The standard Gibbs energy change in kJ mol^{ | vedclass

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Question

(C) The cell reaction is $A_{(s)} + B_{(aq)}^{3+} \longrightarrow A_{(aq)}^{3+} + B_{(s)}$.From the reaction,the number of electrons transferred $(n)$

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$-144.75$

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(C) The cell reaction is $A_{(s)} + B_{(aq)}^{3+} \longrightarrow A_{(aq)}^{3+} + B_{(s)}$.From the reaction,the number of electrons transferred $(n)$ is $3$.The standard emf of the cell $(E^\circ_{cell})$ is $0.5 \ V$.The formula for standard Gibbs energy change is $\Delta G^\circ = -nFE^\circ_{cell}$.Substituting the values: $\Delta G^\circ = -(3) 	imes (96500 \ C \ mol^{-1}) 	imes (0.5 \ V)$.$\Delta G^\circ = -144750 \ J \ mol^{-1}$.Converting to $kJ \ mol^{-1}$: $\Delta G^\circ = -144.75 \ kJ \ mol^{-1}$.

The standard Gibbs energy change in $kJ \ mol^{-1}$ for a galvanic cell $A_{(s)} + B_{(aq)}^{3+} \longrightarrow A_{(aq)}^{3+} + B_{(s)}$ that has a standard emf of $0.5 \ V$ is: $\left(F = 96500 \ C \ mol^{-1}\right)$

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