Explain with a suitable example the relation between the gibbs energy of chemical reaction (G) and the functioning of the electrochemical cell.

Explanation

Electrical work done in one second in an electrochemical cell is equal to the electrical potential multiplied by the total charge passing. If we want to obtain maximum work from a galvanic cell then the charge has to be passed reversibly. The reversible work done by a galvanic call is equal to decrease in its Gibbs energy and therefore, if the emf is E and F is the amount of charge passing and `Delta_(r)G=- n FE_([cell])`
It may be remembered that `E_([cell])` is an intersive parameter but `Delta_(r) G` is an extensive thermodynamic property and the vlaue depends on n. Thus, if we write the reaction
`Zn_((s))+Cu_((aq))^(2+)to Zn _((aq))^(2+)+Cu_((s))`
`Delta_(r)G=- 2FE _((cell))`
but went we write the reaction
`2Zn_([s])+2 Cu_([aq])^(2+)to Zn_([aq])^(2+)+2Cu_([s])`
`Delta_(r)G=- 4FE_([cell])`
If the concentration of each of the rectiong species is unity, then `E_([cell])=E_([cell])^(Θ)` and we have `Delta_(r)G^(Θ)=-nFE_((cell))^(Θ)`
We can calculate equilibeium constant using the equation
`Delta_(r)G^(Θ)=-RTln K_(c)`