To solve the problem, we will use the Nernst equation, which relates the cell potential under non-standard conditions to the standard cell potential. 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 (given as 5.92 V)
- \(R\) = universal gas constant (8.314 J K\(^{-1}\) mol\(^{-1}\))
- \(T\) = temperature in Kelvin (298 K)
- \(n\) = number of moles of electrons transferred in the reaction
- \(F\) = Faraday's constant (96500 C mol\(^{-1}\))
- \(Q\) = reaction quotient
### Step 1: Write the half-reaction
The dissociation of lithium fluoride (LiF) can be represented as:
\[
\text{LiF} \rightarrow \text{Li}^+ + \text{F}^-
\]
### Step 2: Determine the reaction quotient \(Q\)
At 2 M concentration of LiF, after dissociation, we have:
- \([\text{Li}^+] = 2 \, \text{M}\)
- \([\text{F}^-] = 2 \, \text{M}\)
- \([\text{LiF}] = 0 \, \text{M}\)
The reaction quotient \(Q\) is given by:
\[
Q = \frac{[\text{Li}^+][\text{F}^-]}{[\text{LiF}]}
\]
Since \([\text{LiF}] = 0\), we can consider \(Q\) as:
\[
Q = [\text{Li}^+][\text{F}^-] = (2)(2) = 4
\]
### Step 3: Substitute values into the Nernst equation
Using the Nernst equation:
\[
E = E^\circ - \frac{RT}{nF} \ln Q
\]
Here, \(E^\circ = 5.92 \, \text{V}\), \(R = 8.314 \, \text{J K}^{-1} \text{mol}^{-1}\), \(T = 298 \, \text{K}\), \(n = 1\) (since one electron is transferred), and \(F = 96500 \, \text{C mol}^{-1}\).
### Step 4: Calculate the term \(\frac{RT}{nF}\)
\[
\frac{RT}{nF} = \frac{(8.314)(298)}{(1)(96500)} \approx 0.008314 \, \text{V}
\]
### Step 5: Calculate \(\ln Q\)
Since \(Q = 4\):
\[
\ln Q = \ln 4 = 2 \ln 2 \approx 2 \times 0.693 = 1.386
\]
### Step 6: Substitute into the Nernst equation
Now substituting into the Nernst equation:
\[
E = 5.92 - (0.008314)(1.386)
\]
Calculating the value:
\[
E \approx 5.92 - 0.01153 \approx 5.90847 \, \text{V}
\]
### Final Answer
The voltage of the cell with 2 M LiF is approximately:
\[
E \approx 5.89 \, \text{V}
\]
