$\therefore \,\,\,0.2905 = {E^0}_{cell} - \frac{{0.0592}}{2}\log \left( {\frac{{{{10}^{ - 2}}}}{{{{10}^{ - 3}}}}} \right)$
$\therefore \,\,{E^0}_{cell} = {E^0}_{cell} - \frac{{0.0592}}{2}{\log _{10}}10$
સંતુલન માટે $E_{cell} = 0$
$E^{0}_{cell} = 0.2905 + 0.0296$
$E^{0}_{cell} = 0.32 \,V$
હવે ${E_{cell}} = {E^0}_{cell} - \frac{{0.0592}}{n}\log \,{K_c}$
$\therefore 0 = 0.32 - \frac{{0.0592}}{2}\log \,{K_c}$
$\therefore \,\,\,0.32 - \frac{{0.0592}}{2}\log \,{K_c}$
$\therefore \log {K_c} = \frac{{0.32}}{{0.0295}}$
${K_c} = 10\left( {\frac{{0.32}}{{0.0295}}} \right)$
$Cr_{(s)} | Cr^{3+}_{(0.1\,M)} | | Fe^{2+}_{(0.01\,M)} | Fe;$
$E^0_{{cr}^{3+} |Cr} = -0.72 \,V,$ $ E^{0}_{{Fe}^{2+}{| Fe}}$ $= -0.42 \,V$
$[\Lambda_{\mathrm{H}^{+}}^{\circ}=350 \,\mathrm{~S}\, \mathrm{~cm}^{2}\, \mathrm{~mol}^{-1},\Lambda_{\mathrm{CH}_{3} \mathrm{COO}^{-}}^{\circ}=50\, \mathrm{~S}\, \mathrm{~cm}^{2}\, \mathrm{~mol}^{-1}]$