Understanding the Concept:
The equivalent weight of an oxidizing or reducing agent in a redox reaction is calculated by dividing its molecular weight by the total number of electrons gained or lost per molecule during the chemical process:
\[
\text{Equivalent Weight} = \frac{\text{Molecular Weight}}{n\text{-factor}}
\]
where the \(n\)-factor corresponds to the absolute change in the oxidation state of the key atom.
Step 1: Determine the oxidation state in the reactant
In the permanganate ion (\(\text{MnO}_4^-\)), let the oxidation state of Manganese (\(\text{Mn}\)) be represented as \(x\). Oxygen typically possesses an oxidation state of \(-2\):
\[
x + 4(-2) = -1 \quad \Rightarrow \quad x - 8 = -1 \quad \Rightarrow \quad x = +7
\]
Step 2: Determine the oxidation state in an acidic product medium
In a strongly acidic solution (such as in the presence of dilute \(\text{H}_2\text{SO}_4\)), the purple permanganate ion is completely reduced to the colorless, stable divalent manganous ion (\(\text{Mn}^{2+}\)). The oxidation state of manganese here is simply \(+2\).
Step 3: Calculate the change in electrons (\(n\)-factor)
The reduction half-reaction can be formulated explicitly as:
\[
\text{MnO}_4^- + 8\text{H}^+ + 5e^- \longrightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}
\]
Subtracting the oxidation states to find the magnitude of change:
\[
\Delta \text{Oxidation State} = (+7) - (+2) = 5
\]
Thus, exactly 5 electrons are consumed per molecule of \(\text{KMnO}_4\), making the \(n\)-factor equal to 5.
Substituting this into our primary equation:
\[
\text{Equivalent Weight} = \frac{\text{Molecular Weight}}{5}
\]