For the cell, `Pt|Cl_(2)(g, 0.4" bar")|Cl^(-)(aq, 0.1M)||Cl^(-)(aq, 0.01M)|Cl_(2)(g,0.2"bar")|Pt`
The measured potential at 298 K is __________ V. [Use `(2.303RT)/(F)=0.0591`]

Explanation

To calculate the measured potential of the given electrochemical cell, we will follow these steps:

### Step 1: Identify the half-reactions
The half-reactions for the cell are:
- **Anode (oxidation)**: \( 2 \text{Cl}^- \rightarrow \text{Cl}_2 + 2 \text{e}^- \)
- **Cathode (reduction)**: \( \text{Cl}_2 + 2 \text{e}^- \rightarrow 2 \text{Cl}^- \)

### Step 2: Write the overall cell reaction
Combining the half-reactions, we get the overall cell reaction:
\[ \text{Cl}_2(g) + 2 \text{Cl}^- \rightarrow 2 \text{Cl}^- + \text{Cl}_2(g) \]
This shows that the same species is both oxidized and reduced, indicating that this is a concentration cell.

### Step 3: Determine the standard cell potential \( E^\circ \)
Since both half-reactions involve the same species, the standard cell potential \( E^\circ \) is zero:
\[ E^\circ_{\text{cell}} = 0 \, \text{V} \]

### Step 4: Use the Nernst equation
The Nernst equation is given by:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log \left( \frac{[\text{Cl}^-]_{\text{cathode}}^2 \cdot P_{\text{Cl}_2, \text{anode}}}{[\text{Cl}^-]_{\text{anode}}^2 \cdot P_{\text{Cl}_2, \text{cathode}}} \right) \]

### Step 5: Identify the concentrations and pressures
From the problem:
- \( [\text{Cl}^-]_{\text{anode}} = 0.1 \, \text{M} \)
- \( [\text{Cl}^-]_{\text{cathode}} = 0.01 \, \text{M} \)
- \( P_{\text{Cl}_2, \text{anode}} = 0.4 \, \text{bar} \)
- \( P_{\text{Cl}_2, \text{cathode}} = 0.2 \, \text{bar} \)
- Number of electrons \( n = 2 \)

### Step 6: Substitute values into the Nernst equation
Substituting the values into the Nernst equation:
\[ E_{\text{cell}} = 0 - \frac{0.0591}{2} \log \left( \frac{(0.01)^2 \cdot 0.4}{(0.1)^2 \cdot 0.2} \right) \]

### Step 7: Simplify the logarithmic expression
Calculating the argument of the logarithm:
\[ \frac{(0.01)^2 \cdot 0.4}{(0.1)^2 \cdot 0.2} = \frac{0.0001 \cdot 0.4}{0.01 \cdot 0.2} = \frac{0.00004}{0.0002} = 0.2 \]
Now, substituting this back into the equation:
\[ E_{\text{cell}} = -\frac{0.0591}{2} \log(0.2) \]

### Step 8: Calculate the logarithm
Using \( \log(0.2) \approx -0.699 \):
\[ E_{\text{cell}} = -\frac{0.0591}{2} \cdot (-0.699) = \frac{0.0591 \cdot 0.699}{2} \]
Calculating this gives:
\[ E_{\text{cell}} \approx \frac{0.0412}{2} \approx 0.0206 \, \text{V} \]

### Step 9: Final calculation
Now, we can finalize:
\[ E_{\text{cell}} \approx 0.0206 \, \text{V} \]

### Conclusion
The measured potential at 298 K is approximately **0.0206 V**.