To solve the problem step by step, we will use the information provided about the electrolysis process and the stoichiometry of the reactions involved.
### Step 1: Calculate the moles of oxygen produced
Given that the volume of oxygen liberated at NTP (Normal Temperature and Pressure) is 11.2 liters, we can calculate the number of moles of oxygen.
At NTP, 1 mole of gas occupies 22.4 liters. Therefore, the number of moles of oxygen (\( O_2 \)) can be calculated as:
\[
\text{Moles of } O_2 = \frac{\text{Volume of } O_2}{\text{Molar volume at NTP}} = \frac{11.2 \, \text{liters}}{22.4 \, \text{liters/mole}} = 0.5 \, \text{moles}
\]
### Step 2: Calculate the mass of oxygen produced
The molar mass of \( O_2 \) is 32 g/mol. Therefore, the mass of oxygen produced can be calculated as:
\[
\text{Mass of } O_2 = \text{Moles of } O_2 \times \text{Molar mass of } O_2 = 0.5 \, \text{moles} \times 32 \, \text{g/mole} = 16 \, \text{grams}
\]
### Step 3: Determine the equivalents of oxygen
The equivalent weight of \( O_2 \) can be calculated as follows:
\[
\text{Equivalent weight of } O_2 = \frac{\text{Molar mass of } O_2}{\text{n-factor}} = \frac{32 \, \text{g}}{2} = 16 \, \text{g/equiv}
\]
Since we produced 16 grams of \( O_2 \), the number of equivalents of \( O_2 \) produced is:
\[
\text{Equivalents of } O_2 = \frac{\text{Mass of } O_2}{\text{Equivalent weight of } O_2} = \frac{16 \, \text{g}}{16 \, \text{g/equiv}} = 1 \, \text{equiv}
\]
### Step 4: Determine the equivalents of copper deposited
The reaction at the cathode involves the deposition of copper from copper sulfate (\( CuSO_4 \)). The equivalent weight of copper (\( Cu \)) is calculated as:
\[
\text{Equivalent weight of } Cu = \frac{\text{Molar mass of } Cu}{\text{n-factor}} = \frac{63.5 \, \text{g}}{2} = 31.75 \, \text{g/equiv}
\]
### Step 5: Calculate the mass of copper deposited
Using the equivalents of \( O_2 \) produced, we can find the mass of copper deposited using the ratio of equivalents:
\[
\frac{\text{Mass of } Cu}{\text{Equivalent weight of } Cu} = \frac{\text{Mass of } O_2}{\text{Equivalent weight of } O_2}
\]
Let \( x \) be the mass of copper deposited:
\[
\frac{x}{31.75} = \frac{16}{16}
\]
This simplifies to:
\[
x = 31.75 \, \text{grams}
\]
### Conclusion
The amount of copper deposited at the cathode is **31.75 grams**.
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