In two vessels each containing 500 ml water, 0.5m mol of aniline (`K_(b)=10^(-9))` and 25 m mol of HCl are added separately. Two hydrogen electrodes are constructed using these solutions. Calculate the emf of cell made by connecting them appropriately.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the concentration of aniline
Given:
- Amount of aniline = 0.5 mmol
- Volume of solution = 500 ml = 0.5 L

Concentration of aniline (C_aniline) can be calculated as:
\[ C_{\text{aniline}} = \frac{\text{Amount of aniline}}{\text{Volume}} = \frac{0.5 \, \text{mmol}}{500 \, \text{ml}} = \frac{0.5 \times 10^{-3} \, \text{mol}}{0.5 \, \text{L}} = 10^{-3} \, \text{mol/L} \]

### Step 2: Calculate the degree of ionization (α) of aniline
Using the formula for the base dissociation constant (K_b):
\[ K_b = C \cdot \alpha^2 \]
Where:
- \( K_b = 10^{-9} \)
- \( C = 10^{-3} \)

Substituting the values:
\[ 10^{-9} = 10^{-3} \cdot \alpha^2 \]
\[ \alpha^2 = \frac{10^{-9}}{10^{-3}} = 10^{-6} \]
\[ \alpha = 10^{-3} \]

### Step 3: Calculate the concentration of OH⁻ ions in the aniline solution
Using:
\[ [\text{OH}^-] = C \cdot \alpha = 10^{-3} \cdot 10^{-3} = 10^{-6} \, \text{mol/L} \]

### Step 4: Calculate the concentration of H⁺ ions in the aniline solution
Using the ion product of water:
\[ K_w = [\text{H}^+] \cdot [\text{OH}^-] = 10^{-14} \]
Thus,
\[ [\text{H}^+] = \frac{10^{-14}}{[\text{OH}^-]} = \frac{10^{-14}}{10^{-6}} = 10^{-8} \, \text{mol/L} \]

### Step 5: Calculate the concentration of H⁺ ions in the HCl solution
Given:
- Amount of HCl = 25 mmol
- Volume of solution = 500 ml = 0.5 L

Concentration of HCl (C_HCl) can be calculated as:
\[ C_{\text{HCl}} = \frac{25 \, \text{mmol}}{500 \, \text{ml}} = \frac{25 \times 10^{-3} \, \text{mol}}{0.5 \, \text{L}} = 5 \times 10^{-2} \, \text{mol/L} \]
Since HCl is a strong acid, the concentration of H⁺ ions is equal to the concentration of HCl:
\[ [\text{H}^+] = 5 \times 10^{-2} \, \text{mol/L} \]

### Step 6: Calculate the EMF of the hydrogen electrode for aniline
Using the Nernst equation:
\[ E = E^\circ - \frac{0.059}{n} \log [\text{H}^+] \]
For the standard hydrogen electrode, \( E^\circ = 0 \) and \( n = 1 \):
\[ E_{\text{aniline}} = 0 - \frac{0.059}{1} \log(10^{-8}) = 0.059 \cdot 8 = 0.472 \, \text{V} \]

### Step 7: Calculate the EMF of the hydrogen electrode for HCl
Using the same Nernst equation:
\[ E_{\text{HCl}} = 0 - \frac{0.059}{1} \log(5 \times 10^{-2}) \]
Calculating:
\[ \log(5 \times 10^{-2}) = \log(5) + \log(10^{-2}) \approx 0.699 - 2 = -1.301 \]
Thus,
\[ E_{\text{HCl}} = 0.059 \cdot 1.301 \approx 0.077 \, \text{V} \]

### Step 8: Calculate the total EMF of the cell
The total EMF when connecting the two electrodes:
\[ E_{\text{total}} = E_{\text{aniline}} - E_{\text{HCl}} = 0.472 \, \text{V} - 0.077 \, \text{V} = 0.395 \, \text{V} \]

### Final Answer:
The EMF of the cell is approximately **0.395 V**.

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