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

Options

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

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Calculate the number of moles of hydrogen gas produced.
Given that 2.24 L of hydrogen gas is liberated, we can use the molar volume of a gas at STP (Standard Temperature and Pressure), which is 22.4 L/mol.

\[ \text{Number of moles of } H_2 = \frac{\text{Volume of } H_2}{\text{Molar volume at STP}} = \frac{2.24 \, \text{L}}{22.4 \, \text{L/mol}} = 0.1 \, \text{mol} \]

### Step 2: Determine the charge required to liberate this amount of hydrogen.
From Faraday's laws of electrolysis, we know that 1 mole of \( H_2 \) requires 2 moles of electrons (since \( H_2 \) is produced from \( 2H^+ + 2e^- \)).

Thus, for 0.1 moles of \( H_2 \):
\[ \text{Moles of electrons} = 0.1 \, \text{mol} \times 2 = 0.2 \, \text{mol} \]

### Step 3: Calculate the total charge (Q) using Faraday's constant.
Faraday's constant (F) is approximately \( 96500 \, \text{C/mol} \). The total charge can be calculated as:
\[ Q = \text{Moles of electrons} \times F = 0.2 \, \text{mol} \times 96500 \, \text{C/mol} = 19300 \, \text{C} \]

### Step 4: Calculate the equivalent weight of copper.
The equivalent weight of copper (Cu) can be calculated using its atomic mass and the number of electrons involved in its reduction:
\[ \text{Equivalent weight of Cu} = \frac{\text{Molar mass of Cu}}{n} = \frac{63.5 \, \text{g/mol}}{2} = 31.75 \, \text{g/equiv} \]

### Step 5: Calculate the mass of copper deposited using Faraday's second law.
According to Faraday's second law:
\[ \frac{\text{Weight of Cu}}{\text{Equivalent weight of Cu}} = \frac{\text{Charge}}{F} \]
Rearranging gives:
\[ \text{Weight of Cu} = \text{Equivalent weight of Cu} \times \frac{Q}{F} \]
Substituting the values:
\[ \text{Weight of Cu} = 31.75 \, \text{g/equiv} \times \frac{19300 \, \text{C}}{96500 \, \text{C/mol}} = 31.75 \, \text{g/equiv} \times 0.2 \, \text{mol} = 6.35 \, \text{g} \]

### Final Answer:
The mass of copper deposited at the cathode is **6.35 grams**.

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