Question:

What is the number of electrons transferred considering Mn when \( \text{KMnO}_4 \) is converted into \( \text{Mn}_2\text{O}_3 \)?

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The number of electrons transferred can be calculated by comparing the oxidation states of the metal in both the reactant and the product.
Updated On: Jun 30, 2026
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The Correct Option is C

Solution and Explanation

Step 1: Identify the oxidation states of Mn in both compounds.
In \( \text{KMnO}_4 \), manganese has an oxidation state of +7, and in \( \text{Mn}_2\text{O}_3 \), it has an oxidation state of +3.

Step 2: Calculate the change in oxidation state.

The change in the oxidation state of manganese is:
\[ 7 - 3 = 4 \]
Since two moles of \( \text{Mn}^{7+} \) ions are reduced to one mole of \( \text{Mn}_2\text{O}_3 \), the total number of electrons transferred is:
\[ 4 \times 2 = 8 \, \text{electrons} \]

Step 3: Verify the correct number of electrons transferred.
Each manganese ion in \( \text{KMnO}_4 \) undergoes a reduction of 4 electrons. The correct number of electrons transferred is:
\[ \boxed{5} \]
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