Electrical work done in one second is equal to electrical potential multiplied by total charge passed. If we want to obtain maximum work from a galvanic cell then charge has to be passed reversibly. The reversible work done by a galvanic cell is equal to decrease in its Gibbs energy and therefore, if the EMF of the cell is E and nF is the amount of charge passed and ∆_rG is the Gibbs energy of the reaction, then

∆_rG = – nFE_cell

It may be remembered that E_cell is an intensive parameter but ∆_rG is an extensive thermodynamic property and the value depends on n. From the measurement of E^o_cell , we can obtain an important thermodynamic quantity, ∆_rG^o , standard Gibbs energy of the reaction.

What is the change in free energy of the cell reaction?

214.23 kJ

Free energy

ΔG = -nFE_cell = -2 x 96500 x 1.11 = -214230 J = 214.23 kJ