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

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Question

(A) The cell reaction is: $Ag(s) + AgCl(s) \rightarrow Ag^{+}(aq) + Cl^{-}(aq) + Ag(s)$ Simplified net cell reaction: $AgCl(s) \rightarrow Ag^{+}(aq)

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$-0.60 \ V$

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(A) The cell reaction is: $Ag(s) + AgCl(s) ightarrow Ag^{+}(aq) + Cl^{-}(aq) + Ag(s)$ Simplified net cell reaction: $AgCl(s) ightarrow Ag^{+}(aq) + Cl^{-}(aq)$ Calculate $\Delta G^{\circ}_{reaction}$: $\Delta G^{\circ}_{reaction} = [\Delta G_f^{\circ}(Ag^{+}) + \Delta G_f^{\circ}(Cl^{-})] - [\Delta G_f^{\circ}(AgCl)]$ $\Delta G^{\circ}_{reaction} = [78 + (-129)] - (-109) \ kJ/mol$ $\Delta G^{\circ}_{reaction} = -51 + 109 = 58 \ kJ/mol = 58000 \ J/mol$ Using the relation $\Delta G^{\circ} = -nFE^{\circ}_{cell}$: Here,$n = 1$ (as $Ag ightarrow Ag^{+} + e^{-}$ and $AgCl + e^{-} ightarrow Ag + Cl^{-}$). $58000 = -1 	imes 96500 	imes E^{\circ}_{cell}$ $E^{\circ}_{cell} = -\frac{58000}{96500} \approx -0.60 \ V$

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

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