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(B) The formula for standard Gibbs energy change is $\Delta G^{\circ} = -nFE^{\circ}_{cell}$.Here,$n = 2$ (number of electrons transferred in the reac

Vedclass · Accepted Answer

$-212.27 \ kJ \ mol^{-1}$

Vedclass · Answer

(B) The formula for standard Gibbs energy change is $\Delta G^{\circ} = -nFE^{\circ}_{cell}$.Here,$n = 2$ (number of electrons transferred in the reaction).$E^{\circ}_{cell} = 1.1 \ V$.$F = 96487 \ C \ mol^{-1}$.Substituting the values: $\Delta G^{\circ} = -2 	imes 96487 	imes 1.1 \ C \cdot V \ mol^{-1}$.$\Delta G^{\circ} = -212271.4 \ J \ mol^{-1}$.Converting to $kJ \ mol^{-1}$: $\Delta G^{\circ} = \frac{-212271.4}{1000} \approx -212.27 \ kJ \ mol^{-1}$.

Standard electrode potential for the cell with cell reaction $Zn_{(s)} + Cu^{2+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Cu_{(s)}$ is $1.1 \ V$. Calculate the standard Gibbs energy change for the cell reaction. (Given $F = 96487 \ C \ mol^{-1}$)

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