`MnO_(4)^(-)` oxidises `H_(2)O_(2)` to `O_(2)` in acidic medium
`xMnO_(4)^(-)+yH_(2)O_(2)+zH^(+) rarr Mn^(2+)+O_(2)+H_(2)O`
Coefficients `x`, `y` and `z` are respectively

Options

A. (`x=1`, `y=2`, `z=3`)
B. (`x=3`, `y=2`, `z=5`)
C. (`x=2`, `y=6`, `z=5`)
D. (`x=2`, `y=5`, `z=6`)

Explanation

To solve the problem of balancing the redox reaction where \( \text{MnO}_4^{-} \) oxidizes \( \text{H}_2\text{O}_2 \) to \( \text{O}_2 \) in acidic medium, we will follow these steps:

### Step 1: Identify the half-reactions
The first step is to identify the oxidation and reduction half-reactions.

**Oxidation half-reaction:**
\( \text{H}_2\text{O}_2 \) is oxidized to \( \text{O}_2 \).
The oxidation state of oxygen in \( \text{H}_2\text{O}_2 \) is -1 and in \( \text{O}_2 \) it is 0.

**Reduction half-reaction:**
\( \text{MnO}_4^{-} \) is reduced to \( \text{Mn}^{2+} \).
The oxidation state of manganese in \( \text{MnO}_4^{-} \) is +7 and in \( \text{Mn}^{2+} \) it is +2.

### Step 2: Balance the oxidation half-reaction
For the oxidation half-reaction:
\[ \text{H}_2\text{O}_2 \rightarrow \text{O}_2 \]

1. **Balance oxygen:** There are 2 oxygen atoms on both sides, so oxygen is balanced.
2. **Balance hydrogen:** Add \( 2 \text{H}^+ \) to the left side:
\[ \text{H}_2\text{O}_2 + 2 \text{H}^+ \rightarrow \text{O}_2 \]
3. **Balance charge by adding electrons:** The left side has a charge of +2 and the right side is neutral. Therefore, we need to add 2 electrons to the left side:
\[ \text{H}_2\text{O}_2 + 2 \text{H}^+ \rightarrow \text{O}_2 + 2 e^- \]

### Step 3: Balance the reduction half-reaction
For the reduction half-reaction:
\[ \text{MnO}_4^{-} \rightarrow \text{Mn}^{2+} \]

1. **Balance manganese:** Manganese is already balanced.
2. **Balance oxygen:** Add \( 4 \text{H}_2\text{O} \) to the right side:
\[ \text{MnO}_4^{-} \rightarrow \text{Mn}^{2+} + 4 \text{H}_2\text{O} \]
3. **Balance hydrogen:** Add \( 8 \text{H}^+ \) to the left side:
\[ \text{MnO}_4^{-} + 8 \text{H}^+ \rightarrow \text{Mn}^{2+} + 4 \text{H}_2\text{O} \]
4. **Balance charge by adding electrons:** The left side has a charge of +7 and the right side has a charge of +2. Therefore, we need to add 5 electrons to the left side:
\[ \text{MnO}_4^{-} + 8 \text{H}^+ + 5 e^- \rightarrow \text{Mn}^{2+} + 4 \text{H}_2\text{O} \]

### Step 4: Equalize the number of electrons
To combine the two half-reactions, we need to equalize the number of electrons transferred. The oxidation half-reaction has 2 electrons, and the reduction half-reaction has 5 electrons. We can multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2.

**Oxidation half-reaction (multiplied by 5):**
\[ 5 \text{H}_2\text{O}_2 + 10 \text{H}^+ \rightarrow 5 \text{O}_2 + 10 e^- \]

**Reduction half-reaction (multiplied by 2):**
\[ 2 \text{MnO}_4^{-} + 16 \text{H}^+ + 10 e^- \rightarrow 2 \text{Mn}^{2+} + 8 \text{H}_2\text{O} \]

### Step 5: Combine the half-reactions
Now we can add the two half-reactions together:
\[ 5 \text{H}_2\text{O}_2 + 2 \text{MnO}_4^{-} + 16 \text{H}^+ \rightarrow 5 \text{O}_2 + 2 \text{Mn}^{2+} + 8 \text{H}_2\text{O} \]

### Step 6: Identify coefficients
From the balanced equation:
- \( x \) (coefficient of \( \text{MnO}_4^{-} \)) = 2
- \( y \) (coefficient of \( \text{H}_2\text{O}_2 \)) = 5
- \( z \) (coefficient of \( \text{H}^+ \)) = 16

### Final Answer
The coefficients are:
- \( x = 2 \)
- \( y = 5 \)
- \( z = 16 \)