`2IO_(3)^(-)+xI^(-)+12H^(+)rarr6I_(2)+6H_2O`
What is the value of x?

Options

A. `12`
B. `6`
C. `10`
D. `2`

Answer

To find the value of \( x \) in the reaction

\[ 2IO_3^{-} + xI^{-} + 12H^{+} \rightarrow 6I_2 + 6H_2O, \]

we will determine the change in oxidation states of iodine in the reactants and products.

### Step 1: Determine the oxidation states

1. **Oxidation state of iodine in \( IO_3^{-} \)**:
- Let the oxidation state of iodine be \( x \).
- The oxidation state of oxygen is \(-2\).
- The equation becomes:
\[ x + 3(-2) = -1 \implies x - 6 = -1 \implies x = +5. \]
- Therefore, the oxidation state of iodine in \( IO_3^{-} \) is \( +5 \).

2. **Oxidation state of iodine in \( I^{-} \)**:
- The oxidation state of iodine in \( I^{-} \) is \(-1\).

3. **Oxidation state of iodine in \( I_2 \)**:
- The oxidation state of iodine in \( I_2 \) is \( 0 \).

### Step 2: Calculate the change in oxidation states

1. **For \( IO_3^{-} \)**:
- Change from \( +5 \) to \( 0 \):
\[ \Delta = 5 - 0 = 5. \]
- Since there are 2 moles of \( IO_3^{-} \), the total change for \( 2IO_3^{-} \) is:
\[ 2 \times 5 = 10. \]

2. **For \( I^{-} \)**:
- Change from \( -1 \) to \( 0 \):
\[ \Delta = 0 - (-1) = 1. \]
- Let \( x \) be the number of moles of \( I^{-} \). The total change for \( xI^{-} \) is:
\[ x \times 1 = x. \]

### Step 3: Set up the equation for electron transfer

In a balanced redox reaction, the total increase in oxidation states (loss of electrons) must equal the total decrease in oxidation states (gain of electrons). Therefore, we can set up the equation:

\[ 10 = x. \]

### Step 4: Solve for \( x \)

From the equation above, we find that:

\[ x = 10. \]

### Conclusion

Thus, the value of \( x \) is \( 10 \).

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