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Chapter 18: Problem 112

Consider the following half-reactions: $$\begin{array}{rl}{\mathrm{Pt}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt}} & {\mathscr{E}^{\circ}=1.188 \mathrm{V}} \\\ {\mathrm{PtCl}_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt}+4 \mathrm{Cl}^{-}} & {\mathscr{E}^{\circ}=0.755 \mathrm{V}} \\\ {\mathrm{NO}_{3}^{-}+4 \mathrm{H}^{+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{NO}+2 \mathrm{H}_{2} \mathrm{O}} & {\mathscr{E}^{\circ}=0.96 \mathrm{V}}\end{array}$$ Explain why platinum metal will dissolve in aqua regia (a mixture of hydrochloric and nitric acids) but not in either concentrated nitric or concentrated hydrochloric acid individually.

Short Answer

Expert verified
Platinum metal dissolves in aqua regia because the mixture of nitric and hydrochloric acids provides the necessary reaction environment for a complete redox reaction to occur. In aqua regia, chloride ions from HCl create a suitable environment for platinum dissolution, while nitrate ions from HNO鈧� provide the required oxidation. However, when platinum is in contact with either concentrated HCl or HNO鈧� individually, there is no suitable oxidation half-reaction available, preventing the dissolution of platinum.

Step by step solution

01

1. Identifying redox half-reactions

First, we need to identify which half-reactions will occur when platinum metal comes into contact with each acid individually and in aqua regia. Let's consider the first half-reaction: \( \mathrm{Pt}^{2+} + 2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt} \\ \mathscr{E}^{\circ} = 1.188{\mathrm{V}}\) This is a reduction half-reaction where platinum ions will get reduced to solid platinum metal. Now let's consider the second half-reaction: \( \mathrm{PtCl}_{4}^{2-} + 2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt} + 4 \mathrm{Cl}^{-} \\ \mathscr{E}^{\circ} = 0.755\,\mathrm{V} \) In this half-reaction, platinum ions in a complex with chlorine will get reduced to solid platinum while chlorine ions are released. Finally, let's consider the third half-reaction: \( \mathrm{NO}_{3}^{-} + 4 \mathrm{H}^{+} + 3 \mathrm{e}^{-} \longrightarrow \mathrm{NO} + 2 \mathrm{H}_{2} \mathrm{O} \\ \mathscr{E}^{\circ} = 0.96\,\mathrm{V}\) In this half-reaction, nitrate ions will get reduced to nitric oxide while water is produced. Now, let's analyze these half-reactions in the context of different acids.
02

2. Analyzing half-reactions with concentrated HCl

When platinum is in contact with concentrated hydrochloric acid (HCl), the relevant half-reaction can be the second one. It involves chlorine ions (producing platinum and chlorine ion release). However, we need an oxidation half-reaction for a complete redox reaction to happen; just having reductants (like Pt虏鈦� and Cl鈦�) doesn't lead to the dissolution of platinum.
03

3. Analyzing half-reactions with concentrated HNO3

When platinum is in contact with concentrated nitric acid (HNO鈧�), the relevant half-reaction can be the third one. It involves nitrate ions (producing nitric oxide and water). However, just like in the case of HCl, we need an oxidation half-reaction for a complete redox reaction to happen.
04

4. Analyzing half-reactions with aqua regia

Aqua regia is a mixture of nitric acid (HNO鈧�) and hydrochloric acid (HCl). When platinum is exposed to this mixture, we have the combination of half-reactions that involve both chloride and nitrate ions. The chloride ions \( (4\mathrm{Cl}^{-}) \), which are involved in the second half-reaction, act as the environment where platinum can dissolve; meanwhile, the nitrate ions \( (\mathrm{NO}_{3}^{-}) \) in the third half-reaction can provide the necessary oxidation for a complete redox reaction to happen. Since aqua regia is a mixture of both acids, it provides the necessary reaction environment for platinum to dissolve, which does not happen when platinum is in contact with concentrated HCl or HNO鈧� individually.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Platinum Dissolution
Platinum is known for its remarkable resistance to corrosion and chemical attacks, but it can dissolve in a specific type of chemical mixture known as aqua regia. Platinum dissolution is fascinating because it defies the metal's characteristic inertness. Aqua regia, a potent mixture of nitric acid and hydrochloric acid, has the unique ability to dissolve platinum metal. This solution is renowned for its ability to dissolve noble metals such as gold and platinum, which often resist the action of most other acids. The combination of these acids makes a powerful solution that can react with platinum, enabling its dissolution.
Platinum dissolution in aqua regia occurs due to a unique set of redox (reduction-oxidation) reactions that enable the normally resistant platinum to become reactive. This reaction involves both oxidation and reduction processes, allowing platinum to dissolve into the acidic solution as a chloroplatinic acid complex.
Redox Reactions
Redox reactions play a crucial role in the process of platinum dissolution in aqua regia. These reactions involve the transfer of electrons between different chemical species. In the case of platinum dissolution, two half-reactions come into play: reduction of nitrate ions and oxidation involving chloride ions.
  • The reduction half-reaction involves nitrate ions from nitric acid, whereby nitrate ions ( \(\mathrm{NO}_{3}^{-}\)) gain electrons and get reduced to nitric oxide, alongside the production of water.
  • The oxidation half-reaction involves chloride ions from hydrochloric acid, which react with platinum to form \(\mathrm{PtCl}_{4}^{2-}\), a soluble complex, releasing electrons.
In aqua regia, these redox reactions occur simultaneously, causing the normally inert platinum to become soluble. This creates a complete redox system, with platinum undergoing oxidation and losing electrons, whilst the nitrate ions undergo reduction, completing the electronic exchange necessary for reaction.
Nitric Acid
Nitric acid is an important component of aqua regia. On its own, concentrated nitric acid is not sufficient to dissolve platinum, due to the lack of an accompanying simultaneous oxidation process. However, within aqua regia, nitric acid contributes to the dissolution process by primarily acting as an oxidizing agent.
Nitric acid produces nitrate ions (\(\mathrm{NO}_{3}^{-}\)) that play a vital role in the redox reaction. These ions are responsible for oxidizing other molecules, which, when coupled with the presence of hydrochloric acid, ensure that platinum dissolution can occur.
  • Nitric acid contributes the necessary electrons that help reduce nitrate ions, thus fueling the reaction environment.
  • This provides the crucial oxidation step that aids in dissolving platinum metal in the acidic solution.
Hydrochloric Acid
Hydrochloric acid is the other vital component in the aqua regia mixture that enables the dissolution of platinum. Though concentrated hydrochloric acid alone cannot oxidize and dissolve platinum, its presence in aqua regia is essential.
Hydrochloric acid facilitates the formation of chloride ions (\(\mathrm{Cl}^{-}\)). These ions are crucial for forming the chloroplatinic acid complex (\(\mathrm{PtCl}_{4}^{2-}\)), which makes platinum soluble.
  • Chloride ions help create this compound, effectively converting solid platinum into a dissolved form.
  • This interaction sets the stage for the redox reaction to proceed efficiently.
The chloride ions enable the platinum metal to transition into a dissolvable state, thus making hydrochloric acid indispensable in the aqua regia mixture, allowing the dissolution of metals that are otherwise resistant to chemical reactions.

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Most popular questions from this chapter

The ultimate electron acceptor in the respiration process is molecular oxygen. Electron transfer through the respiratory chain takes place through a complex series of oxidationreduction reactions. Some of the electron transport steps use iron-containing proteins called \(c y\) tochromes. All cytochromes transport electrons by converting the iron in the cytochromes from the +3 to the +2 oxidation state. Consider the following reduction potentials for three different cytochromes used in the transfer process of electrons to oxygen (the potentials have been corrected for \(\mathrm{pH}\) and for temperature): cytochrome \(\mathrm{a}\left(\mathrm{Fe}^{3+}\right)+\mathrm{e}^{-} \longrightarrow\) cytochrome \(\mathrm{a}\left(\mathrm{Fe}^{2+}\right)\) $$ \begin{array}{c}\mathscr{C}=0.385 \mathrm{~V}\end{array} $$ cytochrome \(\mathrm{b}\left(\mathrm{Fe}^{3+}\right)+\mathrm{e}^{-} \longrightarrow\) cytochrome \(\mathrm{b}\left(\mathrm{Fe}^{2+}\right)\) $$ \begin{array}{l}\mathscr{E}=0.030 \mathrm{~V}\end{array} $$ cytochrome \(\mathrm{c}\left(\mathrm{Fe}^{3+}\right)+\mathrm{e}^{-} \longrightarrow \mathrm{cytochrome} \mathrm{c}\left(\mathrm{Fe}^{2+}\right)\) $$ \begin{array}{c}\mathscr{C}=0.254 \mathrm{~V}\end{array} $$ In the electron transfer series, electrons are transferred from one cytochrome to another. Using this information, determine the cytochrome order necessary for spontaneous transport of electrons from one cytochrome to another, which eventually will lead to electron transfer to \(\mathrm{O}_{2}\).

The overall reaction and standard cell potential at \(25^{\circ} \mathrm{C}\) for the rechargeable nickel-cadmium alkaline battery is \(\mathrm{Cd}(s)+\mathrm{NiO}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) \(\mathrm{Ni}(\mathrm{OH})_{2}(s)+\mathrm{Cd}(\mathrm{OH})_{2}(s) \qquad \mathscr{E}^{\circ}=1.10 \mathrm{V}\) For every mole of Cd consumed in the cell, what is the maximum useful work that can be obtained at standard conditions?

When aluminum foil is placed in hydrochloric acid, nothing happens for the first 30 seconds or so. This is followed by vigorous bubbling and the eventual disappearance of the foil. Explain these observations.

If the cell potential is proportional to work and the standard reduction potential for the hydrogen ion is zero, does this mean that the reduction of the hydrogen ion requires no work?

Consider the galvanic cell based on the following halfreactions: $$\begin{array}{ll}{\mathrm{Au}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au}} & {\mathscr{E}^{\circ}=1.50 \mathrm{V}} \\ {\mathrm{T} 1^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Tl}} & {\mathscr{E}^{\circ}=-0.34 \mathrm{V}}\end{array}$$ a. Determine the overall cell reaction and calculate \(\mathscr{E}_{\text { cell }}\) b. Calculate \(\Delta G^{\circ}\) and \(K\) for the cell reaction at \(25^{\circ} \mathrm{C}\) c. Calculate \(\mathscr{E}_{\text { cell }}\) at \(25^{\circ} \mathrm{C}\) when \(\left[\mathrm{Au}^{3+}\right]=1.0 \times 10^{-2} \mathrm{M}\) and \(\left[\mathrm{T} 1^{+}\right]=1.0 \times 10^{-4} \mathrm{M}\)
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