In the reaction, `3Br_(2) + 6NaOH rarr NaBrO_(3) + 5NaBr+3H_(2)O` which element loses as well as gains electrons -

Options

A. Na
B. Br
C. O
D. H

Answer

To determine which element loses and gains electrons in the reaction \(3Br_2 + 6NaOH \rightarrow NaBrO_3 + 5NaBr + 3H_2O\), we will analyze the oxidation states of bromine in the reactants and products.

### Step-by-Step Solution:

1. **Identify the Oxidation States:**
- In \(Br_2\) (bromine molecule), the oxidation state of bromine is 0 (as it is in its elemental form).
- In \(NaBrO_3\), we need to determine the oxidation state of bromine. Sodium (Na) is +1, and oxygen (O) is -2. The compound is neutral, so:
\[ +1 + x + 3(-2) = 0 \implies x - 5 = 0 \implies x = +5 \]
Thus, bromine in \(NaBrO_3\) is +5.
- In \(NaBr\), sodium is +1 and bromine must be -1 to balance the charge, so bromine is -1.
- In \(H_2O\), hydrogen is +1 and oxygen is -2.

2. **Determine Changes in Oxidation States:**
- Bromine changes from 0 in \(Br_2\) to +5 in \(NaBrO_3\) (oxidation).
- Bromine also changes from 0 in \(Br_2\) to -1 in \(NaBr\) (reduction).

3. **Identify Oxidation and Reduction:**
- **Oxidation** is defined as the loss of electrons. Here, bromine goes from 0 to +5, indicating it loses 5 electrons.
- **Reduction** is defined as the gain of electrons. Here, bromine goes from 0 to -1, indicating it gains 1 electron.

4. **Conclusion:**
- The element that loses electrons (oxidation) is bromine (from 0 to +5).
- The element that gains electrons (reduction) is also bromine (from 0 to -1).

### Final Answer:
Bromine (Br) is the element that both loses electrons (oxidation) and gains electrons (reduction) in the reaction.