Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Zn gives \(\mathrm{H}_{2}\) gas with \(\mathrm{H}_{2} \mathrm{SO}_{4}\) and \(\mathrm{HCl}\) but not with \(\mathrm{HNO}_{3}\) because (a) \(\mathrm{Zn}\) acts as oxidizing agent when reacts with \(\mathrm{HNO}_{3}\) (b) \(\mathrm{HNO}_{3}\) is weaker acid then \(\mathrm{H}_{2} \mathrm{SO}_{4}\) and \(\mathrm{HCl}\) (c) in electrochemical series \(\mathrm{Zn}\) is above hydrogen (d) \(\mathrm{NO}_{3}^{-}\)is reduced in preference to hydronium ion

Short Answer

Expert verified
(d) NO3^- is reduced in preference to hydronium ion.

Step by step solution

01

Understanding the Reaction with HCl and H2SO4

When zinc (Zn) reacts with hydrochloric acid (HCl) or sulfuric acid (H2SO4), it displaces the hydrogen, resulting in the production of hydrogen gas (H2) due to the position of zinc above hydrogen in the electrochemical series. Thus, for these reactions, Zn acts effectively as a reducing agent that donates electrons to the hydrogen ions, producing H2 gas.
02

Explaining the Reaction with HNO3

In the case of zinc reacting with nitric acid (HNO3), the situation is different. Nitric acid is a powerful oxidizing agent, which means it has a strong tendency to accept electrons. Hence, instead of facilitating the production of hydrogen gas, in this reaction, the nitrate ion (NO3^-) in HNO3 gets reduced preferentially over hydronium ions (H+). This results in a different reaction product and no evolution of hydrogen gas.
03

Analyzing the Electrochemical Series

Zinc (Zn) is above hydrogen in the electrochemical series, indicating that under normal circumstances, Zn can displace hydrogen ions, leading to the evolution of H2. However, this principle mainly applies when the acid in question does not have a component that is more favorably reduced, like NO3^- in HNO3.
04

Conclusion on HNO3 Interaction

Given these findings, we can conclude that the presence of NO3^- in HNO3, being a better electron acceptor, prevents the release of H2 gas by causing preferential reduction over the hydronium (H+) ions.

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 Vaia!

Key Concepts

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

Electrochemical Series
The electrochemical series is a useful tool to determine the ease with which different elements can either donate or accept electrons. This sequence lists metals and ions according to their respective Standard Electrode Potentials. It indicates how likely a metal is to oxidize (lose electrons) or how keen an ion is to reduce (gain electrons).

Zinc (Zn), for instance, is positioned above hydrogen in this series. This tells us that zinc can more readily donate its electrons to ions such as those found in acids. As a result, when Zn interacts with acids like hydrochloric (\(\mathrm{HCl}\)) or sulfuric (\(\mathrm{H}_{2}\mathrm{SO}_{4}\)) acid, it tends to displace hydrogen ions, leading to hydrogen gas (\(\mathrm{H}_{2}\)) production. However, not all reactions behave the same due to the varying nature of the acids involved.
Oxidation-Reduction Reactions
Oxidation-reduction reactions, commonly known as redox reactions, are chemical processes where one substance loses electrons (oxidation) and another gains electrons (reduction). These reactions are fundamental to understanding the behavior of metals like zinc in acidic environments.

In the scenario of zinc with hydrochloric acid or sulfuric acid, zinc serves as the reducing agent. It donates electrons to hydrogen ions, leading to the formation of hydrogen gas. The essence of these reactions is the transfer of electrons, where the zinc oxidizes by losing electrons, and the hydrogen ions reduce by gaining them. Conversely, when zinc reacts with nitric acid, the reduction does not occur with the hydrogen ions but rather with the nitrate ions, preventing hydrogen gas production.
Hydrogen Gas Production
The production of hydrogen gas is a fascinating aspect of chemical reactions involving metals and acids. When zinc reacts with hydrochloric or sulfuric acid, the interactions demonstrate a clear example of hydrogen gas formation. This occurs because zinc reduces the hydrogen ions, converting them into hydrogen molecules, which then form hydrogen gas.

The underlying reason behind this gas production is due to zinc's position in the electrochemical series. Since it is above hydrogen, it effectively donates electrons, facilitating the gas release. However, different acids influence the reaction outcomes, as seen with nitric acid, which prevents the typical hydrogen gas evolution due to its oxidizing properties.
Acid-Base Reactions
In acid-base reactions, acids are substances that release hydrogen ions (\(\mathrm{H}^{+}\)) in solutions, while bases are those that accept hydrogen ions. Metals like zinc, when introduced to acidic solutions, can lead to significant changes in the ionic landscape.

In the interaction between zinc and acids such as hydrochloric acid or sulfuric acid, the hydrogen ions from the acids engage actively with the zinc metal. Zinc, having a higher potential to oxidize, helps transfer electrons to these ions, resulting in hydrogen gas creation.

However, acids have varied behaviors and components. Nitric acid, being a strong oxidizing agent, interacts differently as it prefers to reduce its nitrate component over hydrogen ions, which means no hydrogen gas is produced. Understanding this differential behavior is crucial in predicting the results of acid-metal reactions.

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

Given below are the half-cell reactions: \(\mathrm{Mn}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Mn} ; \mathrm{E}^{\circ}=-1.18 \mathrm{~V}\) \(2\left(\mathrm{Mn}^{3+}+\mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2}+\right) ; \mathrm{E}^{\circ}=+1.51 \mathrm{~V}\) The \(\mathrm{E}^{\circ}\) for \(3 \mathrm{Mn}^{2+} \rightarrow \mathrm{Mn}+2 \mathrm{Mn}^{3+}\) will be: (a) \(-0.33 \mathrm{~V}\); the reaction will not occur (b) \(-0.33 \mathrm{~V}\); the reaction will occur (c) \(-2.69 \mathrm{~V}\); the reaction will not occur (d) \(-2.69 \mathrm{~V}\); the reaction will occur

If the pressure of hydrogen gas is increased from \(1 \mathrm{arm}\) to \(100 \mathrm{~atm}\), keeping the hydrogen ion concentration constant at \(1 \mathrm{M}\), the voltage of the hydrogen half cell at \(25^{\circ} \mathrm{C}\) will be (a) \(-0.059 \mathrm{~V}\) (b) \(+0.059 \mathrm{~V}\) (c) \(5.09 \mathrm{~V}\) (d) \(0.259 \mathrm{~V}\)

For a spontaneous reaction the \(\Delta \mathrm{G}\), equilibrium constant \((\mathrm{K})\) and \(E_{\mathrm{ccll}}^{\circ}\) will be respectively (a) \(-\mathrm{ve},>1,+\mathrm{ve}\) (b) \(+\mathrm{ve},>1,-\mathrm{ve}\) (c) \(-\mathrm{ve},<1,-\mathrm{ve}\) (d) \(-v \mathrm{e},>1,-v e\)

The specific conductance of \(0.1 \mathrm{~N} \mathrm{KCl}\) solution at \(23^{\circ} \mathrm{C}\) is \(0.012 \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}\). The resistance of cell containing the solution at the same temperature was found to be 55 ohm. The cell constant will be (a) \(0.142 \mathrm{~cm}^{-1}\) (b) \(0.616 \mathrm{~cm}^{-1}\) (c) \(6.16 \mathrm{~cm}^{-1}\) (d) \(616 \mathrm{~cm}^{-1}\)

In the ionic equation \(\mathrm{xBrO}_{3}^{-}+\mathrm{yCr}^{3+}+\mathrm{zH}_{2} \mathrm{O} \longrightarrow \mathrm{Br}_{2}+\mathrm{HCrO}_{4}^{-}+\mathrm{H}^{+}\) (a) \(\mathrm{x}=6, \mathrm{y}=8, \mathrm{z}=10\) (b) \(x=10, y=6, z=22\) (c) \(x=6, y=8, z=10\) (d) \(x=6, y=10, z=22\)

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.

Sign-up for free