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 \(2ClO_3^{-} \rightarrow Cl_2\) and determine how many electrons are gained or liberated, we will follow these steps:

### Step 1: Determine the Oxidation States
First, we need to find the oxidation states of chlorine in both \(ClO_3^{-}\) and \(Cl_2\).

- In \(ClO_3^{-}\):
- Let the oxidation state of Cl be \(x\).
- The oxidation state of O is \(-2\).
- The overall charge of the ion is \(-1\).

Therefore, we can set up the equation:
\[ x + 3(-2) = -1 \implies x - 6 = -1 \implies x = +5 \]

- In \(Cl_2\):
- The oxidation state of Cl is \(0\) (elemental form).

### Step 2: Identify the Change in Oxidation State
Now we can see that chlorine is going from an oxidation state of \(+5\) in \(ClO_3^{-}\) to \(0\) in \(Cl_2\). This indicates that chlorine is being reduced.

### Step 3: Write the Half-Reaction
The half-reaction can be written as:
\[ 2ClO_3^{-} \rightarrow Cl_2 \]

### Step 4: Balance the Chlorine Atoms
There are 2 chlorine atoms on the left (from \(2ClO_3^{-}\)) and 2 chlorine atoms on the right (from \(Cl_2\)), so chlorine is already balanced.

### Step 5: Balance the Oxygen Atoms
Next, we need to balance the oxygen atoms. There are 6 oxygen atoms on the left (from \(2ClO_3^{-}\)) and none on the right. To balance the oxygen, we add water molecules to the right side:
\[ 2ClO_3^{-} \rightarrow Cl_2 + 6H_2O \]

### Step 6: Balance the Hydrogen Atoms
Now we have 12 hydrogen atoms on the right (from \(6H_2O\)). To balance the hydrogen, we add \(12H^+\) to the left side:
\[ 2ClO_3^{-} + 12H^+ \rightarrow Cl_2 + 6H_2O \]

### Step 7: Balance the Charges
Now we need to balance the charges. On the left side, we have:
- Charge from \(2ClO_3^{-}\): \(2 \times -1 = -2\)
- Charge from \(12H^+\): \(+12\)

Total charge on the left: \(+12 - 2 = +10\)

On the right side, the charge is \(0\) (neutral \(Cl_2\) and \(H_2O\)). To balance the charges, we add \(10\) electrons to the left side:
\[ 2ClO_3^{-} + 12H^+ + 10e^{-} \rightarrow Cl_2 + 6H_2O \]

### Step 8: Conclusion
In this half-reaction, \(10\) electrons are gained by the \(ClO_3^{-}\) ions during the reduction process.

### Final Answer
**10 electrons are gained in this half-reaction.**

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