The value of n in : `MnO_(4)^(-)+8H^(+)+n erarr Mn^(2+)+4H_(2)O` is

Options

A. 5
B. 4
C. 3
D. 2

Answer

To find the value of \( n \) in the redox reaction:

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

we will follow these steps:

### Step 1: Determine the oxidation states
- In \(\text{MnO}_4^{-}\), manganese (Mn) has an oxidation state of +7.
- In \(\text{Mn}^{2+}\), manganese has an oxidation state of +2.

### Step 2: Calculate the change in oxidation state
- The change in oxidation state for manganese is from +7 to +2, which means it is reduced by 5 units.

### Step 3: Write the half-reaction
- The half-reaction can be written as:
\[ \text{MnO}_4^{-} + 8\text{H}^{+} + n \text{e}^{-} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \]

### Step 4: Balance the charges
- On the left side, the total charge is:
- From \(\text{MnO}_4^{-}\): -1
- From \(8\text{H}^{+}\): +8
- From \(n\) electrons: -n
- Therefore, the total charge on the left side is:
\[ -1 + 8 - n = 7 - n \]

- On the right side, the total charge is:
- From \(\text{Mn}^{2+}\): +2
- From \(4\text{H}_2\text{O}\): 0
- Therefore, the total charge on the right side is:
\[ +2 \]

### Step 5: Set the charges equal
- Set the total charge on both sides equal to each other:
\[ 7 - n = 2 \]

### Step 6: Solve for \( n \)
- Rearranging the equation gives:
\[ n = 7 - 2 \]
\[ n = 5 \]

### Conclusion
The value of \( n \) is \( 5 \).

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