Question

For the following cell reaction
$Ag\left|A{g}^{+}\right|AgCl\left|C{l}^{\ominus }\right|C{l}_2,Pt$
$\Delta G^{\circ }f(AgCl)=-109kJ/mol$
$\Delta G^{\circ }f\left(C{l}^{-}\right)=-129kJ/mol$
$\Delta G^{\circ }f\left(A{g}^{+}\right)78kJ/mol$
$E^{\circ }$ of the cell is

• A. -0.60 V
• B. 0.60 V
• C. 6.0 V
• D. None of these

Answer 1

Answer: (A)

For the given cell,
$Ag\left|A{g}^{+}\right|AgCl\left|C{l}^{-}\right|C{l}_2,Pt$
The cell reactions are as follows
At anode :
$Ag⟶A{g}^{+}+e^{-}$
At cathode :
$AgCl+e^{-}⟶Ag(s)+C{l}^{-}$
Net cell reaction :
$AgCl⟶A{g}^{+}+C{l}^{-}$
$\therefore \Delta G_{\text{reaction }}^{\circ }=\Sigma \Delta G_p^{\circ }-\Sigma \Delta G_R^{\circ }$
$=(78-129)-(-109)$
$=+58kJ/mol$
$\Delta G^{\circ }=-nF{E}^{\circ }$
$58\times 10^{3}J=-1\times 96500\times E^{\circ }$ cell
$\Rightarrow E_{\text{cell }}^{\circ }=\frac{-58\times 1000}{96500}$ $=-0.6V$