$\therefore \,\,\,\log \,\,K = \,\,\frac{{nF{E^o}_{cell}}}{{2.303\,\,\,RT}}\,$
$\therefore \,\,\,\log \,\,K = \frac{{2 \times 96500 \times 0.46}}{{2.303 \times 8.314 \times 298}}\,\,\,\,\,\left[ {\,R\, = \,8.314\,\,joule\,} \right]$
$\therefore \,\log \,K\,= \,15.559\,\,\,\,\,\therefore \,\,K\,\, = \,\,Anti\log \,\,15.559\,\,\,$
$ = \,\,3.625\, \times \,{10^{15}}\,\, \approx \,\,4\, \times \,{10^{15}}$
$Pt|{H_2}_{\left( {1{\mkern 1mu} atm} \right)}|0.1{\mkern 1mu} M{\mkern 1mu} HCl||{\mkern 1mu} {\mkern 1mu} 0.1{\mkern 1mu} M\,C{H_3}COOH|{H_2}_{\left( {1{\mkern 1mu} atm} \right)}|Pt$
$E^oFe^{3+}_{(aq)} + e^{-} \rightarrow Fe^{2+}_{(aq)}$ કેટલા ............ $\mathrm{V}$ થાય?
$Fe^{3+}\,\,_{(aq)} + 3e^{-} \rightarrow Fe_{(s)}$ ; $E^o = -0.036 \,volt; $
$Fe^{2+}_{(aq)} + 2e^{-} \rightarrow Fe_{(s)}$ ; $E^o = -0.440 \,volt$