Question

For the cell : $Pt/H{g}_{(l)}/H{g}_{2}C{l}_{2(s)}/C{l}^{-}(0.1M)||C{l}^{-}(0.01M)/C{l}_2(1atm)/Pt$ $E_{\text{cell }}$ is equal to $1.16V$. What will be the value of $E_{\text{cell}}^{\circ }$

• A. 1.16 V
• B. 1.1 V
• C. 1.22 V
• D. -1.22 V

Answer 1

Answer: (B)

$Hg(l)+2C{l}^{-}$(aq.) $⟶H{g}_{2}C{l}_2(s)+2e^{-}$
$2e^{-}C{l}_2⟶2C{l}^{-}$
$E=E_{\text{cell }}^{\circ }-\frac{0.06}{2}\log \frac{{\left[C{l}^{-}\right]}_C^2}{{\left[C{l}^{-}\right]}_A^2}$
$E=E_{\text{cell }}^{\circ }-\frac{0.06}{2}\log \frac{10^{-2}}{{\left(10^{-2}\right)}^2}$
$1.16=E_{\text{cell }}^{\circ }-\frac{0.06}{2}\times (-2)$
$1.16=0.06=F_{\text{cell }}^{\circ }$
$E_{\text{cell }}^{\circ }=1.1V$