The values of x and y in the following redox reaction
`xCI_(2)+6 OH^(-) rarr CIO_(3)^(3) rarr CIO_(3)^(-)+y CI^(-)+3 H_(2)O` are

Options

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

Answer

To solve for the values of x and y in the given redox reaction:

\[ x \text{Cl}_2 + 6 \text{OH}^- \rightarrow \text{ClO}_3^{3-} + y \text{Cl}^- + 3 \text{H}_2\text{O} \]

we will use the half-reaction method to balance the equation. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the half-reactions

1. **Oxidation half-reaction**: Chlorine (Cl) is oxidized from \( \text{Cl}_2 \) (oxidation state 0) to \( \text{ClO}_3^{-} \) (oxidation state +5).
2. **Reduction half-reaction**: Chlorine (Cl) is reduced from \( \text{Cl}_2 \) (oxidation state 0) to \( \text{Cl}^- \) (oxidation state -1).

### Step 2: Write the oxidation half-reaction

For the oxidation half-reaction:
\[ \text{Cl}_2 \rightarrow 2 \text{ClO}_3^{-} \]

- Each Cl atom goes from 0 to +5, which means it loses 5 electrons. Since there are 2 Cl atoms, the total loss is \( 2 \times 5 = 10 \) electrons.

Thus, the oxidation half-reaction becomes:
\[ \text{Cl}_2 + 12 \text{OH}^- \rightarrow 2 \text{ClO}_3^{-} + 10 e^- + 6 \text{H}_2\text{O} \]

### Step 3: Write the reduction half-reaction

For the reduction half-reaction:
\[ \text{Cl}_2 \rightarrow 2 \text{Cl}^- \]

- Each Cl atom goes from 0 to -1, which means it gains 1 electron. Since there are 2 Cl atoms, the total gain is \( 2 \) electrons.

Thus, the reduction half-reaction becomes:
\[ \text{Cl}_2 + 2 e^- \rightarrow 2 \text{Cl}^- \]

### Step 4: Balance the electrons

To balance the electrons in both half-reactions, we need to multiply the reduction half-reaction by 5 (since 10 electrons are lost in oxidation):
\[ 5 \left( \text{Cl}_2 + 2 e^- \rightarrow 2 \text{Cl}^- \right) \Rightarrow 5 \text{Cl}_2 + 10 e^- \rightarrow 10 \text{Cl}^- \]

### Step 5: Combine the half-reactions

Now, we can add the two half-reactions together:
\[ \text{Cl}_2 + 12 \text{OH}^- \rightarrow 2 \text{ClO}_3^{-} + 10 e^- + 6 \text{H}_2\text{O} \]
\[ 5 \text{Cl}_2 + 10 e^- \rightarrow 10 \text{Cl}^- \]

Adding these gives:
\[ 3 \text{Cl}_2 + 6 \text{OH}^- \rightarrow \text{ClO}_3^{-} + 5 \text{Cl}^- + 3 \text{H}_2\text{O} \]

### Step 6: Identify x and y

From the balanced equation:
- The coefficient of \( \text{Cl}_2 \) (x) is 3.
- The coefficient of \( \text{Cl}^- \) (y) is 5.

Thus, the values of x and y are:
\[ x = 3, \quad y = 5 \]

### Final Answer:
The values of \( x \) and \( y \) are \( 3 \) and \( 5 \), respectively.

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