For the following cell reaction,Ag|Ag^{+}|AgC | vedclass

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(A) The cell representation is $Ag|Ag^{+}|AgCl|Cl^{-}|Cl_2, Pt$. The half-cell reactions are: Anode: $Ag \rightarrow Ag^{+} + e^{-}$ Cathode: $AgCl +

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$-0.60 \ V$

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(A) The cell representation is $Ag|Ag^{+}|AgCl|Cl^{-}|Cl_2, Pt$. The half-cell reactions are: Anode: $Ag ightarrow Ag^{+} + e^{-}$ Cathode: $AgCl + e^{-} ightarrow Ag(s) + Cl^{-}$ Net cell reaction: $AgCl ightarrow Ag^{+} + Cl^{-}$ The standard Gibbs free energy change for the reaction is: $\Delta G_{	ext{reaction}}^{\circ} = \Sigma \Delta G_f^{\circ}(	ext{products}) - \Sigma \Delta G_f^{\circ}(	ext{reactants})$ $\Delta G_{	ext{reaction}}^{\circ} = [\Delta G_f^{\circ}(Ag^{+}) + \Delta G_f^{\circ}(Cl^{-})] - [\Delta G_f^{\circ}(AgCl)]$ $\Delta G_{	ext{reaction}}^{\circ} = [78 + (-129)] - (-109) \ kJ / mol$ $\Delta G_{	ext{reaction}}^{\circ} = -51 + 109 = +58 \ kJ / mol$ Using the relation $\Delta G^{\circ} = -n F E_{	ext{cell}}^{\circ}$,where $n = 1$ and $F = 96500 \ C / mol$: $58 	imes 10^3 \ J / mol = -1 	imes 96500 \ C / mol 	imes E_{	ext{cell}}^{\circ}$ $E_{	ext{cell}}^{\circ} = \frac{-58000}{96500} \ V \approx -0.60 \ V$

For the following cell reaction,$Ag|Ag^{+}|AgCl|Cl^{-}|Cl_2, Pt$
$\Delta G_f^{\circ}(AgCl) = -109 \ kJ / mol$
$\Delta G_f^{\circ}(Cl^{-}) = -129 \ kJ / mol$
$\Delta G_f^{\circ}(Ag^{+}) = 78 \ kJ / mol$
The $E^{\circ}$ of the cell is:

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For the following cell reaction,$Ag|Ag^{+}|AgCl|Cl^{-}|Cl_2, Pt$ $\Delta G_f^{\circ}(AgCl) = -109 \ kJ / mol$ $\Delta G_f^{\circ}(Cl^{-}) = -129 \ kJ / mol$ $\Delta G_f^{\circ}(Ag^{+}) = 78 \ kJ / mol$ The $E^{\circ}$ of the cell is: