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

To the aqueous solution of the salt acidified potassium permanganate is added and its colour is discharged. It indicates the absence of (a) \(\mathrm{Sn}^{2+}\) (b) \(\mathrm{Be}^{2+}\) (c) \(\mathrm{Fe}^{2+}\) (d) \(\mathrm{NO}_{3}^{-}\)

Short Answer

Expert verified
The absence of \( \text{Be}^{2+} \) is indicated.

Step by step solution

01

Understand the Role of Acidified Potassium Permanganate

Acidified potassium permanganate (\( \text{KMnO}_4 \)) is a strong oxidizing agent. Its violet color is discharged (fades) when it is reduced during a reaction with a reducing agent.
02

Identify Reducing Agents Amongst the Options

Review the options to identify which of the ions can act as reducing agents. Reducing agents are substances that lose electrons during reactions (are oxidized) and reduce another substance by giving away electrons.
03

Analyze the Properties of Each Ion

- \( \text{Sn}^{2+} \) is known to be a reducing agent as it can be oxidized to \( \text{Sn}^{4+} \).- \( \text{Be}^{2+} \) is not a reducing agent and typically does not undergo redox reactions.- \( \text{Fe}^{2+} \) is a common reducing agent, capable of being oxidized to \( \text{Fe}^{3+} \).- \( \text{NO}_3^- \) is generally not a reducing agent in acidified solutions.
04

Draw the Conclusion Based on Reducibility

Since \( \text{Be}^{2+} \) does not act as a reducing agent and does not participate in redox reactions that discharge the violet color of potassium permanganate, it indicates the absence of \( \text{Be}^{2+} \).

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.

Oxidizing Agent
In the realm of redox reactions, an oxidizing agent plays a crucial role. It is a substance that accepts electrons during a chemical reaction. By gaining electrons, the oxidizing agent causes another substance to lose electrons, hence undergoing oxidation. This mechanism is fundamental in redox chemistry and is often referred to as "being reduced," as the oxidizing agent itself is reduced. An interesting point is that oxidizing agents often include elements in high oxidation states or compounds with high electronegative elements, like oxygen or fluorine.
  • Common examples of oxidizing agents include \(\text{KMnO}_4\) (potassium permanganate), \(\text{H}_2\text{O}_2\) (hydrogen peroxide), and nitric acid.
  • In the case of the exercise, acidified potassium permanganate acts as the oxidizing agent.
Thus, the vibrant purple color of potassium permanganate solution originates from it functioning in its oxidized form, \(\text{MnO}_4^-\), which fades as it is reduced in reactions with reducing agents.
Reducing Agent
A reducing agent is a fascinating component of redox reactions. Unlike an oxidizing agent, it donates electrons to another substance. This donation results in the reducing agent undergoing oxidation itself. It's essentially a substance that loses electrons and gains a more positive charge. In the solution provided, identifying the reducing agent is key to understanding which ion would not discharge the color of potassium permanganate.
  • Common reducing agents include elements or compounds in lower oxidation states capable of losing electrons, such as \(\text{Sn}^{2+}\) and \(\text{Fe}^{2+}\).
  • For instance, \(\text{Sn}^{2+}\) can donate electrons to become \(\text{Sn}^{4+}\), and likewise, \(\text{Fe}^{2+}\) can oxidize to \(\text{Fe}^{3+}\).
One of the choices, \(\text{Be}^{2+}\), does not act as a reducing agent in this context, explaining its lack of activity in fading the potassium permanganate.
Potassium Permanganate
Potassium permanganate, with the chemical formula \(\text{KMnO}_4\), is renowned for its properties as an oxidizing agent. It is used extensively in industrial and chemistry labs because of its ability to accept electrons and oxidize other substances. When dissolved in water, potassium permanganate gives a distinct purple color due to the presence of \(\text{MnO}_4^-\) ions, which are in a high oxidation state.
  • The solution becomes colorless or slightly brown as it reduces, indicating a successful redox reaction.
  • Its effectiveness as an oxidizing agent is enhanced when acidified, as in the exercise.
Therefore, when the color of potassium permanganate fades, it indicates the reduction of the \(\text{MnO}_4^-\) ion by a reducing agent, confirming its ability to oxidize other substances.
Transition Metal Ions
Transition metal ions have intriguing roles in chemistry due to their ability to exist in various oxidation states. This property allows them to be versatile in redox reactions, often acting as either oxidizing or reducing agents. They are typically characterized by unfilled d electron subshells, which provide opportunities for electron exchange processes.
  • Examples from the given exercise include \(\text{Fe}^{2+}\) and \(\text{Sn}^{2+}\), both capable of oxidation to a higher state (\(\text{Fe}^{3+}\) and \(\text{Sn}^{4+}\), respectively).
  • Their ability to transition between different oxidation states is crucial for their role as reducing agents.
Transition metal ions often lead vibrant chemical reactions, participating actively by gaining or losing electrons, which is central in redox chemistry and many practical applications.

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

\(20 \mathrm{~mL}\) of \(0.1 \mathrm{M} \mathrm{NaOH}\) and \(30 \mathrm{~mL}\) of \(0.12 \mathrm{M} \mathrm{NaHCO}_{3}\) solutions are mixed. \(25 \mathrm{ml}\) of the resulting solution are titrated against \(0.1 \mathrm{M} \mathrm{HCl}\) using phenolphthalein as the indicator. What would be the titre value? (a) \(100 \mathrm{~mL}\) (b) \(10 \mathrm{~mL}\) (c) \(25 \mathrm{~mL}\) (d) \(20 \mathrm{~mL}\)

Dimethyl glyoxime in a suitable solvent was refluxed for 10 minutes with pure pieces of nickel sheet, it will result in (a) blue precipitate (b) yellow precipitate (c) red precipitate (d) no precipitate

A gas ' \(X\) ' is passed through water to form a saturated solution. The aqueous solution on treatment with the \(\mathrm{AgNO}_{3}\) gives a white precipitate. The saturated aqueous solution also dissolves magnesium ribbon with evolution of a colourless gas ' \(\mathrm{Y}\) '. Identify ' \(\mathrm{X}\) ' and ' \(\mathrm{Y}\) '. (a) \(\mathrm{X}=\mathrm{Cl}_{2}, \mathrm{Y}=\mathrm{H}_{2}\) (b) \(\mathrm{X}=\mathrm{H}_{2}, \mathrm{Y}=\mathrm{Cl}_{2}\) (c) \(\mathrm{X}=\mathrm{Cl}_{2}, \mathrm{Y}=\mathrm{CO}_{2}\) (d) \(\mathrm{X}=\mathrm{CO}_{2}, \mathrm{Y}=\mathrm{Cl}_{2}\)

How do we differentiate between \(\mathrm{Fe}^{3+}\) and \(\mathrm{Cr}^{3+}\) in group III? (a) by increasing \(\mathrm{NH}_{4}{ }^{+}\)ion concentration (b) by decreasing \(\mathrm{OH}^{-}\)ion concentration (c) by adding excess of \(\mathrm{NH}_{4} \mathrm{OH}\) solution (d) both (a) and (b)

Consider the following observation \(\mathrm{M}^{\mathrm{n}+}+\mathrm{HCl} \longrightarrow \mathrm{white}\) precipitate \(\stackrel{\Delta}{\longrightarrow}\) Water soluble The metal ion \(\mathrm{M}^{\mathrm{n}+}\) will be (a) \(\mathrm{Pb}^{2+}\) (b) \(\mathrm{Hg}^{2+}\) (c) \(\mathrm{Sn}^{2+}\) (d) \(\mathrm{Ag}^{+}\)

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