In the half reaction :
`2ClO_(3)^(-)rarrCl_(2)`

Options

A. 5 electrons are gained
B. 5 electrons are liberated
C. 10 electrons are gained
D. 10 electrons are liberated

Answer

To solve the half-reaction \( 2 \text{ClO}_3^{-} \rightarrow \text{Cl}_2 \), we need to follow these steps:

### Step 1: Write the unbalanced half-reaction
The given half-reaction is:
\[ 2 \text{ClO}_3^{-} \rightarrow \text{Cl}_2 \]

### Step 2: Balance the chlorine atoms
On the left side, we have 2 chlorine atoms from \( 2 \text{ClO}_3^{-} \) and on the right side, we have 2 chlorine atoms in \( \text{Cl}_2 \). The chlorine atoms are already balanced.

### Step 3: Balance the oxygen atoms
We have 6 oxygen atoms on the left side (from \( 2 \text{ClO}_3^{-} \)). To balance the oxygen atoms, we need to add water molecules to the right side:
\[ 2 \text{ClO}_3^{-} \rightarrow \text{Cl}_2 + 6 \text{H}_2\text{O} \]

### Step 4: Balance the hydrogen atoms
Now we have 6 water molecules on the right side, which means we have 12 hydrogen atoms. To balance the hydrogen atoms, we need to add 12 \( \text{H}^+ \) ions to the left side:
\[ 2 \text{ClO}_3^{-} + 12 \text{H}^+ \rightarrow \text{Cl}_2 + 6 \text{H}_2\text{O} \]

### Step 5: Balance the charges
Now let's balance the charges. On the left side, we have:
- \( 2 \text{ClO}_3^{-} \) contributes -2 charge.
- \( 12 \text{H}^+ \) contributes +12 charge.

Total charge on the left side:
\[ -2 + 12 = +10 \]

On the right side, the charge is 0 (since \( \text{Cl}_2 \) and \( 6 \text{H}_2\text{O} \) are neutral). To balance the charges, we need to add 10 electrons to the left side:
\[ 2 \text{ClO}_3^{-} + 12 \text{H}^+ + 10 e^{-} \rightarrow \text{Cl}_2 + 6 \text{H}_2\text{O} \]

### Conclusion
The balanced half-reaction shows that 10 electrons are gained in this reduction process. Thus, the answer to the question is that 10 electrons are involved.

### Final Answer
10 electrons are gained in the half-reaction \( 2 \text{ClO}_3^{-} \rightarrow \text{Cl}_2 \).

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