Three electrolytic cells X,Y,Z containing solution of NaCl, `AgNO_3` and `CuSO_4` respectively are connected in series combination. During electrolysis 21.6gm of silver deposits at cathode in cell Y. Which is incorrect statement.

Options

A. 6.35 gm copper deposits at cathode in cell z
B. 2.24 litre `Cl_2` is liberated(at 1atm and 273 K) at anode in cell X
C. 2.24 litre `O_2` is liberated(at 1atm and 273 K) at anode in cell Y
D. 2.24 litre `H_2` is liberated(at 1atm and 273 K) at anode in cell X

Answer

To solve the problem, we need to analyze the electrolysis of three electrolytic cells connected in series, containing NaCl, AgNO3, and CuSO4 solutions. We are given that 21.6 grams of silver are deposited at the cathode in cell Y (AgNO3). We need to determine which of the provided statements is incorrect.

### Step-by-Step Solution:

1. **Calculate the amount of electricity used for silver deposition:**
- The molar mass of silver (Ag) is approximately 108 g/mol.
- The equivalent weight of silver is calculated as:
\[ \text{Equivalent weight of Ag} = \frac{\text{Molar mass}}{n} = \frac{108}{1} = 108 \, \text{g/equiv} \]
- Using Faraday's law, we can calculate the amount of electricity (Q) that caused the deposition of 21.6 g of silver:
\[ \text{Equivalent of Ag deposited} = \frac{21.6 \, \text{g}}{108 \, \text{g/equiv}} = 0.2 \, \text{equiv} \]
- The total charge (Q) can be calculated using the formula:
\[ Q = n \times F = 0.2 \, \text{equiv} \times 96500 \, \text{C/equiv} = 19300 \, \text{C} \]

2. **Calculate the amount of copper deposited in cell Z:**
- The molar mass of copper (Cu) is approximately 63.5 g/mol.
- The equivalent weight of copper is:
\[ \text{Equivalent weight of Cu} = \frac{63.5}{2} = 31.75 \, \text{g/equiv} \]
- Using Faraday's law again, we can find the weight of copper deposited (W):
\[ \text{Equivalent of Cu deposited} = \frac{W}{31.75} = \frac{19300}{96500} \]
- Rearranging gives:
\[ W = 31.75 \times \frac{19300}{96500} = 6.35 \, \text{g} \]

3. **Calculate the amount of chlorine gas liberated at the anode:**
- The reaction at the anode for NaCl is:
\[ 2 \text{Cl}^- \rightarrow \text{Cl}_2 + 2 \text{e}^- \]
- The equivalent weight of chlorine gas (Cl2) is:
\[ \text{Equivalent weight of Cl2} = \frac{70.9}{2} = 35.45 \, \text{g/equiv} \]
- Using Faraday's law:
\[ \text{Volume of Cl2} = \frac{Q}{n \times F} = \frac{19300}{2 \times 96500} = 0.2 \, \text{equiv} \]
- The volume of Cl2 at STP (22.4 L/equiv):
\[ \text{Volume of Cl2} = 0.2 \times 22.4 = 4.48 \, \text{L} \]

4. **Determine the incorrect statement:**
- From the calculations:
- **Statement A:** 6.35 g of copper deposited at cathode in cell Z (Correct).
- **Statement B:** 2.24 L of Cl2 is liberated at 1 atm and 273 K at anode (Incorrect, it should be 4.48 L).
- **Statement C:** 4.48 L of O2 is liberated at anode (Correct).
- **Statement D:** 2.24 L of H2 is liberated (Correct).

Thus, the incorrect statement is **B**.