/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 116 Consider the following half-reac... [FREE SOLUTION] | 91影视

91影视

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

Short Answer

Expert verified
Platinum metal dissolves in aqua regia due to the combined effect of nitric and hydrochloric acids that create a favorable redox reaction with a standard cell potential of 0.228 V. Individually, neither nitric acid (E鈧 = 0.228 V) nor hydrochloric acid (E鈧 = 0.433 V) can create a sufficient standard cell potential to dissolve platinum metal.

Step by step solution

01

1. Determine the redox reactions in different conditions:

Here, we will determine the overall redox reactions taking place when Pt metal react with a) aqua regia, b) nitric acid, and c) hydrochloric acid. a) In aqua regia, the half-reactions involved are the reduction of Pt虏鈦 to Pt, and the reduction of NO鈧冣伝 to NO. We will add the half-reactions to obtain the overall redox reaction. b) In nitric acid, the half-reactions involved are the reduction of Pt虏鈦 to Pt, and the reduction of NO鈧冣伝 to NO. c) In hydrochloric acid, the half-reactions involved are the reduction of Pt虏鈦 to Pt, and the reduction of PtCl鈧劼测伝 to Pt and Cl鈦.
02

2. Determine the standard cell potential in different conditions:

In this step, we will calculate the standard cell potentials (E鈧) of the redox reactions to understand which reaction is favorable in each condition by using the Nernst equation: E鈧(cell) = E鈧(reduction) - E鈧(oxidation) a) In aqua regia, E鈧 = E鈧(Pt虏鈦 鈫 Pt) - E鈧(NO鈧冣伝 鈫 NO) = 1.188 V - 0.96 V = 0.228 V b) In nitric acid, E鈧 = E鈧(Pt虏鈦 鈫 Pt) - E鈧(NO鈧冣伝 鈫 NO) = 1.188 V - 0.96 V = 0.228 V c) In hydrochloric acid, E鈧 = E鈧(Pt虏鈦 鈫 Pt) - E鈧(PtCl鈧劼测伝 鈫 Pt) = 1.188 V - 0.755 V = 0.433 V
03

3. Analyze the favorability of the redox reactions:

Based on the standard cell potentials calculated above, we can analyze the favorability of the redox reactions in each condition: a) In aqua regia, the standard cell potential E鈧 = 0.228 V, indicates that the redox reaction is favorable, and platinum metal will dissolve in aqua regia. b) In nitric acid, the standard cell potential E鈧 = 0.228 V, is positive but not large enough to facilitate the dissolution of platinum metal. This indicates that the reaction with nitric acid alone is not sufficient to dissolve platinum-metal. c) In hydrochloric acid, the standard cell potential E鈧 = 0.433 V, indicates that the reaction is not favorable, and platinum metal will not dissolve in hydrochloric acid. In conclusion, platinum metal dissolves in aqua regia due to the combined effect of the nitric and hydrochloric acids that create a favorable redox reaction. Individually, neither nitric nor hydrochloric acid can create a sufficient standard cell potential to dissolve platinum metal.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Standard Cell Potential
Standard cell potential, represented as \( \mathscr{E}^{\circ} \), is a measure of the voltage difference between two half-cells in an electrochemical cell. This potential helps predict whether a reaction will occur spontaneously. If the cell potential \( \mathscr{E}^{\circ} \) is greater than zero, the redox reaction is considered to be thermodynamically favorable, meaning it can proceed on its own without the input of external energy.

In the context of platinum dissolving in aqua regia, we calculate the standard cell potentials by considering the different half-reactions involved. By comparing the standard reduction potentials of these half-reactions, we can determine the overall cell potential for the reactions.

For instance, in aqua regia, the standard cell potential is calculated by subtracting the oxidation potential of nitric oxide from that of platinum (1.188 V for platinum and 0.96 V for NO). This results in a positive value of 0.228 V, indicating a favorable reaction condition allowing platinum to dissolve in aqua regia.
Half-reactions
Half-reactions are critical components of redox reactions and involve either the gain or loss of electrons. They show how electrons are transferred in a reaction, often breaking down complex processes into simpler, more understandable steps.

For the dissolution of platinum in aqua regia, there are two noteworthy half-reactions. The first involves the reduction of platinum ions (\(\mathrm{Pt}^{2+}\)) to platinum metal, capturing electrons in the process. The second half-reaction involves the reduction of nitrate ions (\(\mathrm{NO}_{3}^{-}\)) to nitric oxide (\(\mathrm{NO}\)), also requiring the transfer of electrons.

These half-reactions are crucial because they provide insight into which substances act as oxidizing agents and reducing agents within the solution. Understanding how these agents interact allows chemists to recognize why certain reactions, like that of platinum in aqua regia, are successful or not in various conditions.
Aqua Regia
Aqua Regia is a powerful and corrosive mixture of concentrated nitric acid and hydrochloric acid, typically in a 1:3 ratio. Known for its ability to dissolve noble metals like gold and platinum, aqua regia achieves this by combining the oxidative power of its component acids.

The action of aqua regia is due to its ability to generate free chlorine and nitrosyl chloride, which serve as excellent oxidizing agents. When platinum metal is introduced into aqua regia, the simultaneous oxidizing actions of both acids create conditions that promote the dissolution of platinum, a metal typically resistant to corrosion due to its stable electronic configuration.

Neither concentrated nitric acid nor hydrochloric acid individually can facilitate this reaction. This is because the oxidizing potential of each acid alone isn't sufficient to overcome the inertness of platinum, highlighting the unique and combined reactivity of aqua regia.
Nernst Equation
The Nernst Equation is a critical tool in electrochemistry, allowing us to determine the cell potential at non-standard conditions. It adjusts the standard cell potential by accounting for the concentration of the reactants and products in a cell.

The equation is expressed as: \[E = E^{\circ} - \frac{RT}{nF} \ln Q\]where \(E\) is the cell potential, \(E^{\circ}\) is the standard cell potential, \(R\) is the universal gas constant, \(T\) is the temperature in kelvin, \(n\) is the number of moles of electrons transferred, \(F\) is the Faraday constant, and \(Q\) is the reaction quotient.

Using the Nernst Equation can help explain why reactions that are not favorable under standard conditions might become feasible under different concentrations, temperatures, or pressures. It's crucial for understanding the actual behavior of redox reactions in real-world scenarios, such as how a system like aqua regia can effectively dissolve a typically inert metal like platinum.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

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 $$ \operatorname{CuI}(s)+\mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s)+\mathrm{I}^{-}(a q) $$

The following standard reduction potentials have been det mined for the aqueous chemistry of indium: $$ \begin{array}{cl} \mathrm{In}^{3+}(a q)+2 \mathrm{e}^{-} \longrightarrow \operatorname{In}^{+}(a q) & \mathscr{E}^{\circ}=-0.444 \mathrm{~V} \\ \mathrm{In}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \operatorname{In}(s) & \mathscr{E}^{\circ}=-0.126 \mathrm{~V} \end{array} $$ a. What is the equilibrium constant for the disproportionation reaction, where a species is both oxidized and reduced, shown below? $$ 3 \operatorname{In}^{+}(a q) \longrightarrow 2 \operatorname{In}(s)+\operatorname{In}^{3+}(a q) $$ b. What is \(\Delta G_{\mathrm{f}}^{\circ}\) for \(\operatorname{In}^{+}(a q)\) if \(\Delta G_{\mathrm{f}}^{\circ}=-97.9 \mathrm{~kJ} / \mathrm{mol}\) for \(\mathrm{In}^{3+}(a q) ?\)

What volumes of \(\mathrm{H}_{2}(g)\) and \(\mathrm{O}_{2}(g)\) at STP are produced from the electrolysis of water by a current of \(2.50 \mathrm{~A}\) in \(15.0 \mathrm{~min} ?\)

An aqueous solution of \(\mathrm{PdCl}_{2}\) is electrolyzed for \(48.6\) seconds, and during this time \(0.1064 \mathrm{~g}\) of \(\mathrm{Pd}\) is deposited on the cathode. What is the average current used in the electrolysis?

A chemist wishes to determine the concentration of \(\mathrm{CrO}_{4}^{2}\) electrochemically. A cell is constructed consisting of a saturated calomel electrode (SCE; see Exercise 115 ) and a silver wire coated with \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\). The \(\mathscr{E}^{\circ}\) value for the following half-reaction is \(0.446 \mathrm{~V}\) relative to the standard hydrogen electrode: $$ \mathrm{Ag}_{2} \mathrm{CrO}_{4}+2 \mathrm{c}^{-} \longrightarrow 2 \mathrm{Ag}+\mathrm{CrO}_{4}{ }^{2-} $$ a. Calculate \(\mathscr{E}_{\text {cell }}\) and \(\Delta G\) at \(25^{\circ} \mathrm{C}\) for the cell reaction when \(\left[\mathrm{CrO}_{4}{ }^{2-}\right]=1.00 \mathrm{~mol} / \mathrm{L}\) b. Write the Nernst equation for the cell. Assume that the SCE concentrations are constant. c. If the coated silver wire is placed in a solution (at \(25^{\circ} \mathrm{C}\) ) in which \(\left[\mathrm{CrO}_{4}{ }^{2-}\right]=1.00 \times 10^{-5} M\), what is the expected cell potential? d. The measured cell potential at \(25^{\circ} \mathrm{C}\) is \(0.504 \mathrm{~V}\) when the coated wire is dipped into a solution of unknown \(\left[\mathrm{CrO}_{4}{ }^{2-}\right]\). What is \(\left[\mathrm{CrO}_{4}{ }^{2-}\right]\) for this solution? e. Using data from this problem and from Table \(18.1\), calculate the solubility product \(\left(K_{\mathrm{sp}}\right)\) for \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.