In the concentration cell
`Pt (H_(2)) |(HA (0.1M)),(Na A (1M))||(HA (1 M)),(Na A (1M))| (H_(2)) Pt`
`(pK_(a) " of " HA = 4)`
Cell potential will be :

Options

A. 0.03V
B. 0.06V
C. `-0.06V`
D. `-0.03V`

Explanation

To solve the problem of calculating the cell potential for the given concentration cell, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Components of the Cell:**
The cell is represented as:
\[ \text{Pt (H}_2\text{)} | (\text{HA (0.1M)}, \text{Na A (1M)}) || (\text{HA (1M)}, \text{Na A (1M)}) | \text{(H}_2\text{) Pt} \]
Here, we have two half-cells with different concentrations of the weak acid HA.

2. **Determine the pH of Each Half-Cell:**
- For the anode side (left side), where HA concentration is 0.1 M:
\[ \text{pH}_{\text{anode}} = \text{pK}_a + \log\left(\frac{[\text{Na A}]}{[\text{HA}]}\right) \]
Given that \( \text{pK}_a = 4 \), and \( [\text{Na A}] = 1 \, \text{M} \) and \( [\text{HA}] = 0.1 \, \text{M} \):
\[ \text{pH}_{\text{anode}} = 4 + \log\left(\frac{1}{0.1}\right) = 4 + 1 = 5 \]

- For the cathode side (right side), where HA concentration is 1 M:
\[ \text{pH}_{\text{cathode}} = \text{pK}_a + \log\left(\frac{[\text{Na A}]}{[\text{HA}]}\right) \]
Here, \( [\text{Na A}] = 1 \, \text{M} \) and \( [\text{HA}] = 1 \, \text{M} \):
\[ \text{pH}_{\text{cathode}} = 4 + \log\left(\frac{1}{1}\right) = 4 + 0 = 4 \]

3. **Calculate the Cell Potential (E_cell):**
The formula for the cell potential of a concentration cell is:
\[ E_{\text{cell}} = \frac{0.0591}{n} \log\left(\frac{[\text{H}^+]_{\text{cathode}}}{[\text{H}^+]_{\text{anode}}}\right) \]
Since \( n = 1 \) for the transfer of one proton (H⁺), we can simplify:
\[ E_{\text{cell}} = 0.0591 \log\left(\frac{[\text{H}^+]_{\text{cathode}}}{[\text{H}^+]_{\text{anode}}}\right) \]

The concentrations of H⁺ ions can be derived from pH:
\[ [\text{H}^+]_{\text{anode}} = 10^{-\text{pH}_{\text{anode}}} = 10^{-5} \, \text{M} \]
\[ [\text{H}^+]_{\text{cathode}} = 10^{-\text{pH}_{\text{cathode}}} = 10^{-4} \, \text{M} \]

Plugging these values into the equation:
\[ E_{\text{cell}} = 0.0591 \log\left(\frac{10^{-4}}{10^{-5}}\right) = 0.0591 \log(10) = 0.0591 \times 1 = 0.0591 \, \text{V} \]

4. **Final Result:**
Therefore, the cell potential \( E_{\text{cell}} \) is:
\[ E_{\text{cell}} = 0.0591 \, \text{V} \]