The value of `x` in the partial redox equation
`MnO_(4)^(-)+8H^(+)+xe hArr Mn^(2+)+4H_(2)O` is

Options

A. `5`
B. `3`
C. `1`
D. `0`

Explanation

To find the value of \( x \) in the partial redox equation:

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

we will follow these steps:

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

### Step 2: Calculate the change in oxidation state
- The change in oxidation state from +7 to +2 indicates a reduction.
- The change in oxidation state can be calculated as follows:

\[ \text{Change} = \text{Final oxidation state} - \text{Initial oxidation state} = +2 - (+7) = -5 \]

### Step 3: Determine the number of electrons transferred
- Since the oxidation state decreases by 5, this means that 5 electrons are gained in the reduction process.
- Therefore, \( x \) (the number of electrons transferred) is equal to 5.

### Step 4: Write the final answer
- Thus, the value of \( x \) in the equation is:

\[ \boxed{5} \]

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