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`

Options

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

Answer

To calculate the EMF (Electromotive Force) of the given electrochemical cell, we will follow these steps:

### Step 1: Identify the Anode and Cathode Reactions
- The left side of the cell is the anode, where oxidation occurs.
- The right side of the cell is the cathode, where reduction occurs.

**Anode Reaction:**
The oxidation half-reaction at the anode involves the conversion of chloride ions (Cl⁻) to chlorine gas (Cl₂):
\[ 2 \text{Cl}^- \rightarrow \text{Cl}_2(g) + 2 \text{e}^- \]

**Cathode Reaction:**
The reduction half-reaction at the cathode involves the conversion of chlorine gas (Cl₂) to chloride ions (Cl⁻):
\[ \text{Cl}_2(g) + 2 \text{e}^- \rightarrow 2 \text{Cl}^- \]

### Step 2: Write the Net Cell Reaction
By combining the anode and cathode reactions, we can cancel out the electrons:
\[ 2 \text{Cl}^- + \text{Cl}_2(g) \rightarrow 2 \text{Cl}^- + \text{Cl}_2(g) \]

### Step 3: Use the Nernst Equation
The Nernst equation for the cell is given by:
\[ E = E^0 - \frac{0.059}{n} \log Q \]
Where:
- \( E^0 \) is the standard electrode potential (which is 0 for this cell since both sides have the same species).
- \( n \) is the number of electrons transferred (which is 2).
- \( Q \) is the reaction quotient.

### Step 4: Calculate the Reaction Quotient (Q)
The reaction quotient \( Q \) is calculated as follows:
\[ Q = \frac{[\text{Cl}^-]^2 \cdot P_{\text{Cl}_2 (LHS)}}{[\text{Cl}^-]^2 \cdot P_{\text{Cl}_2 (RHS)}} \]
Substituting the values:
- Concentration of Cl⁻ on the left = 0.1 M
- Concentration of Cl⁻ on the right = 0.01 M
- Pressure of Cl₂ on the left = 0.4 bar
- Pressure of Cl₂ on the right = 0.2 bar

Thus,
\[ Q = \frac{(0.01)^2 \cdot 0.4}{(0.1)^2 \cdot 0.2} \]

### Step 5: Simplify Q
Calculating \( Q \):
\[ Q = \frac{0.0001 \cdot 0.4}{0.01 \cdot 0.2} = \frac{0.00004}{0.002} = 0.02 \]

### Step 6: Substitute into the Nernst Equation
Now substituting into the Nernst equation:
\[ E = 0 - \frac{0.059}{2} \log(0.02) \]

### Step 7: Calculate the EMF
Calculating \( \log(0.02) \):
\[ \log(0.02) \approx -1.699 \]
Now substitute this value:
\[ E = -\frac{0.059}{2} \cdot (-1.699) \]
\[ E = 0.0295 \cdot 1.699 \approx 0.0502 \text{ V} \]

### Final Answer
The EMF of the cell is approximately \( 0.051 \text{ V} \).