The standard free energy change ( in J) for t | vedclass

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(C) The cell reaction is $3 Fe^{2+}_{(aq)} + 2 Cr_{(s)} \rightleftharpoons 2 Cr^{3+}_{(aq)} + 3 Fe_{(s)}$.Here,$Cr$ is oxidized to $Cr^{3+}$ (anode) a

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$-173,700$

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(C) The cell reaction is $3 Fe^{2+}_{(aq)} + 2 Cr_{(s)} ightleftharpoons 2 Cr^{3+}_{(aq)} + 3 Fe_{(s)}$.Here,$Cr$ is oxidized to $Cr^{3+}$ (anode) and $Fe^{2+}$ is reduced to $Fe$ (cathode).The standard cell potential is $E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ} = E_{Fe^{2+}/Fe}^{\circ} - E_{Cr^{3+}/Cr}^{\circ}$.$E_{cell}^{\circ} = -0.44 \, V - (-0.74 \, V) = 0.30 \, V$.The number of electrons transferred $(n)$ is $6$.The standard free energy change is given by $\Delta G^{\circ} = -n F E_{cell}^{\circ}$.$\Delta G^{\circ} = -6 	imes 96500 \, C 	imes 0.30 \, V = -173,700 \, J$.

The standard free energy change $($ in $J)$ for the reaction $3 Fe^{2+}_{(aq)} + 2 Cr_{(s)} \rightleftharpoons 2 Cr^{3+}_{(aq)} + 3 Fe_{(s)}$ given $E_{Fe^{2+}/Fe}^{\circ} = -0.44 \, V$ and $E_{Cr^{3+}/Cr}^{\circ} = -0.74 \, V$ is $(F = 96500 \, C)$.

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