The coefficients of `I^(-), IO_(3)^(-)` and `H^(+)` in the balanced redox reaction,`I^(-) + IO_(3)^(-) + H^(+ )to I_(2) + H_(2)O^(-)`, are respectively

Options

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

Explanation

To balance the redox reaction \( I^- + IO_3^- + H^+ \rightarrow I_2 + H_2O \), we will follow these steps:

### Step 1: Identify Oxidation and Reduction
- **Oxidation**: \( I^- \) (oxidation state -1) is oxidized to \( I_2 \) (oxidation state 0).
- **Reduction**: \( IO_3^- \) (oxidation state +5) is reduced to \( I_2 \) (oxidation state 0).

### Step 2: Write Half-Reactions
1. **Oxidation Half-Reaction**:
\[ 2I^- \rightarrow I_2 + 2e^- \]
Here, 2 moles of \( I^- \) produce 1 mole of \( I_2 \) and release 2 electrons.

2. **Reduction Half-Reaction**:
\[ IO_3^- + 12H^+ + 10e^- \rightarrow I_2 + 6H_2O \]
Here, 1 mole of \( IO_3^- \) is reduced to \( I_2 \) while consuming 10 electrons and producing 6 moles of water.

### Step 3: Balance Electrons
To combine the half-reactions, we need to balance the number of electrons. The oxidation half-reaction produces 2 electrons, while the reduction half-reaction consumes 10 electrons. Therefore, we multiply the oxidation half-reaction by 5 to equalize the electrons:

\[ 5(2I^- \rightarrow I_2 + 2e^-) \Rightarrow 10I^- \rightarrow 5I_2 + 10e^- \]

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

\[ 10I^- + IO_3^- + 12H^+ + 10e^- \rightarrow 5I_2 + 6H_2O + 10e^- \]

### Step 5: Cancel Electrons
The electrons on both sides cancel out:

\[ 10I^- + IO_3^- + 12H^+ \rightarrow 5I_2 + 6H_2O \]

### Step 6: Write the Final Balanced Equation
The final balanced equation is:

\[ 10I^- + IO_3^- + 12H^+ \rightarrow 5I_2 + 6H_2O \]

### Step 7: Identify Coefficients
From the balanced equation, we can identify the coefficients:
- Coefficient of \( I^- \) is **10**.
- Coefficient of \( IO_3^- \) is **1**.
- Coefficient of \( H^+ \) is **12**.

### Final Answer:
The coefficients of \( I^- \), \( IO_3^- \), and \( H^+ \) in the balanced redox reaction are **10**, **1**, and **12**, respectively.

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