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

The best oxidizing agent of the following group of ions is (a) \(\mathrm{Ag}^{+}(\mathrm{aq}) ;\) (b) \(\mathrm{Cl}^{-}(\mathrm{aq}) ;\) (c) \(\mathrm{H}^{+}(\mathrm{aq})\) (d) \(\mathrm{Na}^{+}(\mathrm{aq}) ;\) (e) \(\mathrm{OH}^{-}(\mathrm{aq})\)

Short Answer

Expert verified
The best oxidizing agent among the given ions is Ag+.

Step by step solution

01

Understanding Oxidizing agents

An oxidizing agent causes the oxidation of other substances by taking their electrons and is itself reduced in the process. Thus, it can be evaluated with their reduction potential. The species with the highest reduction potential is typically the strongest oxidizing agent.
02

Identifying the strongest oxidizing agent

Looking to the standard reduction potentials for ions, the ions with more positive reduction potentials are the stronger reducing agents. The given ions are: Ag+, Cl-, H+, Na+ and OH-. Referring to a table of standard reduction potentials, these species have the following potentials: \n\nAg+ + e- -> Ag (Eº = +0.80 V)\n2H+ + 2e- -> H2 (Eº = 0.00 V)\n2Cl- -> Cl2 + 2e- (Eº = -1.36 V)\nNa+ + e- -> Na (Eº = -2.71 V)\n2OH- -> O2 + H2O + 4e- (Eº = -0.40 V)
03

Result

From the reduction potentials, the Ag+ ion has the most positive (hence the highest) reduction potential. Therefore, among the options, it is the best oxidizing agent.

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.

Standard Reduction Potential
In the world of chemistry, understanding the standard reduction potential is crucial for determining the behavior of ions during chemical reactions. The standard reduction potential indicates the tendency of a species to gain electrons, thus becoming reduced.
This potential is measured in volts and often expressed using a half-reaction. For example, the reduction of silver ion is given by:
\[ \mathrm{Ag}^{+} + e^{-} \rightarrow \mathrm{Ag} \quad (E^{\circ} = +0.80 \, \text{V}) \]
A positive potential value suggests that reduction is favorable, making the species an excellent oxidizing agent. The higher the positive value, the greater its ability to pull electrons towards itself.
Reduction Reactions
Reduction reactions are processes where a molecule, atom, or ion gains electrons. This "gain of electrons" can be remembered with the mnemonic "OIL RIG"—Oxidation Is Loss, Reduction Is Gain. During a reduction reaction, the charge on the species is decreased, often making it more stable.
These reactions are the opposite of oxidation reactions, where electrons are lost. In the case of our example ions, - When \(\mathrm{Ag}^{+}\) gains an electron, it becomes silver metal (Ag). - Conversely, \(\mathrm{Na}^{+}\) has a significantly negative reduction potential, making it a poor oxidizing agent.
Reduction reactions are integral to many natural and industrial processes, such as energy production in batteries and metal extraction.
Electron Transfer
Electron transfer is a fundamental process in chemistry where electrons move from one atom or molecule to another. This transfer is critical in redox reactions, where one species is oxidized (loses electrons) and the other is reduced (gains electrons).In our context, the transfer depends heavily on the standard reduction potentials of the involved species:
  • Species with higher (more positive) reduction potential are generally better at gaining electrons.
  • Species with lower (more negative) reduction potential are better at losing electrons.
For instance, silver ion (\(\mathrm{Ag}^{+}\)) has a stronger pull for electrons compared to \(\mathrm{Na}^{+}\), as indicated by their respective potential values. Thus, \(\mathrm{Ag}^{+}\) is more likely to participate in reactions as an oxidizing agent by accepting electrons.

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

In \(\mathrm{ZnO}\), the band gap between the valence and conduction bands is \(290 \mathrm{kJmol}^{-1}\), and in \(\mathrm{CdS}\) it is \(250 \mathrm{kJmol}^{-1} .\) Show that CdS absorbs some visible light but ZnO does not. Explain the observed colors: \(\mathrm{ZnO}\) is white and \(\mathrm{CdS}\) is yellow.

By means of orbital diagrams, write electron configurations for the following transition element atom and ions: \((a) \mathrm{Ti} ;(\mathbf{b}) \mathrm{V}^{3+} ;(\mathrm{c}) \mathrm{Cr}^{2+} ;(\mathrm{d}) \mathrm{Mn}^{4+} ;(\mathrm{e}) \mathrm{Mn}^{2+} ;(\mathrm{f}) \mathrm{Fe}^{3+}\).

Balance the following oxidation-reduction equations. $$\text { (a) } \mathrm{Fe}_{2} \mathrm{S}_{3}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{s})+\mathrm{S}(\mathrm{s})$$ $$\begin{aligned} &\text { (b) } \mathrm{Mn}^{2+}(\mathrm{aq})+\mathrm{S}_{2} \mathrm{O}_{8}^{2-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O} \longrightarrow\\\ &&\mathrm{MnO}_{4}^{-}(\mathrm{aq})+\mathrm{SO}_{4}^{2-}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq}) \end{aligned}$$ $$\begin{aligned} &\text { (c) } \mathrm{Ag}(\mathrm{s})+\mathrm{CN}^{-}(\mathrm{aq})+\mathrm{O}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O} \longrightarrow\\\ &&\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \end{aligned}$$

Why do the atomic radii vary so much more for two main-group elements that differ by one unit in atomic number than they do for two transition elements that differ by one unit?

Nearly all mercury(II) compounds exhibit covalent bonding. Mercury(II) chloride is a covalent molecule that dissolves in warm water. The stability of this compound is exploited in the determination of the levels of chloride ion in blood serum. Typical human blood serum levels range from 90 to \(115 \mathrm{mmol} \mathrm{L}^{-1}\) The chloride concentration is determined by titration with \(\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2} .\) The indicator used in the titration is diphenylcarbazone, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{N}=\mathrm{NCONHNHC}_{6} \mathrm{H}_{5}\) which complexes with the mercury(II) ion after all the chloride has reacted with the mercury(II). Free diphenylcarbazone is pink in solution, and when it is complexed with mercury(II), it is blue. Thus, the diphenylcarbazone acts as an indicator, changing from pink to blue when the first excess of mercury(II) appears. In an experiment, \(\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}(\) aq) solution is standardized by titrating \(2.00 \mathrm{mL}\) of \(0.0108 \mathrm{M} \mathrm{NaCl}\) solution. It takes \(1.12 \mathrm{mL}\) of \(\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})\) to reach the diphenylcarbazone end point. A 0.500 mL serum sample is treated with 3.50 mL water, 0.50 mL of 10\% sodium tungstate solution, and \(0.50 \mathrm{mL}\) of \(0.33 \mathrm{M}\) \(\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) to precipitate proteins. After the proteins are precipitated, the sample is filtered and a \(2.00 \mathrm{mL}\) aliquot of the filtrate is titrated with \(\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}\) solution, requiring \(1.23 \mathrm{mL}\). Calculate the concentration of Cl^- Express your answer in mmol L \(^{-1}\). Does this concentration fall in the normal range?

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