Example 2.3

Example . 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 D r G is the Gibbs energy of the reaction, then D r G = – nFE (cell) ( . ) It may be remembered that E (cell) is an intensive parameter but D r G is an extensive thermodynamic property and the value depends on n . Thus, if we write the reaction Zn(s) + Cu + (aq) ¾® Zn + (aq) + Cu(s) ( . ) D r G = – FE (cell) but when we write the reaction Zn (s) + Cu + (aq) ¾® Zn + (aq) + 2Cu(s) D r G = – FE (cell) If the concentration of all the reacting species is unity, then E (cell) = and we have D r G o = – nF (cell) ( . ) Thus, from the measurement of we can obtain an important thermodynamic quantity, D r G o , standard Gibbs energy of the reaction. From the latter we can calculate equilibrium constant by the equation: D r G o = – RT ln K . . . Electro- chemical Cell and Gibbs Energy of the Reaction Calculate the equilibrium constant of the reaction: Cu(s) + 2Ag + (aq) ® Cu + (aq) + 2Ag(s) = . V = . V log K C = . V or log K C = . . V V × = . K C = . ×