Gibbs energy of cell reactions and cell potential

The electrical work done in a galvanic cell is the electricity (charge) passed multiplied by the cell potential:

The charge of one mole of electrons is $F$ coulombs. For a cell reaction involving $n$ moles of electrons, the charge passed is $nF$ coulombs, and hence

W. Gibbs, in 1878, concluded that the electrical work done in a galvanic cell is equal to the decrease in Gibbs energy, $-\Delta G$, of the cell reaction. It then follows that

Under standard state conditions we write

Eq. (5.28) explains why $E^0_{cell}$ is an intensive property. We know that $\Delta G^0$ is an extensive property, since its value depends on the amount of substance. If the stoichiometric equation of the redox reaction is multiplied by 2 — that is, the amounts of substances oxidised and reduced are doubled — $\Delta G^0$ doubles, and the moles of electrons transferred also double. The ratio

$E^0_{cell} = -\frac{\Delta G^0}{nF} \quad\text{then becomes}\quad E^0_{cell} = -\frac{2\,\Delta G^0}{2\,nF} = -\frac{\Delta G^0}{nF}$ …