In a redox reaction, `H_2 O_2` oxidizes `K_4 [Fe (CN)_6]` into `Fe^(3+), CO_(3)^(2-) and NO_(3)^(-)` ions in acidic medium, then how many moles of `H_2 O_2` will react with 1 mole of `K_4 [Fe (CN)_6]`

Options

A. 5 moles
B. 9 moles
C. 8 moles
D. 30.5 moles

Answer

To solve the problem of how many moles of \( H_2O_2 \) will react with 1 mole of \( K_4[Fe(CN)_6] \) in a redox reaction, we will follow these steps:

### Step 1: Write the balanced redox reaction
In acidic medium, \( H_2O_2 \) oxidizes \( K_4[Fe(CN)_6] \) to \( Fe^{3+} \), \( CO_3^{2-} \), and \( NO_3^{-} \). The half-reaction for the oxidation of \( K_4[Fe(CN)_6] \) can be written as:
\[ H_2O_2 + 2H^+ + K_4[Fe(CN)_6] \rightarrow K_4[Fe(CN)_6]^{3-} + 2H_2O \]

### Step 2: Determine the n-factor of \( H_2O_2 \)
The n-factor of \( H_2O_2 \) in this reaction is 2, as it is reduced to water.

### Step 3: Determine the n-factor of \( K_4[Fe(CN)_6] \)
Next, we need to find the n-factor for \( K_4[Fe(CN)_6] \):
- The potassium ion \( K^+ \) does not change, so its contribution is 0.
- The iron \( Fe^{2+} \) is oxidized to \( Fe^{3+} \), which contributes 1 electron.
- Each \( CN^{-} \) (C in +2 oxidation state) is oxidized to \( C^{4+} \), contributing 2 electrons. Since there are 6 \( CN^{-} \) groups, this contributes \( 2 \times 6 = 12 \) electrons.
- The \( NO_3^{-} \) ions are formed from \( CN^{-} \) which contributes 8 electrons for each \( CN^{-} \) changing from \( CN^{-} \) to \( NO_3^{-} \). For 6 \( CN^{-} \), this contributes \( 8 \times 6 = 48 \) electrons.

Adding these contributions together:
\[ 0 + 1 + 12 + 48 = 61 \]
Thus, the n-factor of \( K_4[Fe(CN)_6] \) is 61.

### Step 4: Calculate the moles of \( H_2O_2 \) required
Using the relationship between n-factor and moles:
- For 1 mole of \( K_4[Fe(CN)_6] \) (n = 61), we can find the moles of \( H_2O_2 \) needed:
\[ \text{Moles of } H_2O_2 = \frac{n \text{ (of } K_4[Fe(CN)_6])}{\text{n-factor of } H_2O_2} = \frac{61}{2} = 30.5 \]

### Final Answer
Thus, **30.5 moles of \( H_2O_2 \)** will react with 1 mole of \( K_4[Fe(CN)_6] \).

---