Question

\(\mathrm{KMnO}_{4}(\mathrm{~mol} . \mathrm{wt}=158)\) oxidizes oxalic acid in acidic medium to \(\mathrm{CO}_{2}\) and water as follows. \(5 \mathrm{C}_{2} \mathrm{O}_{4}^{2-}+2 \mathrm{MnO}_{4}^{-}+16 \mathrm{H}^{+} \longrightarrow 10 \mathrm{CO}_{2}+\) \(2 \mathrm{Mn}^{2+}+8 \mathrm{H}_{2} \mathrm{O}\) 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 \( \mathrm{KMnO}_4 \) is 31.6, which corresponds to option (b).

Step-by-step solution

Identify the reaction type

The reaction given is a redox (oxidation-reduction) reaction, where the \ \( \mathrm{MnO}_4^− \ \) ion is reduced to \ \( \mathrm{Mn}^{2+} \ \) and the oxalate \ \( \mathrm{C}_2\mathrm{O}_4^{2-} \ \) is oxidized to \ \( \mathrm{CO}_2 \ \). We need to determine the change in oxidation state to find the equivalent weight of \ \( \mathrm{KMnO}_4 \ \).

Calculate change in oxidation number

For the manganese atom in \ \( \mathrm{KMnO}_4 \ \), the oxidation state changes from +7 (in \ \( \mathrm{MnO}_4^− \ \)) to +2 (in \ \( \mathrm{Mn}^{2+} \ \)), a change of 5 per atom. Since there are 2 manganese atoms reacting, the total change in oxidation state is 10.

Determine n-factor for \\ \( \mathrm{KMnO}_4 \\ \)

The n-factor of \ \( \mathrm{KMnO}_4 \ \) in this reaction is equal to the total change in moles of electrons, which is 5 electrons per molecule of \ \( \mathrm{KMnO}_4 \ \) as each \ \( \mathrm{MnO}_4^− \ \) accepts 5 electrons to become \ \( \mathrm{Mn}^{2+} \ \). Thus, n-factor = 5.

Calculate the equivalent weight

The equivalent weight of a compound in a redox reaction is calculated using the formula: \[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{n-\text{factor}} \] For \ \( \mathrm{KMnO}_4 \ \), the molecular weight is 158 and the n-factor is 5. Thus, \[ \text{Equivalent weight} = \frac{158}{5} = 31.6 \]

Conclusion

The equivalent weight of \ \( \mathrm{KMnO}_4 \ \) in this reaction is \(31.6\). Therefore, the correct answer is option (b) \(31.6\).

Key concepts

Oxidation Number

The oxidation number is an essential concept in understanding redox reactions. It represents the hypothetical charge that an atom would have if all bonds to atoms of different elements were 100% ionic. This concept helps track how electrons are transferred in reactions, especially redox reactions.
An oxidation number, also called oxidation state, can be positive, negative, or zero, depending on the presence of electrons around the atom.

• Atoms in their elemental state have an oxidation number of 0.
• For monoatomic ions, their oxidation number is equal to their charge. For example, in \( \text{Na}^+ \), sodium has an oxidation number of +1.
• The sum of oxidation numbers in a neutral compound must be 0, while it must equal the charge of the ion in polyatomic ions.

By calculating changes in oxidation numbers, we can determine which elements are oxidized or reduced in a reaction as in the original exercise with \( \text{KMnO}_4 \).

Equivalent Weight

Equivalent weight is an important concept in redox reactions because it helps determine the amount of a substance that reacts or the amount needed to react with a given quantity of another substance.
This is especially relevant when working with titration problems or other stoichiometric calculations.
The formula to calculate equivalent weight in redox reactions is:

• \[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{n-\text{factor}} \]

This relationship shows that the equivalent weight depends on two key values: the molecular weight and the n-factor. In the original exercise, we found the equivalent weight of \( \text{KMnO}_4 \) by dividing its molecular weight by its n-factor. Understanding how equivalent weight is derived helps in accurately performing chemical calculations.

n-factor

The n-factor is a crucial concept when dealing with redox reactions as it indicates how many electrons are involved per molecule in a reaction transformation. It is determined by calculating the total change in oxidation numbers of the elements involved.
To calculate it:

• Identify the change in oxidation state for each atom involved.
• Multiply the change by the number of atoms involved.
• Sum up these values for all atoms and divide by the stoichiometric coefficient if necessary.

For instance, in \( \text{KMnO}_4 \), each \( \text{MnO}_4^- \) ion gains 5 electrons to be reduced to \( \text{Mn}^{2+} \), thus giving an n-factor of 5. Recognizing the n-factor is pivotal as it is directly related to the equivalent weight calculation.

Oxidation-Reduction

Oxidation-reduction (redox) reactions are processes where the oxidation state of molecules, atoms, or ions changes by gaining or losing electrons. Oxidation involves the loss of electrons, resulting in an increase in oxidation state, while reduction involves the gain of electrons, resulting in a decrease in oxidation state.
In the given reaction:

• \( \text{MnO}_4^- \) is reduced to \( \text{Mn}^{2+} \), showcasing a decrease of oxidation state by 5.
• \( \text{C}_2\text{O}_4^{2-} \) is oxidized to \( \text{CO}_2 \), involving an increase in the oxidation state of carbon.

These redox processes are complementary, meaning the loss of electrons by one reactant corresponds to the gain by the other. Mastering this concept is essential in various fields such as biochemistry, environmental science, and industrial applications.