$\therefore \,\,\,{E_1} = {E^0} - \frac{{0.0592}}{2}\,\,\log \,\,\frac{{0.01}}{1}$
$\therefore \,\,{E_2} = {E^0} - \frac{{0.0592}}{2}\,\,\,\log \,\,\frac{{1.0}}{{0.01}}\,\,$
$\,\therefore \,\,{E_1} > {E_2}$
$Zn = Z{n^{2 + }} + 2{e^ - };\,\,{E^o} = + 0.76\,V$
$Fe = F{e^{2 + }} + 2{e^ - };\,\,{E^o} = + 0.41\,V$
નીચેના કોષ પ્રક્રિયા માટે $EMF$ ......... $\mathrm{V}$ છે
$F{e^{2 + }} + Zn\, \to \,Z{n^{2 + }} + Fe$