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(C) The standard Gibbs energy change $\Delta G^{\circ}$ is related to the standard cell potential $E^{\circ}_{	ext{cell}}$ by the equation: $\Delta G

Vedclass · Accepted Answer

$-96.5$

Vedclass · Answer

(C) The standard Gibbs energy change $\Delta G^{\circ}$ is related to the standard cell potential $E^{\circ}_{	ext{cell}}$ by the equation: $\Delta G^{\circ} = -nFE^{\circ}_{	ext{cell}}$.In the given reaction,$Zn$ is oxidized to $Zn^{2+}$ $(n=2)$ and $Ni^{2+}$ is reduced to $Ni$ $(n=2)$,so the number of electrons transferred $n = 2$.The Faraday constant $F \approx 96500 \ C \ mol^{-1}$.Given $E^{\circ}_{	ext{cell}} = 0.5 \ V$.Substituting the values: $\Delta G^{\circ} = -2 	imes 96500 \ C \ mol^{-1} 	imes 0.5 \ V = -96500 \ J \ mol^{-1}$.Converting to $kJ \ mol^{-1}$: $\Delta G^{\circ} = -96.5 \ kJ \ mol^{-1}$.

For the cell involving the following reaction: $Zn_{(s)} + Ni^{2+}_{(aq)} \longrightarrow Zn^{2+}_{(aq)} + Ni_{(s)}$. Given $E^{\circ}_{\text{cell}} = 0.5 \ V$. What is the standard Gibbs energy change of the cell reaction (in $kJ$)?

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