\({E_{cell}}\, = \,\,E_{cell}^ \circ \, + \,\,\frac{{0.059}}{2}\log \,\,\frac{{[F{e^{ + 2}}]}}{{[Z{n^{ + 2}}]}}\)
\(0.2905\,\, = \,\,E_{cell}^ \circ \, + \,\,\frac{{0.059}}{2}\,\log \,\frac{{0.01}}{{0.1}}\)
\(\therefore \,E_{cell}^ \circ \, = \,\,0.2905\,\, + \,\,0.0295\,\, = \,\,0.32\,\,volt\)
\(Now,\,\,\,E_{cell}^ \circ \, = \,\,\frac{{0.059}}{2}\,\,\log \,\,{K_c}\)
\(0.32\,\, = \,\,\frac{{0.059}}{2}\,\log \,\,{K_c}\)
\(KC\,\, = \,\,{10^{0.32/0.0295}}\)
$Pt ( s )\left| H _2( s )( latm )\right| H ^{+}\left( aq ,\left[ H ^{+}\right]=1\right)|| Fe ^{3+}( aq ), Fe ^{2+}( aq ) \mid \operatorname{Pt}( s )$
આપેલ : $E _{ Fe ^{3+} / e ^{2 *}}^0=0.771\,V$ અને $E _{ H ^{+}+\frac{1}{2} H _2}^0=0\,V , T =298\,K$
જો કોષનો પોટેન્શિયલ $0.712\,V$, હોય, તો $Fe ^{-2}$ થી $Fe ^{+3}$ ની સાંદ્રતાની ગુણોત્તર છે.(નજીકનો પૂર્ણાંક)
$Pt|{H_2}_{\left( {1{\mkern 1mu} atm} \right)}|0.1{\mkern 1mu} M{\mkern 1mu} HCl||{\mkern 1mu} {\mkern 1mu} 0.1{\mkern 1mu} M\,C{H_3}COOH|{H_2}_{\left( {1{\mkern 1mu} atm} \right)}|Pt$
$(i)\, Cu^{2+} + 2e^- \rightarrow Cu\,,$ $ E^o = 0.337\, V$
$(ii)\, Cu^{2+} + e^- \rightarrow Cu^+\,,$ $ E^o = 0.153\, V$
તો પ્રક્રિયા $Cu^+ + e^- \rightarrow Cu$ માટે $E^o$........... $V$ થશે.
જો $\Lambda_{{m}}^{\circ}$ $({HA})=190 \,{~S} \,{~cm}^{2} {~mol}^{-1}$, ${HA}$નો આયનીકરણ અચળાંક $\left({K}_{{a}}\right)$ $....\,\times 10^{-6}$ બરાબર છે.