Question

For the following cell reaction $Ag \left| Ag ^{+}\right| AgCl \left| Cl ^{\ominus}\right| Cl _{2}, Pt$ $\Delta G ^{\circ} f ( AgCl )=-109\, kJ / mol$ $\Delta G ^{\circ} f \left( Cl ^{-}\right)=-129\, kJ / mol$ $\Delta G ^{\circ} f \left( Ag ^{+}\right) 78 \,kJ / mol$ $E ^{\circ}$ of the cell is

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

Answer 1

Answer: (A)

-0.60 V

Answer 2

For the given cell,
$Ag \left| Ag ^{+}\right| AgCl \left| Cl ^{-}\right| Cl _{2}, Pt$
The cell reactions are as follows
At anode :
$Ag \longrightarrow Ag ^{+}+e^{-}$
At cathode :
$AgCl +e^{-} \longrightarrow Ag (s)+ Cl ^{-}$
Net cell reaction :
$AgCl \longrightarrow Ag ^{+}+ Cl ^{-}$
$\therefore \Delta G_{\text {reaction }}^{\circ}=\Sigma \Delta G_{p}^{\circ}-\Sigma \Delta G_{R}^{\circ}$
$=(78-129)-(-109)$
$=+58 \,kJ / mol$
$\Delta G^{\circ}=-n F E^{\circ}$
$58 \times 10^{3} J =-1 \times 96500 \times E ^{\circ}$ cell
$\Rightarrow E_{\text {cell }}^{\circ}=\frac{-58 \times 1000}{96500}$
$=-0.6\, V$