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\(\mathrm{KMnO}_{4}\) is a strong oxidizing agent in acid medium. To provide acid medium \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is used instead of HCl. This is because a. Only \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is completely ionized. b. Rate is faster in the presence of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) c. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is a dibasic acid d. \(\mathrm{HCl}\) is oxidized by \(\mathrm{KMnO}_{4}\) to \(\mathrm{Cl}_{2}\) e. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is a stronger acid than \(\mathrm{HCl}\).

Short Answer

Expert verified
d. HCl is oxidized by KMnOâ‚„ to Clâ‚‚, unlike Hâ‚‚SOâ‚„.

Step by step solution

01

Identify Reaction Characteristics

Understand that KMnOâ‚„ is employed in acidic media as a strong oxidizer. In the presence of an acid like Hâ‚‚SOâ‚„ or HCl, it can transfer oxygen to the other reactants, facilitating oxidation.
02

Analyze the Role of Hâ‚‚SOâ‚„

Hâ‚‚SOâ‚„ is a strong acid that completely ionizes in solution, providing a consistent acidic environment for reactions and facilitating the reduction of KMnOâ‚„.
03

Investigate Possible Reactions with HCl

Unlike Hâ‚‚SOâ‚„, HCl can participate in side reactions. In the presence of KMnOâ‚„, HCl might get oxidized to Clâ‚‚ instead of only providing an acidic medium.
04

Explore Oxidation of HCl by KMnOâ‚„

Consider the possibility that KMnOâ‚„, being a strong oxidizing agent, can oxidize HCl to chlorine gas (Clâ‚‚). This prevents HCl from merely serving its intended role as an acid and avoids the undesired production of a toxic byproduct.
05

Conclusion

Since HCl can potentially be oxidized by KMnOâ‚„ to Clâ‚‚, causing unwanted side reactions and complications, Hâ‚‚SOâ‚„ is a preferred choice to maintain an acidic environment without such complications.

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Key Concepts

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

Strong Oxidizing Agent
Potassium permanganate, also known by its chemical formula \(\text{KMnO}_4\), is renowned for being a strong oxidizing agent. This means that \(\text{KMnO}_4\) has the ability to accept electrons from other substances in a chemical reaction. By doing so, it facilitates the oxidation of those substances, essentially causing them to lose electrons.

Such reactions are particularly useful in various chemical processes where oxidation is required. Whether it’s in laboratory settings or larger industrial applications, \(\text{KMnO}_4\) plays a crucial role due to its efficiency and strength as an oxidizing agent.

The power of \(\text{KMnO}_4\) is fully realized in acidic conditions, where it can most effectively transfer oxygen to the reactants. This is why it's often used in tandem with acids like \(\text{H}_2\text{SO}_4\) to create the perfect medium for catalyzing these reactions.
Hâ‚‚SOâ‚„ Acid Medium
Sulfuric acid, \(\text{H}_2\text{SO}_4\), is frequently chosen as the acid medium for redox reactions involving strong oxidizers like \(\text{KMnO}_4\). There are several reasons for its preference over other acids:
  • Complete Ionization: \(\text{H}_2\text{SO}_4\) fully ionizes in solution, providing a stable acidic environment. This constant acidity is essential for the consistent progress of the reaction.
  • Stronger Acid: As a strong acid, \(\text{H}_2\text{SO}_4\) ensures a sufficiently acidic medium without undergoing unwanted side reactions.
  • Dibasic Nature: Being dibasic, it can donate two protons, allowing more flexibility in various kinds of chemical reactions.
By maintaining a controlled acidic environment, \(\text{H}_2\text{SO}_4\) enables \(\text{KMnO}_4\) to act efficiently as an oxidizing agent, facilitating the oxidation of the intended reactants without interference.
KMnOâ‚„ and HCl Reaction
The interaction between \(\text{KMnO}_4\) and \(\text{HCl}\) presents a unique case compared to \(\text{H}_2\text{SO}_4\). Unlike sulfuric acid, hydrochloric acid can undergo oxidation when in contact with a strong oxidizing agent such as \(\text{KMnO}_4\).

Instead of merely providing an acidic setting, \(\text{HCl}\) itself can be converted into chlorine gas (\(\text{Cl}_2\)). This conversion is not typically desired in many chemical processes for reasons such as:
  • Production of Chlorine Gas: The reaction leads to \(\text{Cl}_2\), a toxic substance, thereby introducing potential safety hazards.
  • Unintended Side Reactions: The oxidation of \(\text{HCl}\) could interfere with the primary chemical reaction of interest.
Using \(\text{KMnO}_4\) with \(\text{HCl}\) needs to be carefully managed to prevent the formation of chlorine gas and avoid compromising reaction conditions.
Oxidation of HCl to Clâ‚‚
When \(\text{KMnO}_4\) is mixed with hydrochloric acid, a noteworthy chemical transformation occurs wherein \(\text{HCl}\) is oxidized to chlorine gas \((\text{Cl}_2)\). This happens because \(\text{KMnO}_4\), being a strong oxidizing agent, has the capacity to pull electrons away from \(\text{HCl}\), facilitating its conversion.

However, generating \(\text{Cl}_2\) as a byproduct can be problematic. Chlorine gas is not only toxic but also poses significant safety challenges.
  • Chlorine Gas Concerns: The presence of \(\text{Cl}_2\) can necessitate additional safety measures, such as proper ventilation or containment procedures, to mitigate potential risks.
  • Reaction Efficiency: The formation of \(\text{Cl}_2\) might lead to reduced efficiency in the primary reaction, diverting from the desired pathway of oxidation.
For these reasons, alternative acids like \(\text{H}_2\text{SO}_4\) are often selected to ensure the safe and efficient use of \(\text{KMnO}_4\) as a powerful oxidizing agent in various chemical solutions.

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Most popular questions from this chapter

Use the following data at \(25^{\circ} \mathrm{C}\) for the questions given below \(\mathrm{Ni}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Ni}(\mathrm{s}) \quad \mathrm{E}^{\circ}=-0.28 \mathrm{~V}\) \(\mathrm{Mg}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Mg}(\mathrm{s}) \quad \mathrm{E}^{\circ}=-2.37 \mathrm{~V}\) What is \(\mathrm{K}\) for the equilibrium at \(25^{\circ} \mathrm{C}\) ? a. \(4 \times 10^{-70}\) b. \(4 \times 10^{70}\) c. \(2 \times 10^{-35}\) d. \(2 \times 10^{70}\)

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