Consider the following half-reactions: $$\begin{aligned}\mathrm{Pt}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt} & & \mathscr{E}^{\circ}=1.188 \mathrm{~V} \\ \mathrm{PtCl}_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt}+4 \mathrm{Cl}^{-} & & \mathscr{C}^{\circ}=0.755 \mathrm{~V} \\ \mathrm{NO}_{3}^{-}+4 \mathrm{H}^{+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{NO}+2 \mathrm{H}_{2} \mathrm{O} & & \mathscr{C}^{\circ}=0.96 \mathrm{~V}\end{aligned}$$ 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.

In aqua regia, we have a mixture of hydrochloric acid (HCl) and nitric acid (HNO₃). HCl provides chloride ions (Cl⁻), while HNO₃ provides nitrate ions (NO₃⁻). These ions can act as reducing or oxidizing agents.

There are two half-reactions involving platinum: 1. \(Pt^{2+} + 2 e^- \rightarrow Pt\) with standard reduction potential, \(𝔼^\circ = 1.188 V\). 2. \(PtCl_4^{2-} + 2 e^- \rightarrow Pt + 4 Cl^-\) with standard reduction potential, \(𝔼^\circ = 0.755 V\). Both these reactions involve the reduction of platinum species. The first reaction (Pt²⁺ reduction) involves the reduction of platinum ions to metallic platinum, while the second reaction (PtCl₄²⁻ reduction) involves the reduction of platinum tetrachloride to metallic platinum.

The half-reaction for nitrate ions is: \(NO_3^{-} + 4 H^{+} + 3 e^{-} \rightarrow NO + 2 H_2O\), with standard reduction potential, \(𝔼^\circ = 0.96 V\). This half-reaction involves the reduction of nitrate ions to nitric oxide (NO).

Now, we need to compare the standard reduction potentials of these half-reactions to determine the possible reactions between the species in the presence of aqua regia. For platinum to dissolve, we need to find a reaction where it gets oxidized. Since the reduction of Pt²⁺ and PtCl₄²⁻ has positive standard reduction potentials (1.188 V and 0.755 V), the reverse reactions, i.e., the oxidation of Pt to Pt²⁺ and Pt to PtCl₄²⁻, will have negative standard reduction potentials.

Platinum metal will only dissolve if a counter half-reaction has a standard reduction potential high enough to facilitate its oxidation. If we look at the half-reaction for nitrate ions, the standard reduction potential is +0.96 V. This is not high enough to facilitate the oxidation of platinum metal since the potentials for the reverse reactions (Pt oxidation) are lower than this value (negative reduction potentials). Therefore, platinum will not dissolve in concentrated nitric acid. There is no given half-reaction for chloride ions directly acting as an oxidizing agent, but we may consider that the free chlorine formed from hydrochloric acid doesn't have a high enough standard reduction potential to oxidize platinum directly. Hence, platinum will not dissolve in concentrated hydrochloric acid as well.

However, when we mix nitric and hydrochloric acids to create aqua regia, we form chlorine gas (Cl₂) that acts as an oxidizing agent. The standard reduction potential of this reaction is: \(Cl_2 + 2e^- \rightarrow 2Cl^-\), with a standard reduction potential of \(𝔼^\circ = 1.36 V\). Since the standard reduction potential of Cl₂ is higher than the oxidation half-reaction of platinum (to form Pt²⁺ or PtCl₄²⁻), the platinum metal can dissolve in aqua regia. During this process, chlorine gas reacts with metallic platinum to form platinum chloride complexes, such as PtCl₄²⁻, which are soluble in aqua regia. In conclusion, platinum metal doesn't dissolve in concentrated nitric acid or hydrochloric acid individually because their standard reduction potentials are not sufficient to facilitate the oxidation of platinum. However, in aqua regia, the presence of Cl₂ generated from the mixture of both acids provides a higher standard reduction potential, allowing the oxidation of platinum to soluble platinum chloride complexes.