Question

\(\mathrm{KMnO}_{4}(\) mol. wt. \(=158\) ) oxidizes oxalic acid in acidic medium to \(\mathrm{CO}_{2}\) and water as follows. \(5 \mathrm{C}_{2} \mathrm{O}_{4}^{2-}+2 \mathrm{MnO}_{4}^{-}+16 \mathrm{H}^{+} \rightarrow 10 \mathrm{CO}_{2}+2 \mathrm{Mn}^{2+}+8 \mathrm{H}_{2} \mathrm{O}\) What is the equivalent weight of \(\mathrm{KMnO}_{4}\) ? a. 158 b. \(31.6\) c. \(39.5\) d. 79

Short answer

The equivalent weight of KMnO₄ is 31.6.

Step-by-step solution

Understand the Reaction

First, identify the change in oxidation state for manganese in the reaction. In \( \text{KMnO}_{4} \), manganese is in the +7 oxidation state, and in \( \text{Mn}^{2+} \), it is in the +2 oxidation state.

Calculate Change in Oxidation States

Calculate the change in oxidation state for one manganese atom. It changes from +7 in \( \text{MnO}_{4}^{-} \) to +2 in \( \text{Mn}^{2+} \), which is a change of +5.

Determine Change per Molecule

As each \( \text{MnO}_{4}^{-} \) ion changes by 5 oxidation states, in the balanced reaction, 2 \( \text{MnO}_{4}^{-} \) ions account for a total change of \( 2 \times 5 = 10 \).

Calculate Equivalent Weight

Equivalent weight is calculated by dividing the molecular weight by the n-factor. The n-factor is the total change in oxidation state per molecule. Thus, the equivalent weight of \( \text{KMnO}_{4} \) is \( \frac{158}{5} \).

Compute the Final Result

Perform the division: \( \frac{158}{5} = 31.6 \). Therefore, the equivalent weight of \( \text{KMnO}_{4} \) is 31.6.

Key concepts

Oxidation State

The concept of oxidation states is crucial in understanding redox reactions, as it indicates the number of electrons involved. It is essentially a bookkeeping tool used by chemists to track how electrons are transferred among atoms in a chemical reaction. In our reaction involving \( \mathrm{KMnO}_{4} \), manganese starts in an oxidation state of +7, meaning it has lost 7 electrons compared to its neutral atom form. When it transforms into \( \mathrm{Mn}^{2+} \), it has an oxidation state of +2. This shift indicates a transfer of electrons.

• The decrease from +7 to +2 implies that the manganese ion gains 5 electrons throughout the reaction, a signal of reduction.
• Observing the change in oxidation state gives us insights into which elements are oxidized and which are reduced

Understanding the changes in oxidation state helps in calculating the equivalent weight of a compound, and it is an indispensable part of balanced chemical equations.

Acidic Medium Reaction

The environment in which a reaction takes place can significantly influence the outcome of that reaction. In an acidic medium, hydrogen ions (\( \text{H}^{+} \)) play a critical role. In our reaction, the presence of \( 16 \text{ H}^{+} \) ions is essential for the manganese in \( \text{KMnO}_{4} \) to be reduced. This leads to the formation of water (\( \text{H}_2\text{O} \)) as well as the transformation of oxalate ions.

Certain reactions only proceed efficiently in acidic conditions because the \( \text{H}^{+} \) ions help in the electron transfer process:

• They combine with the \( \text{MnO}_4^{-} \) ions to facilitate the reduction process to \( \text{Mn}^{2+} \).
• These ions also help in breaking down the oxalate ions (\( \text{C}_2\text{O}_4^{2-} \)) into carbon dioxide (\( \text{CO}_2 \)).

Recognizing the necessity of the acidic medium allows one to predict and control the expected products of a redox reaction.

KMnO₄

Potassium permanganate (\( \text{KMnO}_{4} \)) is a well-known oxidizing agent often used in various chemical reactions to drive the transfer of electrons from one substance to another. It is particularly powerful due to its ability to accept electrons readily.

• In our reaction, it acts as the oxidizing agent, meaning it accepts electrons from the oxalic acid.
• The manganese in \( \text{KMnO}_{4} \) changes from +7 to +2, showing a big decrease in its oxidation state as it gains electrons.
• The robust oxidizing property of \( \text{KMnO}_{4} \) not only makes it valuable in this reaction but in various industrial and laboratory applications.

Understanding \( \text{KMnO}_{4} \)'s role helps in predicting how it will react in different circumstances, enhancing practical laboratory skills.

Chemical Reactivity

Chemical reactivity refers to how readily substances undergo chemical reactions. It is crucial in determining the behavior of a compound under specific conditions. The reactivity of \( \text{KMnO}_{4} \) is significant due to its strong oxidizing ability, meaning it can easily gain electrons from other substances.

• In the presence of an acidic medium, this reactivity is accentuated, allowing \( \text{KMnO}_{4} \) to efficiently oxidize other chemicals, such as oxalic acid to carbon dioxide.
• This reaction exemplifies a substantial shift in both the composition and state of reactants and products, governed by the nature and strength of the reactants.

A thorough understanding of chemical reactivity not only aids in laboratory experiments but also helps in anticipating the behaviors of different chemical compounds in various settings.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions based on balanced equations. It is fundamental for measuring how much of a substance is consumed or produced.

In our reaction, stoichiometry guides us in figuring out the ratios of the reactants (oxalic acid and \( \text{KMnO}_{4} \)) and the products formed (carbon dioxide, manganese ion, and water).

• Here, five units of oxalic acid react with two units of \( \text{KMnO}_{4} \); this consistent ratio is vital for obtaining the correct amount of products.
• Stoichiometry ensures conservation of mass and balances the equation, maintaining the integrity of scientific calculations.
• The relationships revealed by stoichiometry allow us to calculate important parameters such as the equivalent weight necessary for reactions.

Mastering stoichiometry simplifies complex chemical calculations and supports successful experimental outcomes.