Problem 114 - A Complete Resource Book in Chemistry for JEE Main

Question

\(\mathrm{KMnO}_{4}\) (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

Step-by-Step Solution

Answer

The equivalent weight of \(\mathrm{KMnO}_{4}\) is 31.6, option (b).

Step 1: Balance the Reaction Equation

The reaction equation is already provided as balanced: \(5 \mathrm{C}_{2} \mathrm{O}_{4}^{2-} + 2 \mathrm{MnO}_{4}^{-} + 16 \mathrm{H}^{+} \rightarrow 10 \mathrm{CO}_{2} + 2 \mathrm{Mn}^{2+} + 8 \mathrm{H}_{2} \mathrm{O}\). Check that each element and charge is balanced.

Step 2: Calculate Change in Oxidation State

In the reaction, manganese \((\mathrm{Mn})\) in \(\mathrm{MnO}_{4}^{-}\) has an oxidation state of +7 and is reduced to +2 in \(\mathrm{Mn}^{2+}\). Therefore, the change in oxidation state per \(\mathrm{Mn}\) atom is \(7 - 2 = 5\).

Step 3: Determine Number of Electrons Transferred

For 2 moles of \(\mathrm{MnO}_{4}^{-}\), each manganese undergoes a change of 5 oxidation states, amounting to \(2 \times 5 = 10\) electrons transferred in total.

Step 4: Define Equivalent Weight Formula

The equivalent weight of a substance in a redox reaction is calculated using the formula: \[ \text{Equivalent weight} = \frac{\text{Molar mass}}{\text{n}} \] where \(n\) is the number of moles of electrons transferred per mole of substance.

Step 5: Calculate Equivalent Weight of \(\mathrm{KMnO}_{4}\)

The molar mass (molecular weight) of \(\mathrm{KMnO}_{4}\) is 158. The value of \(n\) (for the electrons transferred per mole of \(\mathrm{KMnO}_{4}\)) is 5. Therefore, \[ \text{Equivalent weight of \(\mathrm{KMnO}_{4}\)} = \frac{158}{5} = 31.6 \]

Step 6: choose the Correct Option

The calculated equivalent weight of \(\mathrm{KMnO}_{4}\) is 31.6. Hence, the correct answer is (b) 31.6.

Key Concepts

Redox Reactions

Redox reactions, or oxidation-reduction reactions, are chemical processes where the transfer of electrons occurs. These reactions always involve two key components: oxidation and reduction. In oxidation, a substance loses electrons, while in reduction, a substance gains electrons.
In the given reaction, oxalic acid is oxidized to carbon dioxide (CO extsubscript{2}), and permanganate ions (MnO extsubscript{4} extsuperscript{-}) are reduced to Mn extsuperscript{2+}.

• Oxidation: 5 C extsubscript{2}O extsubscript{4} extsuperscript{2-} → 10 CO extsubscript{2} + 10e extsuperscript{-}
• Reduction: 2 MnO extsubscript{4} extsuperscript{-} + 10e extsuperscript{-} → 2 Mn extsuperscript{2+}

These transformations are crucial for determining equivalent weights, which hinges on counting the number of electrons transferred.

Oxidation States

Understanding oxidation states is pivotal in redox reactions. An oxidation state is a numerical representation that is used to keep track of electron transfer in a reaction, helping us know which atoms are oxidized and which are reduced.
In the equation provided, manganese in the permanganate ion (MnO extsubscript{4} extsuperscript{-}) has an oxidation state of +7. During the reaction, manganese is reduced to Mn extsuperscript{2+}, indicating its final oxidation state is +2.
Thus, this change signifies a reduction by 5 electron units per manganese atom. Recognizing these changes helps in calculating electron transfer numbers, critical for solving equivalent weight problems.

Mole Concept

The mole concept is fundamental in stoichiometry, allowing us to quantify the amount of substances involved in a chemical reaction. It helps in understanding the proportions of reactants and products.
In this exercise, 2 moles of MnO extsubscript{4} extsuperscript{-} react with 5 moles of C extsubscript{2}O extsubscript{4} extsuperscript{2-}.
This relationship is crucial when determining how many moles of electrons are transferred, as these counts directly relate to calculating equivalent weights. Understanding moles ensures accurate conversion, which supports precise balance and equivalency calculations.

Electron Transfer

Electron transfer is the heart of redox reactions, involving the movement of electrons from one reactant to another. In our reaction, manganese is the element undergoing a notable change.
Each Mn in MnO extsubscript{4} extsuperscript{-} experiences a reduction by 5 units of oxidation state, resulting in a transfer of 5 electrons per manganese. Given that there are 2 moles of MnO extsubscript{4} extsuperscript{-}, this transfers a total of 10 electrons.
Understanding this electron transfer is key to determining the number of electrons that correspond to one equivalent of the substance, which directly aids in the calculation of equivalent weights.

Chemical Balancing

Balancing chemical equations is essential to ensure that the conservation of mass and charge is maintained throughout a reaction. This process involves making sure both sides of the equation have an equal number of each atom and the same overall charge.
In the given reaction, it starts as balanced: 2 MnO extsubscript{4} extsuperscript{-} + 5 C extsubscript{2}O extsubscript{4} extsuperscript{2-} + 16 H extsuperscript{+} → 2 Mn extsuperscript{2+} + 10 CO extsubscript{2} + 8 H extsubscript{2}O.
This balance allows correct calculation of electron transfers, and thus, the correct determination of the equivalent weight of KMnO extsubscript{4}. Always verify that the atoms and charge are balanced, or else any calculations based on this equation might lead to incorrect results.