Question

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 $2\,M\,LiF$? ($\ln 2=0.693, R=8.314\,\text{J K}^{-1}\text{mol}^{-1}$ and $F=96500\,\text{C mol}^{-1}$)

• A. \(5.90 \, V \)
• B. \(5.937 \, V \)
• C. \(5.88 \, V \)
• D. \(4.9 \, V \)
• E. \(4.8 \, V \)

Hint:

When a salt like \(LiF\) dissociates, both ions contribute to the reaction quotient \(Q\). Don't forget to square the concentration if the stoichiometry dictates it!

Answer 1

The Correct Option is C

Concept: The cell potential under non-standard conditions is calculated using the Nernst Equation: \[ E = E^\circ - \frac{RT}{nF} \ln Q \] The cell reaction is: \( 2Li(s) + F_2(g) \rightarrow 2Li^+(aq) + 2F^-(aq) \). Here, \(n = 2\). The reaction quotient \( Q = [Li^+][F^-] \). Since \(LiF\) is \(2 \, M\), \([Li^+] = 2\) and \([F^-] = 2\).

Step 1: Simplify the Nernst Equation constant.
At \(298 \, K\), \(\frac{RT}{F} \approx 0.0257 \, V\). \[ E = 5.92 - \frac{0.0257}{2} \ln([2][2]) \]

Step 2: Solve for the natural log.
\[ \ln(4) = 2 \ln(2) = 2 \times 0.693 = 1.386 \]

Step 3: Calculate final voltage.
\[ E = 5.92 - (0.01285 \times 1.386) \] \[ E = 5.92 - 0.0178 \approx 5.902 \, V \]