Relationship Between Potential, Gibbs Energy And Equilibrium Constant
Category : JEE Main & Advanced
The electrical work (electrical energy) is equal to the product of the EMF of the cell and electrical charge that flows through the external circuit i.e.,
\[{{W}_{\max }}=nF{{E}_{cell}}\] ......(i)
According to thermodynamics the free energy change \[(\Delta G)\] is equal to the maximum work. In the cell work is done on the surroundings by which electrical energy flows through the external circuit, So
\[-{{W}_{\max ,}}=\Delta G\] ......(ii)
from eq. (i) and (ii) \[\Delta G=-nFE_{cell}^{{}}\]
In standard conditions \[\Delta {{G}^{0}}=-\,nFE_{cell}^{0}\]
Where \[\Delta {{G}^{0}}=\]standard free energy change
But \[E_{cell}^{0}=\frac{2.303}{nF}RT\,\log {{K}_{c}}\]
\[\therefore \]\[\Delta {{G}^{0}}=-nF\times \frac{2.303}{nF}RT\,\log \,{{K}_{c}}\]
\[\Delta {{G}^{0}}=-\text{ 2}\text{.303 RT log }{{\text{K}}_{\text{c}}}\text{ }\]or \[\Delta G=\Delta G{}^\circ +2.303RT\log Q\]
\[\Delta {{G}^{0}}=-RT\,\ln \,{{K}_{c}}\,\,\,\,\,\,\,\,(2.303\,\log X=\ln \,X)\]
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