`IO_(3)^(-)+I^(-)+H^(+) to`

Options

A. 5,1,6
B. 1,5,6
C. 6,1,5
D. 5,6,1

Answer

To solve the redox reaction involving \( IO_3^- \), \( I^- \), and \( H^+ \), we will follow a systematic approach to balance the reaction and determine the coefficients of each species involved.

### Step-by-Step Solution

**Step 1: Identify the oxidation states.**
- For \( IO_3^- \): The oxidation state of iodine (I) can be calculated as follows:
\[ x - 6 = -1 \implies x = +5 \]
- For \( I^- \): The oxidation state is \( -1 \).
- For \( I_2 \): The oxidation state is \( 0 \).

**Step 2: Write the half-reactions.**
- **Oxidation half-reaction** (for \( I^- \) to \( I_2 \)):
\[ 2 I^- \rightarrow I_2 + 2 e^- \]
- **Reduction half-reaction** (for \( IO_3^- \) to \( I_2 \)):
\[ 2 IO_3^- + 10 e^- + 12 H^+ \rightarrow 6 I_2 + 6 H_2O \]

**Step 3: Balance the half-reactions.**
- The oxidation half-reaction is already balanced.
- For the reduction half-reaction, we need to ensure that both iodine atoms and the charges are balanced. We have:
- 2 moles of \( IO_3^- \) giving 6 moles of \( I_2 \).
- The electrons and protons are balanced as well.

**Step 4: Combine the half-reactions.**
- To combine, we need to ensure that the electrons cancel out. We multiply the oxidation half-reaction by 5:
\[ 10 I^- \rightarrow 5 I_2 + 10 e^- \]
- Now, add the two half-reactions:
\[ 2 IO_3^- + 10 I^- + 12 H^+ \rightarrow 6 I_2 + 6 H_2O \]

**Step 5: Write the final balanced equation.**
- The final balanced equation is:
\[ 2 IO_3^- + 10 I^- + 12 H^+ \rightarrow 6 I_2 + 6 H_2O \]

**Step 6: Determine the coefficients.**
- The coefficients for the balanced equation are:
- \( IO_3^- \): 2
- \( I^- \): 10
- \( H^+ \): 12
- \( I_2 \): 6
- \( H_2O \): 6

### Final Answer
The coefficients are:
- \( IO_3^- \): 2
- \( I^- \): 10
- \( H^+ \): 12
- \( I_2 \): 6
- \( H_2O \): 6