248.  For the cell Zn/Zn²⁺ (aq) || Cu²⁺(aq) | Cu, derive the relation between E°_cell and K_c at 298 k.

Gibbs energy of the reaction given by:
${∆}_{r}G$ = – nFE_(cell)
thus the reaction 
Zn(s) + Cu²⁺(aq)---> Zn²⁺(aq) + Cu(s)

${∆}_{r}G$ = – 2FE_(cell)

but when we write the reaction
2 Zn (s) + 2 Cu²⁺----->2 Zn²⁺(aq) + 2Cu(s)

${∆}_{r}G$ = – 4FE_(cell)

If the concentration of all the reacting species is unity, then
E_(cell) = ${\mathrm{E}}_{\mathrm{cell}}^0$
and we have
${∆}_{r}G$ = – nF${\mathrm{E}}_{\mathrm{cell}}^0$

Thus, from the measurement of ${\mathrm{E}}_{\mathrm{cell}}^0$ we can obtain an important thermodynamic quantity, ${∆}_{r}G$, standard Gibbs energy of the reaction.
From the latter we can calculate equilibrium constant by the equation:

${∆}_{r}G$ = –RT ln K_c