In the reaction : `MnO_(4)^(-)+xH^(+)+n e^(-)rarrMn^(2+)+yH_(2)O` What is the value of n :

Options

A. 5
B. 8
C. 6
D. 3

Answer

To determine the value of \( n \) in the redox reaction

\[ \text{MnO}_4^{-} + x \text{H}^{+} + n e^{-} \rightarrow \text{Mn}^{2+} + y \text{H}_2\text{O} \]

we need to balance the reaction in terms of both mass and charge.

### Step 1: Identify the oxidation states
In the reaction, manganese in \( \text{MnO}_4^{-} \) has an oxidation state of +7, while in \( \text{Mn}^{2+} \) it has an oxidation state of +2. This indicates that manganese is being reduced.

### Step 2: Balance the oxygen atoms
The \( \text{MnO}_4^{-} \) contains 4 oxygen atoms. To balance these oxygen atoms, we can add water molecules (\( \text{H}_2\text{O} \)) to the products side. Therefore, we add 4 \( \text{H}_2\text{O} \):

\[ \text{MnO}_4^{-} + x \text{H}^{+} + n e^{-} \rightarrow \text{Mn}^{2+} + 4 \text{H}_2\text{O} \]

### Step 3: Balance the hydrogen atoms
Now we have 8 hydrogen atoms on the products side (from 4 \( \text{H}_2\text{O} \)). To balance the hydrogen atoms, we need to add 8 \( \text{H}^{+} \) ions to the reactants side:

\[ \text{MnO}_4^{-} + 8 \text{H}^{+} + n e^{-} \rightarrow \text{Mn}^{2+} + 4 \text{H}_2\text{O} \]

### Step 4: Balance the charges
Now we need to balance the charges on both sides of the equation.

- On the left side, the total charge is:
- \( 8 \) from \( 8 \text{H}^{+} \)
- \( -1 \) from \( \text{MnO}_4^{-} \)
- Total charge = \( 8 - 1 = +7 \)

- On the right side, the total charge is:
- \( +2 \) from \( \text{Mn}^{2+} \)
- Total charge = \( +2 \)

To balance the charges, we need to add electrons to the left side. Let \( n \) be the number of electrons. The equation for charge balance becomes:

\[ 7 - n = 2 \]

### Step 5: Solve for \( n \)
Rearranging the equation gives:

\[ n = 7 - 2 = 5 \]

### Conclusion
Thus, the value of \( n \) is \( 5 \).

\[ \boxed{5} \]