For a cell reaction involving two electron changes, ${\mathrm{E}}_{\text{cell }}^0=0.3\mathrm{V}\text{ at }{25}^{\circ }\mathrm{C}$ The cell equilibrium constant of the reaction is

E^o   cell and equilibrium constant K_c are related as 

${\mathrm{E}}^{\mathrm{o}} = \frac{0.0591}{\mathrm{n}}{\log }_{10} {\mathrm{K}}_{\mathrm{c}}$

given E^O _cell = 0.3 V

number of electrons involved ,n = 2 

$0.3=\frac{0.0591}{2}{\log }_{10}{\mathrm{K}}_{\mathrm{c}}$

on solving 

log₁₀ K_c= 10 

K_c=Antilog 10

K_c= 10¹⁰