Q 2259723614.     The electrical energy produced by a galvanic cell is given by the product of its electromotive force and the
quantity of electricity which passes. Till about the middle of the nineteenth century it was generally
believed that electrical energy of a reversible cell originated from the decrease in the enthalpy `(-DeltaH)` of
the cell reaction. This view received support from the fact that the enthalpy change of the reaction
`Zn(s) + Cu^(2+)(aq) -> Zn^(2+)(aq) + Cu(s)`
taking place in the Daniell cell is about 210,000 joules which is very close to the electrical energy of the
cell viz., 212,300 joules, as calculated above. However, towards the close of the century, it was pointed
out by Gibbs, and independently by Helmholtz, that the electrical energy of a reversible cell originates
from the free energy decrease `(-DeltaG)` of reaction occurring in the cell. Suppose in a particular cell reaction,
n is the number of electrons liberated at one electrode (or taken up at the other electrode), then evidently,
n faradays (nF coulombs) of electricity will be generated in the complete cell reaction. If, for the sake of
simplicity, the EMF of the cell, viz., `E_(text(cell))` is denoted by E, then
Electrical energy produced by the cell = nFE
Hence, `-DeltaG = nFE`
Read the following paragraph and answer the questions given below:
In the Daniell cell, the EMF is 1.10 volts. The cell reaction involves liberation as well as taking up of 2 moles of electrons. What will be the electrical energy generated in the cell for the complete cell reaction