculate the required voltage (∆E) to achieve 99% separation between the metals: \[\Delta E = E_{\text{Metal 1}} - E_{\text{Metal 2}}\] Compare the calculated ∆E with the difference in the reduction potentials of the second and third metals. If ∆E is less than the difference in the reduction potentials of the second and third metals, the separation is feasible. Remember to perform this analysis for each pair of metals to ensure it is possible to separate all three metals from the solution.

Key concepts

Nernst Equation

The Nernst equation is a fundamental principle in electrochemistry that provides a mathematical relationship between the reduction potential of an electrochemical cell under non-standard conditions and its standard reduction potential, temperature, and concentration of the reacting species. It allows chemists to predict the direction and extent of chemical reactions in an electrochemical cell.

For a half-reaction, the Nernst equation is expressed as:

\(E = E^{°} - \frac{RT}{nF} \ln Q\)

In this equation, \(E\) represents the cell’s reduction potential, \(E^{°}\) is the standard reduction potential, \(R\) is the universal gas constant, \(T\) is the temperature in Kelvin, \(n\) is the number of electrons transferred, \(F\) is Faraday’s constant, and \(Q\) is the reaction quotient reflecting the ratio of the concentrations of the products to the reactants. By plugging in the values for these variables, we can calculate the cell's actual reduction potential. This calculation is critical when considering operations such as electrolysis, where we aim to control the deposition of metals from their ionic forms in solution.

Electrolysis

Electrolysis is an electrochemical process used to induce a chemical reaction through the application of an electric current. This process can cause the decomposition of chemical compounds and is often used for the purification or separation of elements from their ionic compounds in solution or molten form.

During electrolysis, an external voltage is applied to drive a non-spontaneous chemical reaction. Electrodes called anodes and cathodes are immersed in the solution and connected to a power supply. At the cathode, reduction occurs, where positively charged ions (cations) gain electrons. At the anode, oxidation takes place, where electrons are removed from negatively charged ions (anions) or neutral atoms. For metals to be separated by electrolysis, like platinum, palladium, and iridium, specific voltages are applied so that each metal plates out on the cathode at different times based on their individual reduction potentials.

Reduction Potential

Reduction potential, often denoted as \(E\) or \(E^°\), is a measure of the tendency of a chemical species to acquire electrons and thereby be reduced. Each half-reaction in an electrochemical cell has an associated reduction potential; the greater the reduction potential, the more readily the substance is reduced (gains electrons).

The standard reduction potential (\(E^°\)) is measured under standard conditions: 1 M concentration, 1 atm pressure, and a temperature of 25°C (298 K). In the context of our exercise, the reduction potential is crucial for determining the order in which metals will plate out during electrolysis since a metal with a higher reduction potential will plate out before a metal with a lower one.

Reaction Quotient

The reaction quotient (\(Q\)) is a measure of the relative amounts of products and reactants present during a reaction at a given moment. For the reaction quotient, the concentrations of gaseous and aqueous species are typically used. It has an expression similar to the equilibrium constant (\(K\)), but while \(K\) applies at equilibrium, \(Q\) can be used to describe a system that is not at equilibrium.

In an electrochemical cell, the reaction quotient is used in the Nernst equation to determine the cell’s voltage at non-standard conditions. As the reaction progresses towards equilibrium, the value of \(Q\) changes, affecting the cell potential. The reaction quotient is also pivotal in predicting the direction of the reaction: if \(Q < K\), the reaction proceeds forward, and if \(Q > K\), the reaction goes in reverse.