Question

How can we say that the reversible work done by the galvanic cell is equal to decrease in its Gibbs energy?

Answer 1

In electrochemical cell, the chemical energy is converted into electrical energy. The cell potential is related to Gibbs energy change. In an electrochemical cell, the system does work by transferring electrical energy through an electric circuit. Electrical work done in one second is equal to the electrical potential multiplied by the total charge passed. The change in Gibbs energy for a reaction is a measure of the maximum useful work that can be obtained from a chemical reaction.
${\mathrm{\Delta }}_{\mathrm{r}}\mathrm{G}=\mathrm{Maximum}\mathrm{work}$

For a reaction, occurring in an electrochemical cell whose electrodes differ in potential by E_cell, the work done when amount of charge nF is pushed along by the potential of the cell is given by nFE_cell so that,
Maximum work = nFE_cell
where F is the Faraday constant and n is the number of moles of electrons transferred in them. When a voltaic cell operates, work is done on the surroundings, as electrical energy flows through external circuit. Such work by convention is taken as negative. Thus,
${\mathrm{\Delta }}_{\mathrm{r}}\mathrm{G}$ = -nFE_cell