What is the equivalent weight of reductant in the given redox reaction, `Cu_(2)S+(5)/(2)O_(2)rarr2CuO+SO_(3)`

Options

A. `(M)/(5)`
B. `(M)/(8)`
C. `(M)/(10)`
D. `(M)/(4)`

Answer

To calculate the equivalent weight of the reductant in the given redox reaction, we will follow these steps:

### Step 1: Identify the reaction
The given redox reaction is:
\[ \text{Cu}_2\text{S} + \frac{5}{2} \text{O}_2 \rightarrow 2 \text{CuO} + \text{SO}_3 \]

### Step 2: Determine the oxidation states
- In \(\text{Cu}_2\text{S}\):
- Copper (Cu) is in the +1 oxidation state.
- Sulfur (S) is in the -2 oxidation state.

- In \(\text{CuO}\):
- Copper (Cu) is in the +2 oxidation state.

- In \(\text{SO}_3\):
- Sulfur (S) is in the +6 oxidation state.

### Step 3: Calculate the change in oxidation states
- For Copper:
- Change from +1 to +2:
\[ \Delta \text{Oxidation state} = 2 - 1 = 1 \]
- Since there are 2 copper atoms, the total change for copper is:
\[ 2 \times 1 = 2 \]

- For Sulfur:
- Change from -2 to +6:
\[ \Delta \text{Oxidation state} = 6 - (-2) = 6 + 2 = 8 \]
- Since there is 1 sulfur atom, the total change for sulfur is:
\[ 1 \times 8 = 8 \]

### Step 4: Calculate the n-factor
The n-factor is the total change in oxidation states:
\[ n = \text{Total change for Cu} + \text{Total change for S} = 2 + 8 = 10 \]

### Step 5: Calculate the equivalent weight
The equivalent weight (EW) is given by the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molar Mass}}{n} \]
Let \( M \) be the molar mass of the reductant (Cu\(_2\)S). Therefore, the equivalent weight is:
\[ \text{Equivalent Weight} = \frac{M}{10} \]

### Final Answer
The equivalent weight of the reductant in the given reaction is:
\[ \frac{M}{10} \]

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