Gibbs Free Energy of Reaction - Practice Questions & MCQ

Let n faraday charge be involved in a cell generating an emf E, then the magnitude of the work done by the cell will be calculated as:
|Work| = Charge × Potential = $(n F) \times(E)$

Now, the maximum work that can be extracted from the cell is equal to the decrease in the Gibb's free energy.

So, it can be said that
$-\Delta \mathrm{G}=\mathrm{nFE}$

The negative sign is incorporated to include the spontaineity relation between E and Gibb's free Energy

Similarly, maximum obtainable work from the cell at standard condition will be:

$\mathrm{W}_{\max }=\mathrm{nFE}_{\text {cell }}^0 \quad$ where $\mathrm{E}_{\text {cell }}^0=$ standard emf of standard cell potential
$-\Delta \mathrm{G}^{\circ}=\mathrm{nFE} \mathrm{E}_{\text {cell }}^0$

Thus, for a spontaneous cell reaction, 

E > 0 and $\Delta$G < 0