The values of the `x` and `y` in the following redox reaction.
`xCl_(2)+6overset(ө)OHrarrClO_(3)^(ө)+yCl^(ө)+3H_(2)O`

Options

A. `x=2,y=4`
B. `x=5,y=3`
C. `x=3,y=5`
D. `x=4,y=2`

Explanation

To solve the redox reaction and find the values of \( x \) and \( y \) in the equation:

\[ xCl_2 + 6OH^- \rightarrow ClO_3^- + yCl^- + 3H_2O \]

we will follow these steps:

### Step 1: Identify Oxidation and Reduction
In this reaction, chlorine (\( Cl_2 \)) is being converted into chlorate (\( ClO_3^- \)) and chloride ions (\( Cl^- \)).

- The oxidation state of chlorine in \( Cl_2 \) is 0.
- In \( ClO_3^- \), the oxidation state of chlorine is +5.
- In \( Cl^- \), the oxidation state of chlorine is -1.

### Step 2: Write Half-Reactions
1. **Oxidation Half-Reaction**:
- Chlorine is oxidized from 0 to +5.
- The half-reaction can be written as:
\[ Cl_2 \rightarrow 2ClO_3^- + 10e^- \]
(Since each \( Cl \) atom loses 5 electrons, for 2 \( Cl \) atoms, it loses 10 electrons.)

2. **Reduction Half-Reaction**:
- Chlorine is reduced from 0 to -1.
- The half-reaction can be written as:
\[ Cl_2 + 2e^- \rightarrow 2Cl^- \]
(Each \( Cl \) atom gains 1 electron, for 2 \( Cl \) atoms, it gains 2 electrons.)

### Step 3: Balance the Electrons
To balance the overall reaction, we need to ensure that the number of electrons lost in oxidation equals the number of electrons gained in reduction.

- Multiply the reduction half-reaction by 5 to match the 10 electrons lost in the oxidation half-reaction:
\[ 5Cl_2 + 10e^- \rightarrow 10Cl^- \]

### Step 4: Combine the Half-Reactions
Now, we can combine the half-reactions:
\[ 3Cl_2 + 6OH^- \rightarrow ClO_3^- + 5Cl^- + 3H_2O \]

### Step 5: Identify Values of \( x \) and \( y \)
From the balanced equation:
- \( x = 3 \) (the coefficient of \( Cl_2 \))
- \( y = 5 \) (the coefficient of \( Cl^- \))

### Final Answer
Thus, the values of \( x \) and \( y \) are:
- \( x = 3 \)
- \( y = 5 \)