2 moles of `KMnO_(4)` can oxidise x moles of hydrogen peroxide in weakly alkaline medium. Find x

Answer

To solve the problem of how many moles of hydrogen peroxide (H₂O₂) can be oxidized by 2 moles of potassium permanganate (KMnO₄) in a weakly alkaline medium, we will follow these steps:

### Step 1: Write the unbalanced chemical equation
The reaction between KMnO₄ and H₂O₂ can be represented as:
\[ \text{KMnO}_4 + \text{H}_2\text{O}_2 \rightarrow \text{MnO}_2 + \text{O}_2 + \text{H}_2\text{O} + \text{KOH} \]

### Step 2: Determine oxidation states
- In KMnO₄, manganese (Mn) has an oxidation state of +7.
- In MnO₂, manganese has an oxidation state of +4.
- In H₂O₂, the oxidation state of oxygen (O) is -1, and in O₂, it is 0.

### Step 3: Identify electron transfer
- The change in oxidation state for Mn from +7 to +4 indicates a reduction of 3 electrons (7 - 4 = 3).
- The change in oxidation state for O from -1 in H₂O₂ to 0 in O₂ indicates an oxidation of 2 electrons (0 - (-1) = 1 for each O, and there are 2 O atoms in H₂O₂, so 2 electrons total).

### Step 4: Balance the half-reactions
To balance the number of electrons transferred:
- For the reduction half-reaction (KMnO₄ to MnO₂), we have 1 KMnO₄ gaining 3 electrons.
- For the oxidation half-reaction (H₂O₂ to O₂), we need to multiply the oxidation half-reaction by 3 to balance the electrons:
\[ 3 \text{H}_2\text{O}_2 \rightarrow 3 \text{O}_2 + 6 \text{H}^+ + 6 e^- \]

### Step 5: Combine and balance the overall reaction
The balanced overall reaction will be:
\[ 2 \text{KMnO}_4 + 3 \text{H}_2\text{O}_2 \rightarrow 2 \text{MnO}_2 + 3 \text{O}_2 + 4 \text{H}_2\text{O} + 2 \text{KOH} \]

### Step 6: Determine the value of x
From the balanced equation, we see that 2 moles of KMnO₄ oxidize 3 moles of H₂O₂. Therefore, the value of x is:
\[ x = 3 \]

### Final Answer
The value of x is 3.

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