The mass of copper that will be deposited at cathode in electrolysis of `0.2M` solution of copper sulphate when a quantity of electricity equal to that required to liberate `2.24L` of hydrogen from `0.1M` aqueous `H_(2)SO_(4)` is passed `(` atomic mass of `Cu=63.5)` will be
(a)`1.59g`
(b)`3.18g`
(c)`6.35g`
(d)`12.70g`

Options

A. `1.59g`
B. `3.18g`
C. `6.35g`
D. `12.70g`

Explanation

To solve the problem of finding the mass of copper deposited at the cathode during the electrolysis of a 0.2 M solution of copper sulfate, we will follow these steps:

### Step 1: Calculate the amount of electricity required to liberate 2.24 L of hydrogen gas.
Using the ideal gas law, we know that 1 mole of gas occupies 22.4 L at STP. Therefore, we can find the number of moles of hydrogen gas liberated from 2.24 L.

\[ \text{Moles of } H_2 = \frac{2.24 \, \text{L}}{22.4 \, \text{L/mol}} = 0.1 \, \text{mol} \]

### Step 2: Determine the number of equivalents of hydrogen.
Since 1 mole of hydrogen gas corresponds to 2 equivalents (as each H atom contributes one electron), we can calculate the equivalents of hydrogen:

\[ \text{Equivalents of } H_2 = 0.1 \, \text{mol} \times 2 = 0.2 \, \text{equivalents} \]

### Step 3: Relate the equivalents of hydrogen to the equivalents of copper deposited.
In the electrolysis of copper sulfate, 1 equivalent of copper (Cu) is deposited for every 2 equivalents of electrons (since Cu²⁺ + 2e⁻ → Cu). Therefore, the equivalents of copper deposited will be the same as the equivalents of hydrogen:

\[ \text{Equivalents of Cu deposited} = 0.2 \, \text{equivalents} \]

### Step 4: Calculate the mass of copper deposited.
The atomic mass of copper (Cu) is given as 63.5 g/mol. Since 1 equivalent of copper corresponds to its atomic mass divided by its valency (which is 2 for Cu²⁺), we can calculate the mass of copper deposited:

\[ \text{Mass of Cu} = \text{Equivalents of Cu} \times \frac{\text{Atomic mass of Cu}}{\text{Valency}} = 0.2 \times \frac{63.5 \, \text{g/mol}}{2} \]

Calculating this gives:

\[ \text{Mass of Cu} = 0.2 \times 31.75 = 6.35 \, \text{g} \]

### Final Answer:
Thus, the mass of copper deposited at the cathode is **6.35 g**.