Iodate ions `(IO_(3)^(-))` can be reduced to iodine by iodide ions. The half equations which represent the redox reaction are:
`IO_(3)^(-)(aq.)+6H^(+)(aq.)+5e^(-) to (1)/(2)I_(2)(s)+3H_(2)O(l)` . . . (i)
`I^(-)(aq.) to (1)/(2)I_(2)(s)+e^(-)` . . . (ii)
How many moles of iodine are produced for every mole of iodate ions consumed in the reaction?

Options

A. 0.5
B. 1
C. 2.5
D. 3

Answer

To determine how many moles of iodine are produced for every mole of iodate ions consumed in the reaction, we will analyze the provided half-reactions step by step.

### Step 1: Write down the half-reactions
The two half-reactions provided are:
1. \( \text{IO}_3^{-} (aq) + 6 \text{H}^{+} (aq) + 5 e^{-} \rightarrow \frac{1}{2} \text{I}_2 (s) + 3 \text{H}_2\text{O} (l) \) (i)
2. \( \text{I}^{-} (aq) \rightarrow \frac{1}{2} \text{I}_2 (s) + e^{-} \) (ii)

### Step 2: Check if the equations are balanced
- For the first half-reaction:
- Iodine: 1 atom on both sides (1 from \( \text{IO}_3^{-} \) and \( \frac{1}{2} \text{I}_2 \)).
- Oxygen: 3 atoms on both sides (3 from \( \text{IO}_3^{-} \) and 3 from \( 3 \text{H}_2\text{O} \)).
- Hydrogen: 6 atoms on both sides (6 from \( 6 \text{H}^{+} \) and 6 from \( 3 \text{H}_2\text{O} \)).
- Charge: 0 on both sides.

The first equation is balanced.

- For the second half-reaction:
- Iodine: 1 atom on both sides (1 from \( \text{I}^{-} \) and \( \frac{1}{2} \text{I}_2 \)).
- Charge: -1 on the left and 0 on the right (due to the loss of 1 electron).

The second equation is also balanced.

### Step 3: Equalize the number of electrons
The first half-reaction involves 5 electrons, while the second half-reaction involves only 1 electron. To equalize the electrons, we multiply the second half-reaction by 5:

\[ 5 \text{I}^{-} (aq) \rightarrow \frac{5}{2} \text{I}_2 (s) + 5 e^{-} \]

### Step 4: Combine the half-reactions
Now we can add the two half-reactions together:

\[ \text{IO}_3^{-} (aq) + 6 \text{H}^{+} (aq) + 5 e^{-} + 5 \text{I}^{-} (aq) \rightarrow \frac{1}{2} \text{I}_2 (s) + 3 \text{H}_2\text{O} (l) + \frac{5}{2} \text{I}_2 (s) + 5 e^{-} \]

The electrons cancel out:

\[ \text{IO}_3^{-} (aq) + 6 \text{H}^{+} (aq) + 5 \text{I}^{-} (aq) \rightarrow 3 \text{I}_2 (s) + 3 \text{H}_2\text{O} (l) \]

### Step 5: Determine the moles of iodine produced
From the balanced equation, we can see that for every mole of \( \text{IO}_3^{-} \) consumed, 3 moles of \( \text{I}_2 \) are produced.

### Final Answer
Therefore, **3 moles of iodine are produced for every mole of iodate ions consumed** in the reaction.

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