Consider the following equation.
`Cr_2O_7^(2-) (aq) + Br^(-) (aq) + H^(+) (aq) to Br_2(l) + Cr^(3) + (aq) + H_2O(l)`
In this reaction, which one of the following is a reducing agent?

Options

A. `Br_2`
B. `Br^(-)`
C. `H^(+)`
D. `Cr_2O_7^(2-)`

Explanation

To determine the reducing agent in the given reaction, we need to analyze the changes in oxidation states of the elements involved. Here’s a step-by-step solution:

### Step 1: Identify the oxidation states of the reactants and products.
- **Chromium in Cr₂O₇²⁻**: The oxidation state of chromium (Cr) in dichromate (Cr₂O₇²⁻) is +6.
- **Bromine in Br⁻**: The oxidation state of bromine (Br) in bromide (Br⁻) is -1.
- **Hydrogen in H⁺**: The oxidation state of hydrogen (H) in H⁺ is +1.

### Step 2: Identify the oxidation states in the products.
- **Bromine in Br₂**: The oxidation state of bromine in elemental bromine (Br₂) is 0.
- **Chromium in Cr³⁺**: The oxidation state of chromium in Cr³⁺ is +3.
- **Water (H₂O)**: The oxidation state of hydrogen remains +1 and oxygen is -2.

### Step 3: Determine the changes in oxidation states.
- **Bromine**: Changes from -1 (in Br⁻) to 0 (in Br₂). This is an increase in oxidation state, indicating that bromine is being oxidized.
- **Chromium**: Changes from +6 (in Cr₂O₇²⁻) to +3 (in Cr³⁺). This is a decrease in oxidation state, indicating that chromium is being reduced.

### Step 4: Identify the reducing agent.
- The substance that gets oxidized (loses electrons) is the reducing agent. Since bromine (Br⁻) is being oxidized from -1 to 0, it acts as the reducing agent in this reaction.

### Conclusion:
Thus, the reducing agent in the reaction is **Br⁻**.

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