Calculate Delta G^{circ} for the reaction Mg_ | vedclass

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(A) The standard Gibbs free energy change is given by the formula: $\Delta G^{\circ} = -nFE_{cell}^{\circ}$ Here,$n$ is the number of moles of electro

Vedclass · Accepted Answer

$-430.4$

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(A) The standard Gibbs free energy change is given by the formula: $\Delta G^{\circ} = -nFE_{cell}^{\circ}$ Here,$n$ is the number of moles of electrons transferred in the balanced redox reaction. For $Mg + Sn^{2+} \longrightarrow Mg^{2+} + Sn$,$n = 2$. $F$ is the Faraday constant,approximately $96500 \ C \ mol^{-1}$. $E_{cell}^{\circ} = 2.23 \ V$. Substituting the values: $\Delta G^{\circ} = -2 	imes 96500 	imes 2.23 \ J \ mol^{-1}$. $\Delta G^{\circ} = -430390 \ J \ mol^{-1}$. Converting to $kJ \ mol^{-1}$ by dividing by $1000$: $\Delta G^{\circ} = -430.39 \ kJ \ mol^{-1} \approx -430.4 \ kJ \ mol^{-1}$.

Calculate $\Delta G^{\circ}$ for the reaction $Mg_{(s)} + Sn_{(aq)}^{2+} \longrightarrow Mg_{(aq)}^{2+} + Sn_{(s)}$ if $E_{cell}^{\circ}$ is $2.23 \ V$. (in $kJ$)

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