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.

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the concentration of H⁺ ions in the HCl solution.
Given:
- Volume of HCl solution = 500 ml = 0.5 L
- Amount of HCl = 25 mmol = 0.025 mol

Concentration of H⁺ from HCl:
\[ [\text{H}^+] = \frac{\text{moles of HCl}}{\text{volume in L}} = \frac{0.025 \text{ mol}}{0.5 \text{ L}} = 0.05 \text{ M} \]

### Step 2: Calculate the concentration of OH⁻ ions in the aniline solution.
Given:
- Amount of aniline = 0.5 mmol = 0.0005 mol
- Volume of aniline solution = 500 ml = 0.5 L

Concentration of aniline:
\[ [\text{Aniline}] = \frac{0.0005 \text{ mol}}{0.5 \text{ L}} = 0.001 \text{ M} \]

Using the base dissociation constant \( K_b = 10^{-9} \):
\[ K_b = [\text{OH}^-]^2 / [\text{Aniline}] \]
Let \( \alpha \) be the degree of dissociation. Then:
\[ [\text{OH}^-] = C \cdot \alpha = 0.001 \cdot \alpha \]
Substituting into the \( K_b \) expression:
\[ 10^{-9} = \frac{(0.001 \cdot \alpha)^2}{0.001} \]
\[ 10^{-9} = 0.001 \cdot \alpha^2 \]
\[ \alpha^2 = 10^{-6} \implies \alpha = 10^{-3} \]
Thus, the concentration of OH⁻ ions:
\[ [\text{OH}^-] = 0.001 \cdot 10^{-3} = 10^{-6} \text{ M} \]

Now, we can find the concentration of H⁺ ions using the water dissociation constant:
\[ K_w = [\text{H}^+][\text{OH}^-] = 10^{-14} \]
\[ [\text{H}^+] = \frac{10^{-14}}{10^{-6}} = 10^{-8} \text{ M} \]

### Step 3: Calculate the EMF of the hydrogen electrodes.
Using the Nernst equation:
\[ E = E^0 - \frac{0.059}{n} \log \left( \frac{1}{[\text{H}^+]} \right) \]
For both electrodes, \( E^0 = 0 \) and \( n = 2 \).

**For HCl:**
\[ E_{\text{HCl}} = 0 - \frac{0.059}{2} \log(0.05) = -0.0295 \log(0.05) \]
Calculating:
\[ \log(0.05) \approx -1.301 \implies E_{\text{HCl}} = -0.0295 \times -1.301 \approx 0.0384 \text{ V} \]

**For Aniline:**
\[ E_{\text{Aniline}} = 0 - \frac{0.059}{2} \log(10^{-8}) = -0.0295 \log(10^{-8}) \]
Calculating:
\[ \log(10^{-8}) = -8 \implies E_{\text{Aniline}} = -0.0295 \times -8 \approx 0.236 \text{ V} \]

### Step 4: Calculate the overall EMF of the cell.
The overall EMF of the cell when connecting the two electrodes:
\[ E_{\text{cell}} = E_{\text{HCl}} - E_{\text{Aniline}} = 0.0384 - 0.236 \approx -0.1976 \text{ V} \]

### Final Answer:
The EMF of the cell made by connecting the hydrogen electrodes is approximately **-0.1976 V**.