Equivalent weight of `FeS_(2) ` in the half reaction
`FeS_(2) rarr Fe_(2)O_(3) + SO_(2)` is :

Options

A. (a) `(M)/(10)`
B. (b) `(M)/(11)`
C. (c) `(M)/(6)`
D. (d) `(M)/(1)`

Answer

To find the equivalent weight of \( \text{FeS}_2 \) in the half-reaction

\[ \text{FeS}_2 \rightarrow \text{Fe}_2\text{O}_3 + \text{SO}_2, \]

we will follow these steps:

### Step 1: Determine the Molar Mass of \( \text{FeS}_2 \)

The molar mass of \( \text{FeS}_2 \) can be calculated using the atomic masses of iron (Fe) and sulfur (S):
- Atomic mass of Fe = 55.85 g/mol
- Atomic mass of S = 32.07 g/mol

The molar mass of \( \text{FeS}_2 \) is:
\[ \text{Molar mass of } \text{FeS}_2 = 1 \times 55.85 + 2 \times 32.07 = 55.85 + 64.14 = 119.99 \, \text{g/mol} \]

### Step 2: Determine the Change in Oxidation States

Next, we need to find the change in oxidation states for both iron and sulfur in the reaction.

1. **Iron (Fe)**:
- In \( \text{FeS}_2 \), iron is in the +2 oxidation state.
- In \( \text{Fe}_2\text{O}_3 \), iron is in the +3 oxidation state.
- The change in oxidation state for Fe is \( +3 - (+2) = +1 \).

2. **Sulfur (S)**:
- In \( \text{FeS}_2 \), sulfur is in the -2 oxidation state.
- In \( \text{SO}_2 \), sulfur is in the +4 oxidation state.
- The change in oxidation state for one sulfur atom is \( +4 - (-2) = +6 \).
- Since there are 2 sulfur atoms, the total change for sulfur is \( 2 \times 6 = 12 \).

### Step 3: Calculate the Total Change in Oxidation States (n-factor)

The n-factor is the total change in oxidation states for all atoms involved in the reaction:
\[ \text{n-factor} = \text{change for Fe} + \text{change for S} = 1 + 12 = 13 \]

### Step 4: Calculate the Equivalent Weight

The equivalent weight of a compound is given by the formula:
\[ \text{Equivalent weight} = \frac{\text{Molar mass}}{n} \]
Substituting the values we found:
\[ \text{Equivalent weight of } \text{FeS}_2 = \frac{119.99 \, \text{g/mol}}{13} \approx 9.23 \, \text{g/equiv} \]

### Final Answer

The equivalent weight of \( \text{FeS}_2 \) in the half-reaction is approximately \( 9.23 \, \text{g/equiv} \).

---