Calculate Delta_r G^o for the cell reaction:Z | vedclass

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(B) The standard Gibbs free energy change is given by the formula: $\Delta_r G^o = -n F E^o_{cell}$First,calculate the standard cell potential: $E^o_{

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$-2.12 	imes 10^5 \, J \, mol^{-1}$

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(B) The standard Gibbs free energy change is given by the formula: $\Delta_r G^o = -n F E^o_{cell}$First,calculate the standard cell potential: $E^o_{cell} = E^o_{cathode} - E^o_{anode} = 0.34 \, V - (-0.76 \, V) = 1.10 \, V$Here,$n = 2$ (number of electrons transferred) and $F = 96500 \, C \, mol^{-1}$ (Faraday's constant).Substituting the values: $\Delta_r G^o = -2 	imes 96500 \, C \, mol^{-1} 	imes 1.10 \, V$$\Delta_r G^o = -212300 \, J \, mol^{-1} = -2.12 	imes 10^5 \, J \, mol^{-1}$

Calculate $\Delta_r G^o$ for the cell reaction:
$Zn_{(s)} + Cu^{2+}_{(aq)} \to Zn^{2+}_{(aq)} + Cu_{(s)}$
Given: $E^o_{Cu^{2+}/Cu} = 0.34 \, V, E^o_{Zn^{2+}/Zn} = -0.76 \, V$

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Calculate $\Delta_r G^o$ for the cell reaction:$Zn_{(s)} + Cu^{2+}_{(aq)} \to Zn^{2+}_{(aq)} + Cu_{(s)}$Given: $E^o_{Cu^{2+}/Cu} = 0.34 \, V, E^o_{Zn^{2+}/Zn} = -0.76 \, V$