Which will be the value of x, y and z in the following equaton.
`xI_(2)+yOH^(-)rarrIO_(3)^(-)+zI+3H_(2)O`

Options

A. `{:(x,y,z),(6,3,5):}`
B. `{:(x,y,z),(3,2,3):}`
C. `{:(x,y,z),(3,6,5):}`
D. `{:(x,y,z),(3,3,3):}`

Explanation

To solve the equation \( xI_2 + yOH^- \rightarrow IO_3^- + zI + 3H_2O \), we need to determine the values of \( x \), \( y \), and \( z \) by balancing the redox reaction.

### Step-by-Step Solution:

1. **Identify the Oxidation and Reduction Half-Reactions**:
- The oxidation half-reaction involves the conversion of \( I_2 \) to \( IO_3^- \).
- The reduction half-reaction involves the conversion of \( I_2 \) to \( I^- \).

2. **Write the Oxidation Half-Reaction**:
\[ I_2 + OH^- \rightarrow IO_3^- + H_2O \]
- Balance iodine: \( I_2 \) has 2 I atoms, so we will need 2 \( IO_3^- \).
- Balance oxygen and hydrogen by adding \( OH^- \) and \( H_2O \).

3. **Balance the Oxidation Half-Reaction**:
\[ I_2 + 12OH^- \rightarrow 2IO_3^- + 6H_2O + 10e^- \]
- This balances the charges and atoms.

4. **Write the Reduction Half-Reaction**:
\[ I_2 + 2e^- \rightarrow 2I^- \]

5. **Balance the Reduction Half-Reaction**:
- The reduction half already has balanced iodine and charge.

6. **Combine the Half-Reactions**:
- To combine, we need to equalize the number of electrons. Multiply the reduction half by 5:
\[ 5(I_2 + 2e^- \rightarrow 2I^-) \]
- This gives:
\[ 5I_2 + 10e^- \rightarrow 10I^- \]

7. **Add the Two Half-Reactions**:
\[ (I_2 + 12OH^- \rightarrow 2IO_3^- + 6H_2O + 10e^-) + (5I_2 + 10e^- \rightarrow 10I^-) \]
- This results in:
\[ 6I_2 + 12OH^- \rightarrow 2IO_3^- + 10I^- + 6H_2O \]

8. **Simplify the Equation**:
- Divide the entire equation by 2:
\[ 3I_2 + 6OH^- \rightarrow IO_3^- + 5I^- + 3H_2O \]

9. **Identify Values of x, y, and z**:
- From the balanced equation:
- \( x = 3 \) (coefficient of \( I_2 \))
- \( y = 6 \) (coefficient of \( OH^- \))
- \( z = 5 \) (coefficient of \( I^- \))

### Final Answer:
- The values are:
- \( x = 3 \)
- \( y = 6 \)
- \( z = 5 \)