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Among the following, identify the correct statement. a. Chloride ion is oxidized by \(\mathrm{O}_{2}\) b. \(\mathrm{Fe}^{2+}\) is oxidized by iodine c. Iodide ion is oxidized by chlorine d. \(\mathrm{Mn}^{2+}\) is oxidized by chlorine

Short Answer

Expert verified
Option c: Iodide ion is oxidized by chlorine.

Step by step solution

01

Analyzing the Given Options

We are provided with a list of statements about oxidation reactions involving ions and elements. We need to determine which pair undergoes a spontaneous oxidation-reduction (redox) reaction.
02

Understanding Oxidation and Reduction

In a redox reaction, oxidation refers to the loss of electrons, whereas reduction refers to the gain of electrons. An element or ion that causes another species to oxidize is called the oxidizing agent.
03

Reviewing Each Option

Let's examine each statement: - **Option a:** Chloride ion (Cl鈦) cannot be oxidized by O鈧 under normal conditions because O鈧 is not a strong enough oxidizing agent for Cl鈦. - **Option b:** Fe虏鈦 being oxidized by iodine (I鈧) is incorrect because Fe虏鈦 has a higher reduction potential than I鈧. - **Option c:** Iodide ion (I鈦) can indeed be oxidized by chlorine (Cl鈧) as chlorine is a stronger oxidizing agent. - **Option d:** Mn虏鈦 is less likely to be oxidized by chlorine under typical lab conditions without specific catalysts or conditions.
04

Selecting the Correct Statement

After analyzing each option, the statement that involves a feasible redox reaction is the option where iodide ions are oxidized by chlorine. This can be represented as: 2 I鈦 + Cl鈧 鈫 2 Cl鈦 + I鈧. Chlorine can oxidize the iodide ion to iodine.

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Key Concepts

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

Oxidation
Oxidation is a fundamental concept in chemistry that involves the loss of electrons by a substance. In a redox (reduction-oxidation) reaction, when one substance undergoes oxidation, another substance must be simultaneously reduced to maintain electron balance. The species losing electrons is said to be oxidized. For example, in the reaction where iodide ions ( I^鈦 ) are converted to iodine ( I鈧 ) by chlorine ( Cl鈧 ), iodide ions lose electrons. This process can be observed in the chemical equation: 2 I^鈦 + Cl鈧 鈫 2 Cl鈦 + I鈧 . Here, iodide ions lose electrons to chlorine, demonstrating the principle of oxidation.
  • Electron loss characterizes oxidation.
  • Always occurs alongside reduction in redox reactions.
  • Examples include metal corrosion and cellular respiration.
Recognizing oxidation helps in understanding important chemical processes and their applications in everyday life.
Oxidizing Agent
In chemical reactions, an oxidizing agent is a substance that induces oxidation in another chemical species by gaining electrons. It is the compound that undergoes reduction in the process. Oxidizing agents are crucial for facilitating redox reactions because they enable certain substances to lose electrons while they themselves gain them, achieving electron transfer balance. For example, chlorine is a common oxidizing agent. In the reaction 2 I^鈭 + Cl鈧 鈫 2 Cl鈦 + I鈧 , chlorine ( Cl鈧 ) acts as the oxidizing agent, facilitating the oxidation of iodide ions by accepting the electrons released by them.
  • Oxidizing agents accept electrons and are reduced.
  • They are essential for energy transfer processes.
  • Common examples include oxygen, chlorine, and hydrogen peroxide.
Understanding oxidizing agents is vital for studying chemical reactivity and designing chemical processes.
Reduction
Reduction is an essential redox concept and refers to the gain of electrons by a molecule, atom, or ion. This process is always accompanied by oxidation in a redox reaction. To understand reduction, consider where chlorine ( Cl鈧 ) is involved with iodide ions ( I^- ). In the conversion: 2 I^鈭 + Cl鈧 鈫 2 Cl鈦 + I鈧 , chlorine gains electrons and gets converted to chloride ions ( Cl鈦 ), thereby undergoing reduction. The term 'reduction' is derived from the Latin word 'reductio,' meaning "to lead back," referencing the decrease in oxidation state.
  • Also termed electron gain.
  • Accompanies oxidation for redox balance.
  • Vital in processes like photosynthesis and metabolism.
Reduction is a cornerstone concept in physical sciences, especially in understanding energy transfer mechanisms.
Reduction Potential
Reduction potential is a measure of the tendency of a chemical species to acquire electrons and thereby be reduced. It is often expressed as voltage under standard conditions. The higher the reduction potential, the greater the species' affinity for electrons and its likelihood to act as an oxidizing agent. For instance, a higher reduction potential indicates more robust oxidizing power, while a lower reduction potential implies readiness to donate electrons, often acting as a reducing agent instead. Standard values (E鈦) help predict the spontaneity of redox reactions. For example, the standard reduction potential for chlorine is higher than that for iodide, meaning chlorine is more likely to gain electrons than iodide. This explains why chlorine is a suitable agent to oxidize iodide ions.
  • Higher values indicate stronger oxidizing ability.
  • Used to predict electron flow direction in electrochemical cells.
  • Essential in fields like electrochemistry and biology.
Comprehension of reduction potential aids in predicting and manipulating chemical reactions in practical applications.

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Most popular questions from this chapter

In the electrolytic cell, flow of electrons is from: a. cathode to anode in solution. b. cathode to anode through external supply. c. cathode to anode through internal supply. d. anode to cathode through internal supply.

Electrolysis of dilute aqueous \(\mathrm{NaCl}\) solution was carried out by passing \(10 \mathrm{~mA}\) current. The time required to liberate \(0.01\) mole of \(\mathrm{H}_{2}\) gas at the cathode is \(\left(1 \text { Faraday }=96500 \mathrm{C} \mathrm{mol}^{-1}\right)^{-}\)) a. \(9.65 \times 10^{4} \mathrm{~s}\) b. \(19.3 \times 10^{4} \mathrm{~s}\) c. \(28.95 \times 10^{4} \mathrm{~s}\) d. \(38.6 \times 10^{4} \mathrm{~s}\)

In the redox reaction, \(\mathrm{MnO}_{4}^{\theta}+\mathrm{I}_{2}+\mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+}+\mathrm{IO}_{3}^{\theta}+\mathrm{H}_{2} \mathrm{O}\) \(\mathrm{MnO}_{4}^{\theta}\) and \(\mathrm{I}_{2}\) would react in the molar ratio of: a. \(3: 1\) b. \(1: 3\) c. \(2: 1\) d. \(1: 2\)

In the following question two statements (Assertion) A and Reason (R) are given. Mark a. if \(\mathrm{A}\) and \(\mathrm{R}\) both are correct and \(\mathrm{R}\) is the correct explanation of \(\mathrm{A}\); b. if \(A\) and \(R\) both are correct but \(R\) is not the correct explanation of A; c. \(\mathrm{A}\) is true but \(\mathrm{R}\) is false; d. A is false but \(\mathrm{R}\) is true, e. A and \(R\) both are false. (A): The correct order of equivalent conductance at infinite dilution is \(\mathrm{KCl}>\mathrm{NaCl}>\mathrm{LiCl}\) \((\mathbf{R}): \mathrm{KCl}\) is stronger electrolyte than \(\mathrm{NaCl}\) which is stronger than \(\mathrm{LiCl}\).

In the following question two statements (Assertion) A and Reason (R) are given. Mark a. if \(\mathrm{A}\) and \(\mathrm{R}\) both are correct and \(\mathrm{R}\) is the correct explanation of \(\mathrm{A}\); b. if \(A\) and \(R\) both are correct but \(R\) is not the correct explanation of A; c. \(\mathrm{A}\) is true but \(\mathrm{R}\) is false; d. A is false but \(\mathrm{R}\) is true, e. A and \(R\) both are false. (A): In brown ring complex, \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{NO}\right] \mathrm{SO}_{4}\), the oxidation number of iron is \(+1\). ( \(\mathbf{R}\) ): Due to charge transfer the one unpaired electron of NO shifts to \(\mathrm{Fe}^{2+}\), thereby \(\mathrm{Fe}^{2+}\) converts into \(\mathrm{Fe}^{+}\).

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