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 reaction:

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

we will follow these steps:

### Step 1: Calculate the total charge on the left-hand side (LHS)

The charges on the left-hand side are:
- The charge of \(\text{MnO}_4^{-}\) is \(-1\).
- The charge of \(8\text{H}^{+}\) is \(+8\).
- The charge due to \(n\) electrons is \(-n\).

Thus, the total charge on the LHS can be calculated as:

\[ \text{Charge on LHS} = -1 + 8 - n = 7 - n \]

### Step 2: Calculate the total charge on the right-hand side (RHS)

The charges on the right-hand side are:
- The charge of \(\text{Mn}^{2+}\) is \(+2\).
- The charge of \(4\text{H}_2\text{O}\) is \(0\) (since water is neutral).

Thus, the total charge on the RHS is:

\[ \text{Charge on RHS} = +2 + 0 = 2 \]

### Step 3: Set the charges on both sides equal to each other

For the reaction to be balanced, the total charge on the left-hand side must equal the total charge on the right-hand side:

\[ 7 - n = 2 \]

### Step 4: Solve for \( n \)

Now, we can solve for \( n \):

\[ 7 - n = 2 \\ -n = 2 - 7 \\ -n = -5 \\ n = 5 \]

### Conclusion

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

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