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

• A. 5.90 V
• B. 5.937 V
• C. $ 5 . 88 V $
• D. $4.9 V $
• E. $ 4 .8 V$

Answer 1

Answer: (C)

Now, the cell reaction is

$Li (s)+\frac{ 1 }{2} F _{2}(g) \longrightarrow Li ^{+}+ F ^{-}$

We know that,

$E_{\text {cell }} =E_{\text {cell }}^{\circ}-\frac{R T}{n F} \ln \frac{\text { [Product }]}{[\text { Reactant }]}$

$=E_{\text {cell }}^{\circ}-\frac{2.303 R T}{n F} \log \left[ Li ^{+}\right]\left[ F ^{-}\right]$

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

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

$=5.92-0.035=5.88\, V$