Copper sulphate solution is electrolysed between two platinum electrodes. A current is passed unit 1.6 g of oxygen is liberated at anode. The amount of copper deposited at the cathode during the same period is:

Options

A. `6.36 g`
B. `63.6g`
C. `12.7 g`
D. `3.2g`

Answer

To solve the problem of determining the amount of copper deposited at the cathode during the electrolysis of copper sulfate solution, we can follow these steps:

### Step 1: Determine the moles of oxygen liberated
We know that the molar mass of oxygen (O2) is 32 g/mol. Given that 1.6 g of oxygen is liberated, we can calculate the number of moles of oxygen produced.

\[ \text{Moles of } O_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{1.6 \, \text{g}}{32 \, \text{g/mol}} = 0.05 \, \text{mol} \]

### Step 2: Determine the charge required to liberate oxygen
From the electrolysis reaction at the anode, we have:

\[ 2 \text{O}^{2-} + 4 \text{e}^- \rightarrow \text{O}_2 \]

This indicates that 4 moles of electrons are required to produce 1 mole of O2. Therefore, for 0.05 moles of O2, the number of moles of electrons required is:

\[ \text{Moles of electrons} = 0.05 \, \text{mol} \times 4 = 0.2 \, \text{mol} \]

### Step 3: Calculate the total charge (in coulombs)
Using Faraday's constant (approximately 96500 C/mol), we can calculate the total charge:

\[ \text{Charge (Q)} = \text{moles of electrons} \times \text{Faraday's constant} = 0.2 \, \text{mol} \times 96500 \, \text{C/mol} = 19300 \, \text{C} \]

### Step 4: Determine the amount of copper deposited at the cathode
The reduction reaction at the cathode involves the deposition of copper ions:

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

This means that 2 moles of electrons are required to deposit 1 mole of copper. Therefore, the moles of copper deposited can be calculated from the moles of electrons:

\[ \text{Moles of Cu} = \frac{\text{moles of electrons}}{2} = \frac{0.2 \, \text{mol}}{2} = 0.1 \, \text{mol} \]

### Step 5: Calculate the mass of copper deposited
The molar mass of copper (Cu) is approximately 63.5 g/mol. Thus, the mass of copper deposited is:

\[ \text{Mass of Cu} = \text{moles of Cu} \times \text{molar mass of Cu} = 0.1 \, \text{mol} \times 63.5 \, \text{g/mol} = 6.35 \, \text{g} \]

### Final Answer:
The amount of copper deposited at the cathode during the electrolysis is **6.35 g**.

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