`I^(-)` reduces `HNO_(2)` to :

Explanation

To solve the question of what \( I^- \) reduces \( HNO_2 \) to, we will follow these steps:

### Step 1: Identify the Reactants and Products
The reactants are \( I^- \) (iodide ion) and \( HNO_2 \) (nitrous acid). We need to determine the products of the reaction.

**Hint:** Recognize that \( I^- \) is a reducing agent and will undergo oxidation, while \( HNO_2 \) will undergo reduction.

### Step 2: Write the Unbalanced Reaction
The unbalanced reaction can be written as:
\[ I^- + HNO_2 \rightarrow I_2 + N_2 + H_2O \]

**Hint:** Start with the basic skeleton of the reaction before balancing.

### Step 3: Determine Oxidation States
Next, we need to assign oxidation states to identify which species are oxidized and reduced:
- For \( I^- \), the oxidation state is -1.
- For \( HNO_2 \), nitrogen has an oxidation state of +3.
- In the products, \( I_2 \) has an oxidation state of 0 (elemental iodine), and \( N_2 \) has an oxidation state of 0 (elemental nitrogen).

**Hint:** Compare the oxidation states of reactants and products to identify oxidation and reduction.

### Step 4: Write Half-Reactions
1. **Oxidation Half-Reaction:**
\[ 2 I^- \rightarrow I_2 + 2 e^- \]
(Iodide is oxidized to iodine)

2. **Reduction Half-Reaction:**
\[ 2 HNO_2 + 6 H^+ + 6 e^- \rightarrow N_2 + 4 H_2O \]
(Nitrous acid is reduced to nitrogen gas)

**Hint:** Write separate half-reactions for oxidation and reduction to clarify the electron transfer.

### Step 5: Balance the Half-Reactions
- The oxidation half-reaction is already balanced.
- For the reduction half-reaction, ensure that the number of atoms and charge are balanced.

**Hint:** Check that both mass and charge are balanced in each half-reaction.

### Step 6: Combine the Half-Reactions
To combine the half-reactions, we need to ensure that the electrons cancel out. We can multiply the oxidation half-reaction by 3 to match the 6 electrons in the reduction half-reaction:
\[ 3(2 I^- \rightarrow I_2 + 2 e^-) \]
This gives:
\[ 6 I^- \rightarrow 3 I_2 + 6 e^- \]

Now, combine the two half-reactions:
\[ 6 I^- + 2 HNO_2 + 6 H^+ \rightarrow 3 I_2 + N_2 + 4 H_2O \]

**Hint:** Make sure to multiply the half-reactions appropriately to balance the electrons.

### Step 7: Write the Final Balanced Equation
The final balanced equation is:
\[ 6 I^- + 2 HNO_2 + 6 H^+ \rightarrow 3 I_2 + N_2 + 4 H_2O \]

### Conclusion
From the balanced equation, we can see that \( HNO_2 \) is reduced to \( N_2 \) gas.

**Final Answer:** \( I^- \) reduces \( HNO_2 \) to \( N_2 \) gas.

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