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

\(\mathrm{Fe}^{3+}\) is reduced to \(\mathrm{Fe}^{2+}\) by using (A) \(\mathrm{H}_{2} \mathrm{O}_{2}\) in presence of \(\mathrm{NaOH}\) (B) \(\mathrm{Na}_{2} \mathrm{O}_{2}\) in water (C) \(\mathrm{H}_{2} \mathrm{O}_{2}\) in presence of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (D) \(\mathrm{Na}_{2} \mathrm{O}_{2}\) in presence of \(\mathrm{H}_{2} \mathrm{SO}_{4}\)

Short Answer

Expert verified
Option (A) \(\mathrm{H}_2\mathrm{O}_2\) in presence of \(\mathrm{NaOH}\) is the correct reducing agent for turning \(\mathrm{Fe}^{3+}\) into \(\mathrm{Fe}^{2+}\).

Step by step solution

01

Analyzing Reducing Agents

To solve this exercise, one must understand the chemistry of reduction-oxidation (redox) reactions. In a redox reaction, the reduction of one species occurs concurrently with the oxidation of another. For the reduction of \(\mathrm{Fe}^{3+}\) to \(\mathrm{Fe}^{2+}\), we are looking for a reducing agent that can donate an electron to \(\mathrm{Fe}^{3+}\). Among the Options (A) and (B), both \(\mathrm{H}_2\mathrm{O}_2\) and \(\mathrm{Na}_2\mathrm{O}_2\) can act as oxidizing as well as reducing agents, depending on the conditions. Option (C) shows \(\mathrm{H}_2\mathrm{O}_2\) in an acidic medium, which generally favors its role as an oxidizing agent. Option (D) shows \(\mathrm{Na}_2\mathrm{O}_2\) in an acidic medium, specifically with \(\mathrm{H}_2\mathrm{SO}_4\), which will release \(\mathrm{O}_2\) that can potentially oxidize \(\mathrm{Fe}^{2+}\) back to \(\mathrm{Fe}^{3+}\).
02

Determining the Suitable Conditions

Reducing \(\mathrm{Fe}^{3+}\) to \(\mathrm{Fe}^{2+}\) is more favorable in an alkaline solution where the reducing potential of \(\mathrm{H}_2\mathrm{O}_2\) increases. Therefore, Option (A), which uses \(\mathrm{H}_2\mathrm{O}_2\) in the presence of \(\mathrm{NaOH}\) (an alkaline condition), is the best choice because it provides the necessary conditions for \(\mathrm{H}_2\mathrm{O}_2\) to act as a reducing agent. This will lead to the reduction of \(\mathrm{Fe}^{3+}\) ions to \(\mathrm{Fe}^{2+}\) ions.
03

Conclusion

After analyzing all the given reactants and the conditions under which they are used, Option (A) \(\mathrm{H}_2\mathrm{O}_2\) in presence of \(\mathrm{NaOH}\) is the correct choice for the reduction of \(\mathrm{Fe}^{3+}\) to \(\mathrm{Fe}^{2+}\).

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.

Reducing Agents
In the world of chemistry, reducing agents play a pivotal role in redox reactions. A reducing agent, by definition, is a substance that donates electrons to another species, causing the reduction of the latter. In the simplification of our textbook problem, the quest was to identify an agent capable of reducing Fe^3+ to Fe^2+.

Reducing agents can come in various forms and their efficacy is influenced by the chemical environment. For instance, hydrogen peroxide (H2O2) can function both as a reducing and an oxidizing agent. Factors such as the pH of the solution determine which role hydrogen peroxide will take. In an alkaline medium like sodium hydroxide (NaOH), it tends more towards being a reducing agent, which is why option (A) is our correct answer for the exercise.

Understanding the dual nature of such chemicals is crucial for students tackling reduction-oxidation concepts in their JEE Advanced preparation.
Oxidation States
The oxidation state, or oxidation number, is a key concept in the study of redox reactions. It allows us to track the transfer of electrons during chemical reactions by assigning a hypothetical charge to atoms in a molecule, given certain rules. In our example, the oxidation state changes from +3 in Fe^3+ to +2 in Fe^2+, indicative of a gain of one electron by the iron ion, hence a reduction process.

During JEE Advanced preparations, it is imperative for students to be adept at determining oxidation states to predict the flow of electrons. For instance, Fe^3+ is in a higher oxidation state compared to Fe^2+, and thus, can be reduced by a suitable reducing agent. Recognizing the changes in oxidation states in chemical reactions leads to a better understanding of the overall redox process.
Chemical Reactions
Chemical reactions are the bread and butter of chemistry, and understanding how different reactants interact in various conditions is a fundamental skill for any JEE Advanced aspirant. Redox reactions are a specific type of chemical reaction where oxidation and reduction occur simultaneously.

Our guiding example involves the reduction of iron ions in different scenarios. Reactions can be influenced by several factors, including the presence of a catalyst, temperature, concentration, and pH, as seen with the role of NaOH in promoting the reducing capability of hydrogen peroxide. By grasping the basics of chemical reactions and the conditions that affect them, students can solve complex problems by breaking them down into simpler parts and applying critical thinking and knowledge of chemical principles.

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

Match the nuclear processes given in column I with the appropriate option(s) in column II. Column I (A) Nuclear fusion (B) Fission in a nuclear reactor (C) \(\beta\) -decay (D) \(\gamma\) -ray emission Column II (P) Absorption of thermal neutrons by \({ }_{92}^{235} \mathrm{U}\) (Q) \({ }_{27}^{60} \mathrm{C}\) o nucleus (R) Energy production in stars via hydrogen conversion to helium (S) Heavy water (T) Neutrino emission

The total number of lone pairs of electrons in \(\mathrm{N}_{2} \mathrm{O}_{3}\) is

A bullet is fired vertically upwards with velocity \(v\) from the surface of a spherical planet. When it reaches its maximum height, its acceleration due to the planet's gravity is \(1 / 4^{\text {th }}\) of its value at the surface of the planet. If the escape velocity from the planet is \(v_{\text {esc }}=v \sqrt{N}\), then the value of \(N\) is (ignore energy loss due to atmosphere)

Among the triatomic molecules/ions, \(\mathrm{BeCl}_{2}, \mathrm{~N}_{3}^{-}, \mathrm{N}_{2} \mathrm{O}, \mathrm{NO}_{2}^{+}, \mathrm{O}_{3}, \mathrm{SCl}_{2}, \mathrm{ICl}_{2}^{-}, \mathrm{I}_{3}^{-}\) and \(\mathrm{XeF}_{2}\), the total number of linear molecule(s)/ion(s) where the hybridization of the central atom does not have contribution from the \(d\) -orbital(s) is [Atomic number : \(\mathrm{S}=16, \mathrm{Cl}=17, \mathrm{I}=53\) and \(\mathrm{Xe}=54]\)

Let \(y(x)\) be a solution of the differential equation \(\left(1+e^{x}\right) y^{\prime}+y e^{x}=1 .\) If \(y(0)=2\), then which of the following statements is (are) true? (A) \(y(-4)=0\) (B) \(\quad y(-2)=0\) (C) \(y(x)\) has a critical point in the interval \((-1,0)\) (D) \(y(x)\) has no critical point in the interval \((-1,0)\)

See all solutions

Recommended explanations on English 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