In the electolysis of a `CuSO_4` solution, how many grams of Cu are plated out on the cathode in the time that it takes to liberate 5.6 litre of `O_2`(g) , measured at 1 atm and 273 K, at the node?

Options

A. 31.75
B. 14.2
C. 4.32
D. None of these

Answer

To solve the problem of how many grams of copper (Cu) are plated out on the cathode during the electrolysis of a copper sulfate (CuSO₄) solution, given that 5.6 liters of oxygen (O₂) are liberated at the anode, we can follow these steps:

### Step 1: Determine the number of moles of O₂ liberated
At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters. Therefore, to find the number of moles of oxygen gas liberated, we can use the formula:

\[ \text{Number of moles of } O_2 = \frac{\text{Volume of } O_2}{\text{Molar volume at STP}} = \frac{5.6 \text{ L}}{22.4 \text{ L/mol}} = 0.25 \text{ moles} \]

### Step 2: Determine the number of equivalents of O₂
The reaction at the anode for the liberation of oxygen from water is:

\[ 2 H_2O \rightarrow 4 H^+ + 4 e^- + O_2 \]

From this reaction, we see that 1 mole of O₂ corresponds to the transfer of 4 moles of electrons (4 equivalents). Therefore, the number of equivalents of O₂ liberated can be calculated as:

\[ \text{Equivalents of } O_2 = \text{Number of moles of } O_2 \times \text{N-factor} = 0.25 \text{ moles} \times 4 = 1 \text{ equivalent} \]

### Step 3: Relate equivalents of Cu deposited to equivalents of O₂ liberated
According to Faraday's second law of electrolysis, the number of equivalents of copper deposited at the cathode will be equal to the number of equivalents of oxygen liberated at the anode. Therefore:

\[ \text{Equivalents of Cu deposited} = 1 \text{ equivalent} \]

### Step 4: Determine the number of moles of Cu deposited
Copper (Cu) is deposited from the Cu²⁺ ions at the cathode according to the reaction:

\[ Cu^{2+} + 2 e^- \rightarrow Cu \]

Here, 1 mole of Cu corresponds to 2 equivalents (since it requires 2 electrons to deposit 1 mole of Cu). Therefore, the number of moles of Cu deposited can be calculated as:

\[ \text{Moles of Cu} = \frac{\text{Equivalents of Cu}}{2} = \frac{1}{2} = 0.5 \text{ moles} \]

### Step 5: Calculate the mass of Cu deposited
The molar mass of copper (Cu) is approximately 63.5 g/mol. To find the mass of copper deposited, we can use the formula:

\[ \text{Mass of Cu} = \text{Number of moles of Cu} \times \text{Molar mass of Cu} = 0.5 \text{ moles} \times 63.5 \text{ g/mol} = 31.75 \text{ grams} \]

### Final Answer
The amount of copper plated out on the cathode is **31.75 grams**.

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