The equivalent weights of \(\mathrm{KMnO}_{4}\) in an acidic, a neutral and a strong alkaline medium respectively are \((\mathrm{M}=\) molecular weight \()\) : (a) \(\mathrm{M} / 5, \mathrm{M} / 2, \mathrm{M}\) (b) \(\mathrm{M} / 5, \mathrm{M} / 3, \mathrm{M} / 2\) (c) \(\mathrm{M} / 5, \mathrm{M} / 3, \mathrm{M}\) (d) \(\mathrm{M} / 3, \mathrm{M}, \mathrm{M} / 5\)

Understanding the changes in oxidation states is central to many aspects of chemistry, including the calculations involving equivalent weight. These changes are integral to balancing redox reactions. In the context of \textbf{potassium permanganate} (\textbf{KMnO}\(_4\)), it acts as an oxidizing agent, meaning that it gains electrons and reduces the other substance. The oxidation state of manganese (\textbf{Mn}) in \textbf{KMnO}\(_4\) is +7. Depending on the medium, this can be reduced to +2, +4, or +6, reflecting different levels of electron gain.

For example, in an acidic solution, \textbf{Mn} goes from +7 to +2, gaining 5 electrons and undergoing a substantial change in oxidation state. These electrons are provided by the other reactant in the process. To effectively utilize this concept, one must apply the concept of electron transfer to determine the equivalent weight of the substance which is crucial for stoichiometric calculations in redox reactions, particularly when preparing for competitive exams in chemistry.

Chemical reactivity can significantly vary with the medium in which a reaction occurs. A striking example is the reactivity of \textbf{KMnO}\(_4\) in acidic, neutral, and alkaline environments. This variability is not just a peculiarity; rather, it is a reminder that the properties of the medium—such as its pH—affect the speciation and the redox potential of chemical compounds.

Acidic conditions often facilitate more extensive reductions, as shown by the reduction of \textbf{Mn} from +7 to +2 when \textbf{KMnO}\(_4\) is in an acidic medium. In comparison, neutral or slightly alkaline solutions result in smaller reductions, such as going from +7 to only +4. The change is even less in a strongly alkaline environment, as manganese's oxidation state is reduced from +7 to just +6. Grasping these changes can be particularly beneficial when studying chemistry for competitive exams, where such details often form the basis for more complex questions.

Electrochemical reactions find a special place in chemistry because they involve the direct conversion between chemical energy and electrical energy. These reactions are redox processes in which oxidation and reduction occur, transferring electrons from one species to another. A fundamental concept within these reactions is the number of electrons transferred, which is directly related to the equivalent weight calculation.

The change in manganese's oxidation state in \textbf{KMnO}\(_4\), as discussed earlier, represents the number of electrons that are transferred in the electrochemical process. In an acidic medium, the acceptance of 5 electrons corresponds to reduction, and in neutral and alkaline media, the acceptance decreases to 3 and 1 electron, respectively. Understanding these electrochemical principles is crucial for interpreting reaction mechanisms and performing accurate calculations in chemistry.

Competitive exams in chemistry often test students' understanding of fundamental principles and their ability to apply these principles to solve problems. Hence, grasping the underlying concepts, such as oxidation state changes and the significance of medium reactivity, is essential.

Such exams may present scenarios requiring the calculation of equivalent weights, as seen with the \textbf{KMnO}\(_4\) example. Understanding how to calculate equivalent weight based on the number of electrons transferred during a reaction will be a skill that can set students apart. For those preparing for these exams, it is advisable to practice a wide variety of problems, especially those related to electrochemical reactions and redox processes, to gain proficiency in this key area of chemistry.