Question

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

Step-by-step solution

Identify the Reaction Change

In the reaction, \[ \mathrm{KMnO}_{4} \rightarrow \mathrm{Mn^{2+}} \] the oxidation state of manganese changes from +7 to +2, decreasing by 5 units per manganese atom.

Determine Equivalents of Reaction

The change in oxidation state is used to determine the equivalents in a redox reaction. With a 5-electron change per manganese atom, each molecule of \( \mathrm{KMnO}_{4} \) undergoes a 5-equivalent change.

Calculate Equivalent Weight

The equivalent weight of a substance is found by dividing the molecular weight by the number of equivalents. For \( \mathrm{KMnO}_{4} \):\[ \text{Equivalent Weight} = \frac{158}{5} = 31.6 \]

Finalize the Result

The equivalent weight of \( \mathrm{KMnO}_{4} \) in the given reaction is 31.6, which corresponds to option (b).

Key concepts

Equivalent Weight Calculation

When dealing with redox reactions, the concept of equivalent weight is important. It represents how much of a substance will react with or supply one mole of electrons. To calculate this, we divide the molecular weight by the number of electrons gained or lost per molecule in the reaction.
In the given exercise with permanganate ion (\( \text{KMnO}_4 \)), manganese goes from an oxidation state of +7 to +2, indicating a change over 5 electrons per molecule. Here is how you calculate it:

• The molecular weight of \( \text{KMnO}_4 \) is 158 g/mol.
• For this reaction, there is a 5-electron change per molecule.

To find the equivalent weight, use the formula: \[\text{Equivalent Weight} = \frac{\text{Molecular Weight}}{\text{Number of Electrons Involved}} = \frac{158}{5} = 31.6 \, \text{g/equiv}\] This helps us understand how much of \( \text{KMnO}_4 \) corresponds to one equivalent of electrons.

Oxidation State Changes

Oxidation and reduction are key concepts in chemistry. When an atom undergoes an oxidation state change, it either gains or loses electrons.
In the redox reaction provided, we'll look at the oxidation state of manganese within the permanganate ion.
Manganese starts at an oxidation state of +7, represented in \( \text{MnO}_4^- \). During the reaction, it transitions to an oxidation state of +2, found in \( \text{Mn}^{2+} \). This means manganese is reduced by 5 units, as it gains electrons.

• Initial Oxidation State: +7
• Final Oxidation State: +2
• Change in Oxidation State: +7 to +2 (gains 5 electrons)

When solving redox reactions, understanding the changes in oxidation states helps in determining how electrons are transferred between reactants.

Balancing Chemical Equations

Balancing chemical equations ensures that the same number of every type of atom is present on both sides of the equation. This reflects the law of conservation of mass. For redox reactions, balancing involves not just atoms, but also the charge through electron exchanges.
Take the reaction from the exercise:\[ 5\,\text{C}_2\text{O}_4^{2-} + 2\,\text{MnO}_4^- + 16\,\text{H}^+ \rightarrow 10\,\text{CO}_2 + 2\,\text{Mn}^{2+} + 8\,\text{H}_2\text{O} \]To balance this equation, follow these steps:

• Balance the atoms of each element. Start with metals (Mn). Here, manganese is already balanced.
• Balance the oxygens by adding water molecules (\( \text{H}_2\text{O} \)).
• Next, balance the hydrogens by adding \( \text{H}^+ \) ions, since this reaction occurs in an acidic medium.
• Finally, balance the charges by ensuring the number of electrons lost equals the number gained.

This systematic approach ensures the equation accurately represents the chemical change occurring in the reaction.