Calculate the equivalent weight of `K_(4) [Fe(CN)_(6)]` for the following reaction
`K_(4) [Fe(CN)_(6)] rarr K^(o+) + Fe^(3+) + CO_(2) + NO_(3)^(-) (M = M`. Wt. of `K_(4) [Fe( CN)_(6) ])`

Options

A. `M//1`
B. `M//13`
C. `M//61`
D. `M//48`

Answer

To calculate the equivalent weight of \( K_4[Fe(CN)_6] \) for the reaction:

\[ K_4[Fe(CN)_6] \rightarrow K^+ + Fe^{3+} + CO_2 + NO_3^- \]

we will follow these steps:

### Step 1: Determine the oxidation states of the elements in \( K_4[Fe(CN)_6] \)

1. **Potassium (K)**: Each potassium ion has an oxidation state of \( +1 \).
2. **Iron (Fe)**: To find the oxidation state of iron in \( K_4[Fe(CN)_6] \):
- The complex can be broken down as follows:
\[ 4(+1) + x + 6(-1) = 0 \]
- Solving for \( x \) (the oxidation state of Fe):
\[ 4 + x - 6 = 0 \implies x = +2 \]

3. **Carbon (C) in \( CN^- \)**: The oxidation state of carbon in cyanide can be determined as follows:
- Each \( CN^- \) has a total charge of \( -1 \):
\[ x + (-3) = -1 \implies x = +3 \]
- Therefore, in \( K_4[Fe(CN)_6] \), carbon has an oxidation state of \( +3 \).

4. **Nitrogen (N) in \( CN^- \)**: The oxidation state of nitrogen is \( -1 \).

### Step 2: Identify the products and their oxidation states

1. **For \( CO_2 \)**:
- The oxidation state of carbon in \( CO_2 \) is \( +4 \).

2. **For \( NO_3^- \)**:
- The oxidation state of nitrogen in \( NO_3^- \) is \( +5 \).

### Step 3: Calculate the changes in oxidation states

- **Potassium (K)**: Remains \( +1 \) (no change).
- **Iron (Fe)**: Changes from \( +2 \) to \( +3 \) (increase of 1).
- **Carbon (C)**: Changes from \( +3 \) to \( +4 \) (increase of 1).
- **Nitrogen (N)**: Changes from \( -1 \) in \( CN^- \) to \( +5 \) in \( NO_3^- \) (increase of 6).

### Step 4: Calculate the total change in oxidation states (n-factor)

The total change in oxidation states (n-factor) can be calculated as follows:

\[ \text{n-factor} = \Delta \text{(Fe)} + \Delta \text{(C)} + \Delta \text{(N)} \]

- Change for Fe: \( 1 \)
- Change for C: \( 1 \times 6 = 6 \) (since there are 6 cyanide ions)
- Change for N: \( 6 \times 1 = 6 \) (since there are 6 cyanide ions)

Thus, the total n-factor is:

\[ n = 1 + 6 + 6 = 13 \]

### Step 5: Calculate the equivalent weight

The equivalent weight (EW) is given by the formula:

\[ \text{Equivalent Weight} = \frac{\text{Molar Weight}}{\text{n-factor}} \]

Let \( M \) be the molar weight of \( K_4[Fe(CN)_6] \). Therefore, the equivalent weight is:

\[ \text{Equivalent Weight} = \frac{M}{13} \]

### Conclusion

The equivalent weight of \( K_4[Fe(CN)_6] \) for the given reaction is:

\[ \frac{M}{13} \]