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: Identify the oxidation states
First, we need to determine the oxidation states of the elements in \( K_4[Fe(CN)_6] \):
- The oxidation state of iron (Fe) in \( K_4[Fe(CN)_6] \) can be calculated as follows:
- Let the oxidation state of Fe be \( x \).
- The CN group has a charge of -1, and there are 6 CN groups, giving a total of -6.
- The overall charge of the complex ion is -4 (since it is \( K_4[Fe(CN)_6] \)).
- Therefore, we have the equation:
\[
x - 6 = -4 \implies x = +2
\]
- Thus, Fe is in the +2 oxidation state.
### Step 2: Determine the change in oxidation states
Next, we analyze the changes in oxidation states during the reaction:
- Iron (Fe) is oxidized from +2 to +3, which is a change of +1. Thus, the change in oxidation number for Fe is 1.
- The carbon in the CN group (C) can be analyzed similarly:
- In \( CN^- \), C is at +2 (since N is -3).
- In the products, C in \( CO_3^{2-} \) is at +4.
- Therefore, the change for C is +2 (from +2 to +4).
- There are 6 carbon atoms, so the total change for carbon is \( 6 \times 2 = 12 \).
### Step 3: Analyze nitrogen oxidation states
- In the CN group, N is at -3, and in \( NO_3^- \), N is at +5.
- The change for nitrogen is \( 5 - (-3) = 8 \).
- There are 6 nitrogen atoms, so the total change for nitrogen is \( 6 \times 8 = 48 \).
### Step 4: Calculate the total change in oxidation states
Now, we sum the changes:
- Total change for Fe: 1
- Total change for C: 12
- Total change for N: 48
- Therefore, the total change (n-factor) for \( K_4[Fe(CN)_6] \) is:
\[
n = 1 + 12 + 48 = 61
\]
### Step 5: Determine the n-factor for \( H_2O_2 \)
In \( H_2O_2 \), the oxidation state of oxygen is -1. When it is converted to water (H2O), the oxidation state of oxygen changes to -2. The change for each oxygen atom is:
\[
-1 \to -2 \quad \text{(change of 1)}
\]
Since there are 2 oxygen atoms in \( H_2O_2 \), the n-factor for \( H_2O_2 \) is:
\[
n = 2
\]
### Step 6: Calculate the moles of \( H_2O_2 \) required
Using the relationship between the n-factors and the moles of reactants:
\[
\text{Moles of } H_2O_2 = \frac{\text{Total n-factor of } K_4[Fe(CN)_6]}{\text{n-factor of } H_2O_2}
\]
Substituting the values:
\[
\text{Moles of } H_2O_2 = \frac{61}{2} = 30.5
\]
### Final Answer
Thus, 30.5 moles of \( H_2O_2 \) will react with 1 mole of \( K_4[Fe(CN)_6] \).
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