Question

Consider the following half-reactions: $$\begin{array}{ll}{\mathrm{IrCl}_{6}^{3-}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Ir}+6 \mathrm{Cl}^{-}} & {\mathscr{E}^{\circ}=0.77 \mathrm{V}} \\\ {\mathrm{PtCl}_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt}+4 \mathrm{Cl}^{-}} & {\mathscr{E}^{\circ}=0.73 \mathrm{V}} \\\ {\mathrm{PdCl}_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pd}+4 \mathrm{Cl}^{-}} & {\mathscr{E}^{\circ}=0.62 \mathrm{V}}\end{array}$$ 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\(M\) 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

It is not feasible to separate the three metals (Ir, Pt, Pd) from the hydrochloric acid solution by electrolysis, as their calculated reduction potentials (0.70V for Ir, 0.68V for Pt, and 0.57V for Pd) are too close to each other. This closeness in potentials would cause the reduction of one metal to simultaneously trigger the reduction of the others, preventing the desired 99% plating out of a metal before another begins to plate out.

Step-by-step solution

Analyze the given half-reactions and their potentials

The half-reactions given and their standard reduction potentials (E°) are: 1. IrCl₆³⁻ + 3e⁻ → Ir + 6Cl⁻, E° = 0.77V 2. PtCl₄²⁻ + 2e⁻ → Pt + 4Cl⁻, E° = 0.73V 3. PdCl₄²⁻ + 2e⁻ → Pd + 4Cl⁻, E° = 0.62V These half-reactions correspond to the reduction of complex ions to their respective metal ions.

Analyze the given concentrations

The concentration of chloride ions in the solution is 1.0M, while the concentration of each complex ion is 0.020M.

Determine the separation feasibility using the Nernst equation

The Nernst equation helps us calculate the reduction potential for each of the half-reactions, considering the given concentrations. The Nernst equation is: E = E° - \( \frac{RT}{nF} \) * ln(Q) Where: - E is the reduction potential - E° is the standard reduction potential - R is the gas constant, 8.314 J/(mol·K) - T is the temperature, in K (assuming the experiment takes place at room temperature, T = 298K) - n is the number of electrons transferred in the half-reaction - F is the Faraday's constant, 96,485 C/mol - Q is the reaction quotient, which equals ([product concentrations]/[reactant concentrations]) We can calculate the reduction potential for each half-reaction using the given concentrations.

Calculate the reduction potential for the Ir half-reaction

For the Ir half-reaction: E° = 0.77V n = 3 electrons Q = ([Ir][Cl⁻]⁶) / ([IrCl₆³⁻]) = (0.020M * (1.0M)⁶) / (0.020M) = (1.0M)⁶ E = 0.77V - (2.303 * \( \frac {8.314 J/(mol·K) * 298 K} {3 * 96485 C/mol} \)) * log((1.0M)⁶) = 0.70V

Calculate the reduction potential for the Pt half-reaction

For the Pt half-reaction: E° = 0.73V n = 2 electrons Q = ([Pt][Cl⁻]⁴) / ([PtCl₄²⁻]) = (0.020M * (1.0M)⁴) / (0.020M) = (1.0M)⁴ E = 0.73V - (2.303 * \( \frac {8.314 J/(mol·K) * 298 K} {2 * 96485 C/mol} \)) * log((1.0M)⁴) = 0.68V

Calculate the reduction potential for the Pd half-reaction

For the Pd half-reaction: E° = 0.62V n = 2 electrons Q = ([Pd][Cl⁻]⁴) / ([PdCl₄²⁻]) = (0.020M * (1.0M)⁴) / (0.020M) = (1.0M)⁴ E = 0.62V - (2.303 * \( \frac {8.314 J/(mol·K) * 298 K} {2 * 96485 C/mol} \)) * log((1.0M)⁴) = 0.57V

Determine the separation feasibility

Now we have the following reduction potentials: 1. Ir half-reaction: E = 0.70V 2. Pt half-reaction: E = 0.68V 3. Pd half-reaction: E = 0.57V Since the reduction potentials are close to each other and we want to plate out 99% of a metal before another begins to plate out, it is not feasible to separate the three metals from this solution by electrolysis. The reason is that the reduction of one metal would cause the reduction of the others due to their close potentials.

Key concepts

Nernst Equation

The Nernst Equation is a powerful tool in electrochemistry used to calculate the actual reduction potential for electrochemical reactions under non-standard conditions. This equation helps us understand the effect of concentration on the electrochemical cell’s potential. The basic formula is: \[ E = E^\circ - \frac{RT}{nF} \ln(Q) \] - **E** is the actual reduction potential. - **E°** is the standard reduction potential. - **R** is the universal gas constant \(8.314 \: \text{J/(mol·K)}\). - **T** is the temperature in Kelvin, often taken as \(298 \: K\) if not specified otherwise. - **n** represents the number of moles of electrons transferred in the half-reaction. - **F** is Faraday's constant \(96,485 \: \text{C/mol}\). - **Q** is the reaction quotient, the ratio of the product concentrations to the reactant concentrations, each raised to the power of its stoichiometric coefficient. The Nernst Equation adjusts the standard potential based on the actual conditions of the reaction, providing insight into practical systems where concentrations are not at standard \(1 \: M\). It is essential when determining how changes in concentration can shift the balance of redox reactions during electrochemical processes.

Reduction Potential

Reduction potential, also known as electrode potential, indicates a substance's tendency to gain electrons and be reduced. Each half-reaction in an electrochemical cell has its own reduction potential, often expressed in volts (V). The standard reduction potential \(E^\circ\) is measured under standard conditions, which include \(1 \: M\) concentrations and a temperature of \(25°C\) (298 K). - **Positive Reduction Potential**: Indicates that a reaction is more likely to gain electrons (be reduced) and act as a good oxidizing agent. - **Negative Reduction Potential**: Implies that the substance is less willing to be reduced and is a better reducing agent. When comparing two half-reactions, the one with the higher reduction potential will tend to be reduced, while the other will be oxidized if they are connected in an electrochemical cell. In electrochemical separations, metals with higher reduction potentials will plate out first. Understanding reduction potentials is crucial for predicting the direction of redox reactions and designing processes such as electrolysis for material separation.

Electrolysis Feasibility

Electrolysis is a process in which electrical energy causes a non-spontaneous chemical reaction to occur, typically used to decompose compounds or separate elements. Electrolysis feasibility is determined by the comparison of reduction potentials of the species involved in the redox reactions. - **Separation of Metals by Electrolysis**: When attempting to separate metals from a solution, as in the given exercise, it’s vital to ensure that their potentials are significantly different. Ideally, there should be a clear gap, often of at least a few tenths of a volt, to favor the plating of one metal before another. - **99% Plating Requirement**: This refers to the need for significant completion of one metal’s deposition before another starts to prevent co-deposition. If reduction potentials are too close, distinguishing a specific metal is more challenging, making pure separations unfeasible. In the discussed problem, the proximity of the reduction potentials for iridium, platinum, and palladium means organic co-deposition is likely, leading to an inseparable plating process. Therefore, the feasibility is low when the reduction potentials are too similar, limiting practical electrochemical separation.