Question

1M \( \text{KMnO}_4 \) solution is equal to _______ N \( \text{KMnO}_4 \) solution.

• A. \( 0.2 \text{N} \)
• B. \( 1 \text{N} \)
• C. \( 2 \text{N} \)
• D. \( 5 \text{N} \)

Hint:

Remember to look at the change in oxidation state to find the n-factor. For \( \text{KMnO}_4 \) in acid, Mn goes from +7 to +2, a change of 5.

Answer 1

The Correct Option is D

The normality (N) of a solution is related to its molarity (M) by the equation: $$ \text{Normality (N)} = \text{Molarity (M)} \times \text{n-factor} $$ where the n-factor (equivalent factor) is the number of equivalents of the substance per mole. For \( \text{KMnO}_4 \) acting as an oxidizing agent in acidic medium, the half-reaction is: $$ \text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} $$ In this reaction, each mole of \( \text{KMnO}_4 \) gains 5 moles of electrons. Therefore, the n-factor for \( \text{KMnO}_4 \) in acidic medium is 5. Given that the molarity of the \( \text{KMnO}_4 \) solution is 1 M, its normality will be: $$ \text{Normality (N)} = 1 \text{ M} \times 5 = 5 \text{ N} $$