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Chapter 18: Problem 138

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

Expert verified
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

01

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.
02

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.
03

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.
04

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
05

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
06

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
07

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.

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Most popular questions from this chapter

The solubility product for \(\operatorname{CuI}(s)\) is \(1.1 \times 10^{-12}\) . Calculate the value of \(\mathscr{E}^{\circ}\) for the half-reaction $$\mathrm{CuI}(s)+\mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s)+\mathrm{I}^{-}(a q)$$

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In the electrolysis of an aqueous solution of $\mathrm{Na}_{2} \mathrm{SO}_{4},$ what reactions occur at the anode and the cathode (assuming standard conditions)? $$\begin{array}{ll} {\text{}} & \quad{ \mathscr{E}^{\circ} } \\ \hline {\mathrm{S}_{2} \mathrm{O}_{8}^{2-}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{SO}_{4}^{2-}} & {2.01 \mathrm{V}} \\ {\mathrm{O}_{2}+4 \mathrm{H}^{+}+4 \mathrm{e}^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}} & {1.23 \mathrm{V}} \\ {2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}+2 \mathrm{OH}^{-}} & {-0.83 \mathrm{V}} \\\ {\mathrm{Na}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Na}} & {-2.71 \mathrm{V}}\end{array}$$

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