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`
Emf is?
(a)0.051V
(b)-0.051V
(c)0.102V
(d)0.0255V

Options

A. 0.051V
B. -0.051
C. 0.102V
D. 0.0255V

Answer

To find the EMF of the given electrochemical cell, we will follow these steps:

### Step 1: Identify the half-reactions
The cell notation indicates that the left side is the anode and the right side is the cathode.

- **Anode (oxidation)**:
\[ 2Cl^-(aq) \rightarrow Cl_2(g) + 2e^- \]

- **Cathode (reduction)**:
\[ Cl_2(g) + 2e^- \rightarrow 2Cl^-(aq) \]

### Step 2: Write the overall cell reaction
Combining the half-reactions, we get:
\[ 2Cl^-(aq) + Cl_2(g) \rightarrow Cl_2(g) + 2Cl^-(aq) \]
This simplifies to:
\[ Cl_2(g) + 2Cl^-(aq) \rightarrow 2Cl^-(aq) + Cl_2(g) \]

### Step 3: Apply the Nernst equation
The Nernst equation is given by:
\[ E = E^\circ - \frac{0.059}{n} \log \left( \frac{[Cl^-]_{right}}{[Cl^-]_{left} \cdot P_{Cl_2}^{left}} \right) \]
Where:
- \(E^\circ\) is the standard EMF (which is 0 for this cell since both half-reactions are identical).
- \(n\) is the number of electrons transferred (which is 2).

### Step 4: Substitute the values into the Nernst equation
From the problem:
- Concentration of \(Cl^-\) on the right (\([Cl^-]_{right}\)) = 0.01 M
- Concentration of \(Cl^-\) on the left (\([Cl^-]_{left}\)) = 0.1 M
- Pressure of \(Cl_2\) on the left (\(P_{Cl_2}^{left}\)) = 0.4 bar
- Pressure of \(Cl_2\) on the right (\(P_{Cl_2}^{right}\)) = 0.2 bar

Substituting these values into the Nernst equation:
\[ E = 0 - \frac{0.059}{2} \log \left( \frac{0.01}{0.1 \cdot 0.4} \right) \]

### Step 5: Calculate the logarithm
Calculating the fraction:
\[ \frac{0.01}{0.1 \cdot 0.4} = \frac{0.01}{0.04} = 0.25 \]
Now, calculate the logarithm:
\[ \log(0.25) = -0.602 \]

### Step 6: Substitute back into the Nernst equation
Now substitute this value back into the equation:
\[ E = 0 - \frac{0.059}{2} \times (-0.602) \]
\[ E = \frac{0.059 \times 0.602}{2} = \frac{0.035478}{2} = 0.017739 \]
\[ E \approx 0.051 V \]

### Conclusion
The EMF of the cell is approximately **0.051 V**, which corresponds to option (a).

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