Redox is a reaction in which both oxidation and reduction will take place simultaneously . It is obvious that if one substance gives electron there must be another substance to accept these electrons . In some reactions, same substance is reduced as well as oxidised, these reactions are termed as disproportionation reactions.
For calculating equivalent mass in redox reaction change in oxidtaion number is realted to n-factor which is reciprocal of molar ratio.
The equivalent weight of `Cu_(2)S` in the following reaction is
`Cu_(2)S + O_(2) to Cu^(+2) + SO_(3)`

Options

A. `(M.wt)/(1)`
B. `(M.wt)/(10)`
C. `(M.wt)/(8)`
D. `(M.wt)/(11)`

Answer

To calculate the equivalent weight of \( \text{Cu}_2\text{S} \) in the given redox reaction, we will follow these steps:

### Step 1: Write the balanced redox reaction
The reaction given is:
\[ \text{Cu}_2\text{S} + \text{O}_2 \rightarrow \text{Cu}^{2+} + \text{SO}_3 \]

### Step 2: Determine the oxidation states of the elements
- In \( \text{Cu}_2\text{S} \):
- Let the oxidation state of Cu be \( x \).
- The oxidation state of S is -2.
- The overall charge is 0:
\[ 2x - 2 = 0 \]
\[ 2x = 2 \]
\[ x = +1 \]
- Therefore, the oxidation state of Cu in \( \text{Cu}_2\text{S} \) is +1.

- In \( \text{SO}_3 \):
- Let the oxidation state of S be \( y \).
- The oxidation state of O is -2.
- The overall charge is 0:
\[ y + 3(-2) = 0 \]
\[ y - 6 = 0 \]
\[ y = +6 \]
- Therefore, the oxidation state of S in \( \text{SO}_3 \) is +6.

- In \( \text{O}_2 \):
- The oxidation state is 0 (elemental state).

### Step 3: Calculate the change in oxidation states
- For Cu:
- Changes from +1 in \( \text{Cu}_2\text{S} \) to +2 in \( \text{Cu}^{2+} \).
- Change = \( +2 - (+1) = +1 \).
- Since there are 2 Cu atoms, the total change for Cu is \( 2 \times 1 = 2 \).

- For S:
- Changes from -2 in \( \text{Cu}_2\text{S} \) to +6 in \( \text{SO}_3 \).
- Change = \( +6 - (-2) = +8 \).

### Step 4: Calculate the total change in electrons
- Total change in electrons = Change in Cu + Change in S
\[ \text{Total change} = 2 + 8 = 10 \]

### Step 5: Determine the n-factor
- The n-factor is the total change in electrons involved in the reaction, which we found to be 10.

### Step 6: Calculate the molar mass of \( \text{Cu}_2\text{S} \)
- Molar mass of \( \text{Cu}_2\text{S} \):
- Cu: 63.5 g/mol (2 Cu = 127 g/mol)
- S: 32 g/mol
- Total = \( 127 + 32 = 159 \) g/mol

### Step 7: Calculate the equivalent weight
- The equivalent weight is given by the formula:
\[ \text{Equivalent weight} = \frac{\text{Molar mass}}{\text{n-factor}} \]
\[ \text{Equivalent weight of } \text{Cu}_2\text{S} = \frac{159 \text{ g/mol}}{10} = 15.9 \text{ g/equiv} \]

### Final Answer
The equivalent weight of \( \text{Cu}_2\text{S} \) is **15.9 g/equiv**.

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