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 EMF of the cell if E^o_cell is 1.10 V?

1.11 V

EMF of the cell,

E_cell = E^o_cell - \(\frac{0.0591}{2}\)log\(\frac{Zn^{2+}}{Cu^{2+}}\)

E_cell = 1.10 - \(\frac{0.0591}{2}\)log\(\frac{0.1}{0.2}\)

E_cell = 1.1 - \(\frac{0.0591}{2}\)log\(\frac{1}{2}\) = 1.1 + \(\frac{0.0591}{2}\)log\(\frac{2}{1}\)

E_cell = 1.1 + \(\frac{0.0591×0.3010}{2}\) = 1.11 V