Question

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

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

Answer 1

Answer: (C)

Now, the cell reaction is
$Li(s)+\frac{1}{2}{F}_2(g)\to (Li{)}^{+}+F^{-}$
We know that,
$E_{\text{cell }}=E_{\text{cell }}^o-\frac{RT}{nF}\ln \frac{\text{ Product }}{\text{ Reactant }}$
$=E_{\text{cell }}^o-\frac{2.303RT}{nF}\log \left[L{i}^{+}\right]\left[F^{-}\right]$
$=5.92-\frac{2.303\times 8.314\times 298}{1\times 96500}\log (2\times 2)$
$=5.92-\frac{0.059}{1}\times 2\log 2$
$=5.92-0.035=5.887V$