In a titration, `H_(2)O_(2)` is oxidised to `O_(2)` by `MnO_(4)^(-)`. 24 mL of 0.1M `H_(2)O_(2)` requires 16 mL of 0.1M `MnO_(4)^(-)` solution. Hence `MnO_(4)^(-)` changes to :

Explanation

To solve the problem, we need to analyze the titration reaction between hydrogen peroxide (H₂O₂) and permanganate ion (MnO₄⁻). Let's break down the steps:

### Step 1: Write the balanced equation for the reaction
In the titration, hydrogen peroxide (H₂O₂) is oxidized to oxygen (O₂) by permanganate ion (MnO₄⁻). The half-reaction for the oxidation of H₂O₂ can be written as:
\[ \text{H}_2\text{O}_2 \rightarrow \text{O}_2 + 2\text{H}^+ + 2\text{e}^- \]

### Step 2: Determine the change in oxidation states
In H₂O₂, the oxidation state of oxygen is -1, and in O₂, it is 0. The change in oxidation state for each oxygen atom is:
\[ -1 \rightarrow 0 \]
Thus, the change in oxidation state is +1 for each oxygen atom. Since there are 2 oxygen atoms in H₂O₂, the total change in oxidation state (n-factor) for H₂O₂ is:
\[ n_{\text{H}_2\text{O}_2} = 2 \times 1 = 2 \]

### Step 3: Calculate the moles of H₂O₂ used
Using the molarity and volume of H₂O₂, we can calculate the moles:
\[ \text{Moles of H}_2\text{O}_2 = \text{Molarity} \times \text{Volume} = 0.1 \, \text{M} \times 0.024 \, \text{L} = 0.0024 \, \text{mol} \]

### Step 4: Relate moles of H₂O₂ to moles of MnO₄⁻
From the stoichiometry of the reaction, we know that the n-factor for H₂O₂ is 2. Therefore, the equivalent moles of H₂O₂ are:
\[ \text{Equivalent moles of H}_2\text{O}_2 = \text{Moles} \times n_{\text{H}_2\text{O}_2} = 0.0024 \, \text{mol} \times 2 = 0.0048 \, \text{equiv} \]

### Step 5: Calculate the moles of MnO₄⁻ used
Now, we can relate the equivalents of H₂O₂ to the equivalents of MnO₄⁻ used:
\[ \text{Equivalent moles of MnO}_4^- = \text{Equivalent moles of H}_2\text{O}_2 \]
Thus, the equivalent moles of MnO₄⁻ is also 0.0048 equiv.

### Step 6: Calculate the n-factor for MnO₄⁻
We know the volume and molarity of MnO₄⁻ used:
\[ \text{Moles of MnO}_4^- = 0.1 \, \text{M} \times 0.016 \, \text{L} = 0.0016 \, \text{mol} \]
The n-factor can be calculated as:
\[ \text{n-factor of MnO}_4^- = \frac{\text{Equivalent moles of MnO}_4^-}{\text{Moles of MnO}_4^-} = \frac{0.0048}{0.0016} = 3 \]

### Step 7: Determine the change in oxidation state of Mn
In MnO₄⁻, the oxidation state of manganese (Mn) is +7. The n-factor we calculated is 3, indicating that the oxidation state of Mn changes by 3. Therefore, the final oxidation state of Mn after the reaction is:
\[ +7 - 3 = +4 \]

### Step 8: Identify the product formed
The compound formed with manganese in the +4 oxidation state is manganese dioxide (MnO₂).

### Final Answer:
Thus, the MnO₄⁻ changes to MnO₂ in the reaction.

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