1. Tardigrade
  2. Question
  3. Chemistry
  4. The voltage of the cell consisting of Li(s) and F2(g) electrodes is 5.92 V at standard condition at 298 K . What is the voltage if the electrolyte consists of 2 M LiF . Given that, ln 2=0.693, R=8.314 J K- 1mol- 1F=96500 C mol- 1

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

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

NTA AbhyasNTA Abhyas 2022

Solution:

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