$Zn | Zn^{2+}(0.01\,M) | | Fe^{2+}(0.001\, M) | Fe$ માટે $25^o$ સે તાપમાને $E_{cell} (emf) = 0.2905\, V $છે, તો સંતુલન અચળાંક $K_c =$ ......

Question

Diffcult

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Solution

b
${E_{cell}} = {E^0}_{cell} - \frac{{0.0592}}{n}\log \frac{{[Z{n^{2 + }}]}}{{[F{e^{2 + }}]}}$

$\therefore \,\,\,0.2905 = {E^0}_{cell} - \frac{{0.0592}}{2}\log \left( {\frac{{{{10}^{ - 2}}}}{{{{10}^{ - 3}}}}} \right)$

$\therefore \,\,{E^0}_{cell} = {E^0}_{cell} - \frac{{0.0592}}{2}{\log _{10}}10$

સંતુલન માટે $E_{cell} = 0$

$E^{0}_{cell} = 0.2905 + 0.0296$

$E^{0}_{cell} = 0.32 \,V$

હવે ${E_{cell}} = {E^0}_{cell} - \frac{{0.0592}}{n}\log \,{K_c}$

$\therefore 0 = 0.32 - \frac{{0.0592}}{2}\log \,{K_c}$

$\therefore \,\,\,0.32 - \frac{{0.0592}}{2}\log \,{K_c}$

$\therefore \log {K_c} = \frac{{0.32}}{{0.0295}}$

${K_c} = 10\left( {\frac{{0.32}}{{0.0295}}} \right)$

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