$\therefore \,\,\,0.2905 = {E^0}_{cell} - \frac{{0.0592}}{2}\log \left( {\frac{{{{10}^{ - 2}}}}{{{{10}^{ - 3}}}}} \right)$
$\therefore \,\,{E^0}_{cell} = {E^0}_{cell} - \frac{{0.0592}}{2}{\log _{10}}10$
સંતુલન માટે $E_{cell} = 0$
$E^{0}_{cell} = 0.2905 + 0.0296$
$E^{0}_{cell} = 0.32 \,V$
હવે ${E_{cell}} = {E^0}_{cell} - \frac{{0.0592}}{n}\log \,{K_c}$
$\therefore 0 = 0.32 - \frac{{0.0592}}{2}\log \,{K_c}$
$\therefore \,\,\,0.32 - \frac{{0.0592}}{2}\log \,{K_c}$
$\therefore \log {K_c} = \frac{{0.32}}{{0.0295}}$
${K_c} = 10\left( {\frac{{0.32}}{{0.0295}}} \right)$
$\mathrm{M}\left|\mathrm{M}^{2+}\right||\mathrm{X}| \mathrm{X}^{2-}$
ધારોકે $\mathrm{E}_{\left(\mathrm{M}^{2+} / \mathrm{M}\right)}^0=0.46 \mathrm{~V}$ અને $\mathrm{E}_{\left(\mathrm{x} / \mathrm{X}^{2-}\right)}^0=0.34 \mathrm{~V}$.
નીચે આપેલામાંથી ક્યું સાયું છે ?
આપેલ,
$\Lambda_{ m }^{\alpha}=133.4\,\left( AgNO _{3}\right) ; \Lambda_{ m }^{\alpha}=149.9( KCl )$
$\Lambda_{ m }^{\alpha}=144.9\, S\, cm ^{2} \,mol ^{-1}\left( KNO _{3}\right)$