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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) \(\mathrm{M} / 3, \mathrm{M}, \mathrm{M} / 5\)

Short Answer

Expert verified
Correct option is (c): \(M/5, M/3, M\).

Step by step solution

01

Determine Equivalent Weight in Acidic Medium

In an acidic medium, the reaction for the reduction of KMnO鈧 involves a change in oxidation state of Mn from +7 to +2. This involves a transfer of 5 electrons: \[ \mathrm{MnO}_4^- + 8\mathrm{H}^+ + 5\mathrm{e}^- \rightarrow \mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O} \] \[ \text{Equivalent weight} = \frac{\text{Molecular weight (M)}}{n} = \frac{M}{5} \]
02

Determine Equivalent Weight in Neutral Medium

In a neutral medium, the manganate ion reduces from Mn in the +7 oxidation state to Mn in the +4 oxidation state (forming MnO鈧), involving the transfer of 3 electrons: \[ \mathrm{MnO}_4^- + 2\mathrm{H}_2\mathrm{O} + 3\mathrm{e}^- \rightarrow \mathrm{MnO}_2 + 4\mathrm{OH}^- \] \[ \text{Equivalent weight} = \frac{M}{3} \]
03

Determine Equivalent Weight in Strong Alkaline Medium

In a strong alkaline medium, KMnO鈧 is reduced from Mn in the +7 state to Mn in the +6 state (forming manganate ion MnO鈧劼测伝), involving the transfer of 1 electron: \[ \mathrm{MnO}_4^- + \mathrm{e}^- \rightarrow \mathrm{MnO}_4^{2-} \] \[ \text{Equivalent weight} = \frac{M}{1} = M \]
04

Identify Correct Option

Based on the equivalent weights calculated, in acidic the ratio is \(M/5\), in neutral \(M/3\), and in alkaline \(M\). The correct option matching these calculations is (c): \(\frac{M}{5}, \frac{M}{3}, M\).

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Key Concepts

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

Oxidation State
The oxidation state is a concept that reflects the number of electrons an atom gains or loses when forming chemical compounds or ions. Understanding this is crucial when dealing with redox reactions and substances like potassium permanganate (KMnO鈧). An element's oxidation state is denoted with a numeral, such as +7 in KMnO鈧 for manganese. This indicates manganese is losing 7 electrons in its highest oxidized state.

Calculating the oxidation state typically involves looking at known oxidation numbers of other elements within the compound. For example, in KMnO鈧, oxygen typically has an oxidation state of -2. By calculating the total charge, you can determine that manganese must be at +7 so the overall charge of the molecule sums to zero. This understanding plays directly into predicting how KMnO鈧 will behave in various chemical reactions.

For students, grasping oxidation states helps in balancing chemical equations in redox reactions. It helps identify which elements are oxidized and which are reduced, critical in understanding reactions in different mediums such as in acidic, neutral, or alkaline solutions. Use these insights to predict reaction outcomes and conditions under which equivalent weights vary.
Redox Reactions
Redox reactions, short for reduction-oxidation reactions, are chemical processes that involve the transfer of electrons between two chemical species. In these reactions, one species is oxidized (loses electrons) and another is reduced (gains electrons). Understanding this transfer is crucial when studying substances such as KMnO鈧, where the manganese undergoes changes in its oxidation state.

When analyzing a redox reaction by KMnO鈧, it acts as an oxidizing agent. For example, in acidic solutions, the MnO鈧勨伝 ion is reduced to Mn虏鈦 while reducing 5 electrons, indicating a complete oxidation-reduction process. Similarly, in neutral media, MnO鈧勨伝 reduces to MnO鈧 transferring 3 electrons, and in alkaline, it goes to MnO鈧劼测伝 with just one electron transferred. This sequential change highlights how the environment impacts electron transfer and equivalent weight is highly dependent on this factor.

It's beneficial to practice writing out half-reactions for clarity, identifying the precise flow of electrons. These insights deepen one's understanding of how compounds like KMnO鈧 achieve balance through electron redistribution, which is central to predicting their reactions and calculating equivalent weights accurately.
Potassium Permanganate (KMnO鈧)
Potassium Permanganate, or KMnO鈧, is a widely used chemical compound known for its powerful oxidizing properties. It is composed of potassium (K), manganese (Mn), and oxygen (O). In many reactions, it serves as an oxidizing agent that converts harmful compounds into less toxic forms.

KMnO鈧 is special because its manganese center changes oxidation states readily, making it effective in various chemical reactions. For educational purposes, students can understand this through its calculations of equivalent weight, which is a measure of the reactive capacity of KMnO鈧 depending on the medium. For example, in acidic conditions, Mn changes from +7 to +2, in neutral from +7 to +4, and in alkaline from +7 to +6. These shifts impact how the compound interacts with other chemicals, aiding in measuring purity and reactions.

In practical terms, KMnO鈧 is used in water treatment, disinfection, and organic synthesis due to its oxidative power. Students should appreciate its versatility and learn how the medium affects its oxidizing behavior, which in turn controlled study of equivalent weights and practical applications in real-world chemical scenarios.

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