Q31 / 160 · AP EAMCET 2009 · Chemistry
For the following cell reaction,
$\begin{gathered}
\mathrm{Ag}\left|\mathrm{Ag}^{+}\right| \mathrm{AgCl}\left|\mathrm{Cl}^{\ominus}\right| \mathrm{Cl}_2, \mathrm{Pt} \\
\Delta G_f^{\circ}(\mathrm{AgCl})=-109 \mathrm{~kJ} / \mathrm{mol} \\
\Delta G_f^{\circ}\left(\mathrm{Cl}^{\ominus}\right)=-129 \mathrm{~kJ} / \mathrm{mol} \\
\Delta G_f^{\circ}\left(\mathrm{Ag}^{+}\right)=78 \mathrm{~kJ} / \mathrm{mol}
\end{gathered}$
$E^{\circ}$ of the cell is
- Correct answerA–0.60 V
- B0.60 V
- C6.0 V
- DNone of these
Solution
For the given cell,
$\mathrm{Ag}\left|\mathrm{Ag}^{+}\right| \mathrm{AgCl}\left|\mathrm{Cl}^{\ominus}\right| \mathrm{Cl}_2, \mathrm{Pt}$
the cell reactions are as follows
$\begin{aligned} \therefore \quad \Delta G_{\text {reaction }}^{\circ} & =\Sigma \Delta G_p^{\circ}-\Sigma \Delta G_R^{\circ} \\ & =(78-129)-(-109) \\ & =+58 \mathrm{~kJ} / \mathrm{mol} \\ \Delta G^{\circ} & =-n F E^{\circ} \\ 58 \times 10^3 \mathrm{~J} & =-1 \times 96500 \times E_{\text {cell }}^{\circ} \\ E_{\text {cell }}^{\circ} & =\frac{-58 \times 1000}{96500} \\ & =-0.6 \mathrm{~V}\end{aligned}$
$\begin{aligned} \therefore \quad \Delta G_{\text {reaction }}^{\circ} & =\Sigma \Delta G_p^{\circ}-\Sigma \Delta G_R^{\circ} \\ & =(78-129)-(-109) \\ & =+58 \mathrm{~kJ} / \mathrm{mol} \\ \Delta G^{\circ} & =-n F E^{\circ} \\ 58 \times 10^3 \mathrm{~J} & =-1 \times 96500 \times E_{\text {cell }}^{\circ} \\ E_{\text {cell }}^{\circ} & =\frac{-58 \times 1000}{96500} \\ & =-0.6 \mathrm{~V}\end{aligned}$
Hence, the correct answer is option: 'A'.