The product of oxidation of `I^(-)` with `MnO_(4)^(-)` in alkaline medium is:

Options

A. `I_(2)`
B. `IO_(3)^(-)`
C. `IO^(-)`
D. `IO_(4)^(-)`

Answer

To solve the question regarding the product of oxidation of \( I^- \) with \( MnO_4^- \) in alkaline medium, we will follow these steps:

### Step 1: Identify the Redox Reaction
In this reaction, \( MnO_4^- \) acts as an oxidizing agent and \( I^- \) is oxidized. We need to write the half-reactions for both the oxidation and reduction processes.

### Step 2: Write the Reduction Half-Reaction
The reduction half-reaction involves \( MnO_4^- \) being reduced. In alkaline medium, \( MnO_4^- \) is reduced to \( MnO_2 \). The balanced half-reaction is:
\[ MnO_4^- + 4OH^- + 2H_2O + 3e^- \rightarrow MnO_2 + 6OH^- \]

### Step 3: Write the Oxidation Half-Reaction
The oxidation half-reaction involves \( I^- \) being oxidized to \( IO_3^- \). The balanced half-reaction is:
\[ I^- \rightarrow IO_3^- + 6e^- \]

### Step 4: Balance the Electrons
To combine the half-reactions, we need to ensure that the number of electrons lost in the oxidation half-reaction equals the number of electrons gained in the reduction half-reaction. The oxidation half-reaction produces 6 electrons, while the reduction half-reaction consumes 3 electrons. Thus, we multiply the reduction half-reaction by 2:
\[ 2MnO_4^- + 8OH^- + 4H_2O + 6e^- \rightarrow 2MnO_2 + 12OH^- \]

### Step 5: Combine the Half-Reactions
Now, we can add the two half-reactions together:
\[ 2MnO_4^- + I^- + 4H_2O \rightarrow 2MnO_2 + IO_3^- + 8OH^- \]

### Step 6: Simplify the Equation
After combining, we can simplify the equation:
\[ 2MnO_4^- + I^- \rightarrow 2MnO_2 + IO_3^- + 4OH^- \]

### Conclusion
The product of the oxidation of \( I^- \) with \( MnO_4^- \) in alkaline medium is \( IO_3^- \).

### Final Answer
The product of oxidation of \( I^- \) with \( MnO_4^- \) in alkaline medium is \( IO_3^- \).

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