Question

Discuss the relation between free energy and EMF.

Answer 1

Hint:Brush up the thermodynamics concept. Think what happens when reversible electric work is done in a galvanic cell. Gibbs free energy is an extensive property which depends on the amount of substance. Try to relate these two points and develop the answer.

Complete step by step solution:
- The electrical work done in a voltaic cell is the product of charge passed and the cell potential. Therefore, it is represented as, $\text{Electrical work = amount of charge }\times \text{ cell potential}$
\[{{W}_{electrical}}=nF{{E}_{cell}}\]
where \[{{W}_{electrical}}\] is the electrical work done, nF coulombs is the amount of electricity passed and \[{{E}_{cell}}\] is the cell potential.
- Maximum work is done by the cell when the process occurs reversibly.
- A galvanic cell can be made to behave in a reversible manner by balancing the external potential supplied to the cell, so that no current flows.
- The difference of potential, between the electrodes when a voltaic cell is balanced against an external source of potential, is called zero current potential or emf of the cell.
- The reversible work done in a voltaic cell by cell reaction is equal to the decrease in Gibbs energy. Hence,
\[Electrical\,Work\,=\,-\Delta G\]
-Therefore, from both the above equations we can write,
\[-\Delta G=nF{{E}_{cell}}\] or $\Delta G=-nF{{E}_{cell}}$
Therefore, the relation between free energy and EMF is given by,
\[\Delta G=-nF{{E}_{cell}}\]

Note: Remember EMF is an intensive property and Gibbs free energy is an extensive property. The decrease in Gibbs free energy is the NEGATIVE product of the amount of charge (nF) and EMF of the cell.