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 HNO₃ provide the required oxidation. However, when platinum is in contact with either concentrated HCl or HNO₃ 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 Pt²⁺ and Cl⁻) 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 (HNO₃), 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 (HNO₃) 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 HNO₃ individually.

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Most popular questions from this chapter

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?

When copper reacts with nitric acid, a mixture of \(\mathrm{NO}(g)\) and \(\mathrm{NO}_{2}(g)\) is evolved. The volume ratio of the two product gases depends on the concentration of the nitric acid according to the equilibrium $$2 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q)+\mathrm{NO}(g) \rightleftharpoons 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)$$ Consider the following standard reduction potentials at $25^{\circ} \mathrm{C} :$ $$3 \mathrm{e}^{-}+4 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ $$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \mathscr{E}^{\circ}=0.957 \mathrm{V}$$ $$\mathrm{e}^{-}+2 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ $$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \mathscr{E}^{\circ}=0.775 \mathrm{V}$$ a. Calculate the equilibrium constant for the above reaction. b. What concentration of nitric acid will produce a NO and NO \(_{2}\) mixture with only 0.20\(\% \mathrm{NO}_{2}\) (by moles) at \(25^{\circ} \mathrm{C}\) and 1.00 atm? Assume that no other gases are present and that the change in acid concentration can be neglected.

Calculate \(\mathscr{E}^{\circ}\) and \(\Delta G^{\circ}\) for the reaction $$2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)$$ at 298 \(\mathrm{K}\) . Using thermodynamic data in Appendix \(4,\) estimate \(\mathscr{E}^{\circ}\) and \(\Delta G^{\circ}\) at \(0^{\circ} \mathrm{C}\) and $90 .^{\circ} \mathrm{C}\( . Assume \)\Delta H^{\circ}\( and \)\Delta S^{\circ}$ do not depend on temperature.

In 1973 the wreckage of the Civil War ironclad USS Monitor was discovered near Cape Hatteras, North Carolina. [The Monitor and the CSS Virginia (formerly the USS Merrimack) fought the first battle between iron-armored ships.] In 1987 investigations were begun to see if the ship could be salvaged. It was reported in Time (June 22, 1987) that scientists were considering adding sacrificial anodes of zinc to the rapidly corroding metal hull of the Monitor. Describe how attaching zinc to the hull would protect the Monitor from further corrosion.

Calculate \(\mathscr{E}^{\circ}\) values for the following cells. Which reactions are spontaneous as written (under standard conditions)? Balance the equations. Standard reduction potentials are found in Table 18.1. a. $\mathrm{MnO}_{4}^{-(a q)}+\mathrm{I}^{-}(a q) \longrightarrow \mathrm{I}_{2}(a q)+\mathrm{Mn}^{2+}(a q)$ b. $\mathrm{MnO}_{4}^{-}(a q)+\mathrm{F}^{-}(a q) \longrightarrow \mathrm{F}_{2}(g)+\mathrm{Mn}^{2+}(a q)$

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