Calculate Delta G^{circ} for the following ce | vedclass

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(B) The cell reaction is: $Zn_{(s)} + Ag_2O_{(s)} + H_2O_{(l)} \longrightarrow Zn^{2+}_{(aq)} + 2Ag_{(s)} + 2OH^{-}_{(aq)}$The standard cell potential

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$-301 \ kJ/mol$

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(B) The cell reaction is: $Zn_{(s)} + Ag_2O_{(s)} + H_2O_{(l)} \longrightarrow Zn^{2+}_{(aq)} + 2Ag_{(s)} + 2OH^{-}_{(aq)}$The standard cell potential is calculated as: $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 0.80 \ V - (-0.76 \ V) = 1.56 \ V$The number of electrons transferred in the reaction is $n = 2$.Using the formula $\Delta G^{\circ} = -nFE^{\circ}$:$\Delta G^{\circ} = -2 	imes 96500 \ C/mol 	imes 1.56 \ V$$\Delta G^{\circ} = -301080 \ J/mol = -301.08 \ kJ/mol \approx -301 \ kJ/mol$

Calculate $\Delta G^{\circ}$ for the following cell reaction.
$Zn_{(s)} + Ag_2O_{(s)} + H_2O_{(l)} \longrightarrow Zn^{2+}_{(aq)} + 2Ag_{(s)} + 2OH^{-}_{(aq)}$
Given: $E^{\circ}_{Ag^{+}/Ag} = +0.80 \ V$ and $E^{\circ}_{Zn^{2+}/Zn} = -0.76 \ V$

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Calculate $\Delta G^{\circ}$ for the following cell reaction.$Zn_{(s)} + Ag_2O_{(s)} + H_2O_{(l)} \longrightarrow Zn^{2+}_{(aq)} + 2Ag_{(s)} + 2OH^{-}_{(aq)}$Given: $E^{\circ}_{Ag^{+}/Ag} = +0.80 \ V$ and $E^{\circ}_{Zn^{2+}/Zn} = -0.76 \ V$