The voltage of the cell consisting of `Li(s) and F_(2)(g)` electrodes is 5.92 V at standard condition at 298 K. What is the voltage if the electrolyte consists of 2 M LiF.
`(ln 2=0.693, R="8.314 JK"^(-1)"mol"^(-1) and F = "96500 C mol"^(-1))`

Options

A. 5.90 V
B. 5.937 V
C. 5.88 V
D. 4.9 V

Answer

To solve the problem, we need to find the voltage of the cell consisting of lithium (Li) and fluorine (F2) electrodes when the electrolyte is 2 M LiF. The standard cell potential (E°) is given as 5.92 V. We will use the Nernst equation to find the new cell potential (E).

### Step-by-Step Solution:

1. **Identify the Cell Reaction**:
The cell reaction for the lithium and fluorine electrodes can be written as:
\[ \text{Li(s)} + \frac{1}{2} \text{F}_2(g) \rightarrow \text{Li}^+(aq) + \text{F}^-(aq) \]

2. **Determine the Number of Electrons (n)**:
In this reaction, one mole of Li loses one electron to form Li⁺. Thus, the number of electrons transferred (n) is:
\[ n = 1 \]

3. **Use the Nernst Equation**:
The Nernst equation is given by:
\[ E = E^\circ - \frac{RT}{nF} \ln Q \]
where:
- \(E\) = cell potential under non-standard conditions
- \(E^\circ\) = standard cell potential (5.92 V)
- \(R\) = universal gas constant (8.314 J/(K·mol))
- \(T\) = temperature in Kelvin (298 K)
- \(n\) = number of moles of electrons (1)
- \(F\) = Faraday's constant (96500 C/mol)
- \(Q\) = reaction quotient

4. **Calculate the Reaction Quotient (Q)**:
For the dissociation of LiF in a 2 M solution:
\[ Q = \frac{[\text{Li}^+][\text{F}^-]}{[\text{F}_2]} = \frac{(2)(2)}{1} = 4 \]
(Assuming the pressure of F2 is 1 atm at standard conditions.)

5. **Substitute Values into the Nernst Equation**:
Substitute the known values into the Nernst equation:
\[ E = 5.92 - \frac{(8.314)(298)}{(1)(96500)} \ln(4) \]

6. **Calculate the Natural Logarithm**:
Since \( \ln(4) = 2 \ln(2) = 2 \times 0.693 = 1.386 \):
\[ E = 5.92 - \frac{(8.314)(298)}{96500} \times 1.386 \]

7. **Calculate the Constant**:
First, calculate \( \frac{(8.314)(298)}{96500} \):
\[ \frac{(8.314)(298)}{96500} \approx 0.0257 \text{ V} \]

8. **Final Calculation**:
Now substitute this back into the equation:
\[ E = 5.92 - (0.0257 \times 1.386) \approx 5.92 - 0.0356 \approx 5.8844 \text{ V} \]

9. **Round the Answer**:
Rounding to two decimal places gives:
\[ E \approx 5.88 \text{ V} \]

### Final Answer:
The voltage of the cell with 2 M LiF electrolyte is approximately **5.88 V**.