`MnO_(2)+2KOH+(1)/(2)O_(2) to K_(2)MnO_(4)+H_(2)O`

Options

A. For disproportionation reaction.
B. For comproportionation reaction.
C. For either intermolecular redox reaction or displacement reaction
D. For either thermal combination redox reaction or thermal decomposition redox reaction.

Answer

To determine the type of reaction represented by the equation:

\[ \text{MnO}_2 + 2\text{KOH} + \frac{1}{2}\text{O}_2 \rightarrow \text{K}_2\text{MnO}_4 + \text{H}_2\text{O} \]

we will follow these steps:

### Step 1: Identify the oxidation states of the elements in the reactants.
- **Manganese (Mn)** in MnO₂:
- Oxygen has an oxidation state of -2.
- Let the oxidation state of Mn be \( x \).
- The equation is: \( x + 2(-2) = 0 \)
- Thus, \( x = +4 \).

- **Oxygen (O)** in \( \frac{1}{2}O_2 \):
- The oxidation state of O in \( O_2 \) is 0.

### Step 2: Identify the oxidation states of the elements in the products.
- **Manganese (Mn)** in \( K_2MnO_4 \):
- The oxidation state of O is -2 (4 oxygens).
- Let the oxidation state of Mn be \( y \).
- The equation is: \( 2(+1) + y + 4(-2) = 0 \)
- Thus, \( y = +6 \).

- **Oxygen (O)** in \( H_2O \):
- The oxidation state of O is -2.

### Step 3: Analyze the changes in oxidation states.
- **Mn** changes from +4 in MnO₂ to +6 in \( K_2MnO_4 \) (oxidation).
- **O** changes from 0 in \( \frac{1}{2}O_2 \) to -2 in \( H_2O \) (reduction).

### Step 4: Determine the type of reaction.
Since we have both oxidation (Mn going from +4 to +6) and reduction (O going from 0 to -2) occurring simultaneously, this is classified as a redox reaction.

### Step 5: Identify the specific type of redox reaction.
- The reaction involves different species undergoing oxidation and reduction, which indicates that it is an **intermolecular redox reaction**.

### Conclusion:
The reaction \( \text{MnO}_2 + 2\text{KOH} + \frac{1}{2}\text{O}_2 \rightarrow \text{K}_2\text{MnO}_4 + \text{H}_2\text{O} \) is an **intermolecular redox reaction**.

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