In the redox reaction `MnO_(4)^(-) +8H^(+)+Br^(-)rarrMn^(2+)+4H_(2)O+5//2br_(2)` which one is the reducing agent ?

Options

A. `H^(+)`
B. `MnO_(4)^(-)`
C. `Br^(-)`
D. `Mn^(2+)`

Explanation

To determine the reducing agent in the given redox reaction, we need to analyze the oxidation states of the elements involved in the reaction. Let's break down the steps:

### Step-by-Step Solution:

1. **Identify the Reaction**: The reaction given is:
\[ \text{MnO}_4^{-} + 8\text{H}^{+} + \text{Br}^{-} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} + \frac{5}{2}\text{Br}_2 \]

2. **Assign Oxidation States**:
- For **MnO4⁻**:
- Oxygen (O) has an oxidation state of -2.
- There are 4 oxygen atoms, contributing -8.
- Let the oxidation state of Mn be \( x \).
- The overall charge is -1, so:
\[ x - 8 = -1 \implies x = +7 \]
- For **Mn²⁺**: The oxidation state of Mn is +2.

- For **Br⁻**: The oxidation state of Br is -1.
- For **Br₂**: The oxidation state of Br is 0 (as it is in elemental form).

3. **Determine Changes in Oxidation States**:
- **Manganese**: Changes from +7 in MnO4⁻ to +2 in Mn²⁺. This is a reduction (gain of electrons).
- **Bromine**: Changes from -1 in Br⁻ to 0 in Br₂. This is an oxidation (loss of electrons).

4. **Identify the Reducing Agent**:
- A reducing agent is a substance that donates electrons and is oxidized in the process. In this reaction, Br⁻ is the species that loses electrons (is oxidized) and therefore acts as the reducing agent.

5. **Conclusion**:
- The reducing agent in the reaction is **Br⁻**.

### Summary:
The reducing agent in the redox reaction is **Br⁻** because it undergoes oxidation by losing electrons, while MnO4⁻ undergoes reduction.