Balance the following half-reactions in acidic medium:
(a) `IO_(3)^(Ө)(aq) rarr I_(3)^(Ө)(aq)`

Explanation

To balance the half-reaction \( IO_3^{-} \rightarrow I_3^{-} \) in acidic medium, we will follow these steps:

### Step 1: Write the unbalanced half-reaction
The unbalanced half-reaction is:
\[ IO_3^{-} \rightarrow I_3^{-} \]

### Step 2: Balance the iodine atoms
On the left side, we have 1 iodine atom in \( IO_3^{-} \) and on the right side, we have 3 iodine atoms in \( I_3^{-} \). To balance the iodine atoms, we need to have 3 \( IO_3^{-} \) on the left side:
\[ 3 IO_3^{-} \rightarrow I_3^{-} \]

### Step 3: Balance the oxygen atoms
Now, we have 9 oxygen atoms on the left side (from 3 \( IO_3^{-} \)). To balance the oxygen atoms, we will add water molecules to the right side. Since there are 9 oxygen atoms on the left, we add 9 \( H_2O \) to the right side:
\[ 3 IO_3^{-} \rightarrow I_3^{-} + 9 H_2O \]

### Step 4: Balance the hydrogen atoms
Now we have 18 hydrogen atoms on the right side (from 9 \( H_2O \)). To balance the hydrogen atoms, we will add 18 \( H^{+} \) ions to the left side:
\[ 3 IO_3^{-} + 18 H^{+} \rightarrow I_3^{-} + 9 H_2O \]

### Step 5: Balance the charges
Now, let's calculate the charges. The left side has:
- Charge from \( 3 IO_3^{-} \): \( 3 \times (-1) = -3 \)
- Charge from \( 18 H^{+} \): \( +18 \)
- Total charge on the left: \( -3 + 18 = +15 \)

The right side has:
- Charge from \( I_3^{-} \): \( -1 \)
- Total charge on the right: \( -1 \)

To balance the charges, we need to add electrons to the left side. The left side has a total charge of +15 and the right side has -1, so we need to add 16 electrons to the left side:
\[ 3 IO_3^{-} + 18 H^{+} + 16 e^{-} \rightarrow I_3^{-} + 9 H_2O \]

### Final Balanced Half-Reaction
The balanced half-reaction in acidic medium is:
\[ 3 IO_3^{-} + 18 H^{+} + 16 e^{-} \rightarrow I_3^{-} + 9 H_2O \]

---