The value of n in the following half equation is `MnO4^(-) + 2H_2O + ne rarr MnO_2 + 4O'H`

Options

A. 3
B. 5
C. 4
D. 6

Answer

To find the value of \( n \) in the half-reaction given, we need to analyze the oxidation states of manganese in the reactants and products.

### Step-by-Step Solution:

1. **Identify the oxidation states of Mn in the reactants and products:**
- In the reactant \( \text{MnO}_4^{-} \), we need to determine the oxidation state of Mn.
- The formula for calculating the oxidation state is:
\[ \text{Oxidation state of Mn} + 4 \times (\text{Oxidation state of O}) = \text{Charge} \]
Here, the oxidation state of O is \(-2\), and the overall charge is \(-1\):
\[ x + 4(-2) = -1 \implies x - 8 = -1 \implies x = +7 \]
- Therefore, the oxidation state of Mn in \( \text{MnO}_4^{-} \) is \( +7 \).

2. **Determine the oxidation state of Mn in the product \( \text{MnO}_2 \):**
- Using the same method:
\[ \text{Oxidation state of Mn} + 2 \times (\text{Oxidation state of O}) = 0 \]
\[ y + 2(-2) = 0 \implies y - 4 = 0 \implies y = +4 \]
- Thus, the oxidation state of Mn in \( \text{MnO}_2 \) is \( +4 \).

3. **Calculate the change in oxidation state:**
- The change in oxidation state from \( +7 \) to \( +4 \) is:
\[ \text{Change} = \text{Final oxidation state} - \text{Initial oxidation state} = +4 - (+7) = -3 \]
- Since we are looking for the value of \( n \) (which represents the number of electrons transferred), we take the absolute value of the change:
\[ n = 7 - 4 = 3 \]

4. **Conclusion:**
- The value of \( n \) in the half-reaction is \( 3 \).

### Final Answer:
The value of \( n \) is \( 3 \).

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