When `Cu_(2)S` is converted into `Cu^(2+)` & `SO_(2)` then equivalent weight of `Cu_(2)S` will be (M=mol. Wt. of `Cu_(2)S`)

Options

A. M
B. `M/2`
C. `M/4`
D. `M/8`

Answer

To find the equivalent weight of \( \text{Cu}_2\text{S} \) when it is converted into \( \text{Cu}^{2+} \) and \( \text{SO}_2 \), we can follow these steps:

### Step 1: Write the balanced reaction
The reaction can be represented as:
\[ \text{Cu}_2\text{S} + \text{O}_2 \rightarrow 2\text{Cu}^{2+} + \text{SO}_2 \]

### Step 2: Determine the oxidation states
- In \( \text{Cu}_2\text{S} \):
- The oxidation state of sulfur (S) is \(-2\).
- Let the oxidation state of copper (Cu) be \(x\). Since there are 2 copper atoms, we have:
\[ 2x - 2 = 0 \implies 2x = 2 \implies x = +1 \]
- Therefore, the oxidation state of Cu in \( \text{Cu}_2\text{S} \) is \( +1 \).

- In \( \text{SO}_2 \):
- The oxidation state of sulfur (S) is \( +4 \).

### Step 3: Calculate the change in oxidation states
- For copper:
- Change from \( +1 \) to \( +2 \) (for 2 Cu):
\[ \Delta \text{Oxidation state for Cu} = 2 \times (2 - 1) = 2 \]

- For sulfur:
- Change from \( -2 \) to \( +4 \):
\[ \Delta \text{Oxidation state for S} = 4 - (-2) = 6 \]

### Step 4: Total change in oxidation states
- Total change in oxidation states:
\[ \text{Total change} = 2 + 6 = 8 \]

### Step 5: Calculate the equivalent weight
- The equivalent weight (EW) is given by the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molecular Weight}}{\text{x factor}} \]
- Here, the molecular weight of \( \text{Cu}_2\text{S} \) is \( M \) and the x factor (total change in oxidation states) is \( 8 \):
\[ \text{Equivalent Weight} = \frac{M}{8} \]

### Final Answer
The equivalent weight of \( \text{Cu}_2\text{S} \) is \( \frac{M}{8} \).

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