Question

The voltage of the cell consisting of $Li(s)$ and $F_2(g)$ electrodes is $5.92V$ at standard condition at $298K.$ What is the voltage if the electrolyte consists of $2MLiF.(ln2=0.693,R=8.314J{K}^{-1}mo{l}^{-1}$ and $F=96500Cmo{l}^{-1})$

• A. 5.90 V
• B. 5.937 V
• C. $5.88V$
• D. $4.9V$
• E. $4.8V$

Answer 1

Answer: (C)

Now, the cell reaction is

$Li(s)+\frac{1}{2}{F}_2(g)⟶L{i}^{+}+F^{-}$

We know that,

$E_{\text{cell }}=E_{\text{cell }}^{\circ }-\frac{RT}{nF}\ln \frac{\text{ [Product }]}{[\text{ Reactant }]}$

$=E_{\text{cell }}^{\circ }-\frac{2.303RT}{nF}\log \left[L{i}^{+}\right]\left[F^{-}\right]$

$=5.92-\frac{2.303\times 8.314\times 298}{1\times 96500}\log (2\times 2)$

$=5.92-\frac{0.059}{1}\times 2\log 2$

$=5.92-0.035=5.88V$