Where, \(\Lambda^{o}\) and \(\Lambda^{\infty}\) are equivalent conductances at a given concentration and at infinite dilution respectively. \(\alpha=\frac{8.0}{400}=2 \times 10^{-2}\)
From Ostwald's dilution law (for weak monobasic acid)
\(\mathrm{K}_{\mathrm{a}}=\frac{\mathrm{C} \alpha^{2}}{(1-\alpha)}\)
\(=\mathrm{C} \alpha^{2} \quad(\because 1>>\alpha)\)
\(=\frac{1}{32}\left(2 \times 10^{-2}\right)^{2}\)
\(=1.25 \times 10^{-5}\)
$Pt|{H_2}_{\left( {1{\mkern 1mu} atm} \right)}|0.1{\mkern 1mu} M{\mkern 1mu} HCl||{\mkern 1mu} {\mkern 1mu} 0.1{\mkern 1mu} M\,C{H_3}COOH|{H_2}_{\left( {1{\mkern 1mu} atm} \right)}|Pt$
$Sn ^{2+}+2 e ^{-} \rightarrow Sn$
$Sn ^{4+}+4 e ^{-} \rightarrow Sn$
ઈલેક્ટ્રોન (વિદ્યુતધ્રુવ) પોટેન્શિયલ ની $E _{ Sn ^{2+} / Sn }^{\circ}=-0.140 V$ અને $E _{ Sn ^{4+} / Sn }^{\circ}=0.010 V$ છે. $Sn ^{4+} / Sn ^{2+}$
$E^{o} _{ Sn ^{4+} / Sn ^{2+}}$માટે પ્રમાણિત ઈલેક્ટ્રોડ (વિદ્યુતધ્રુવ) પોંટેન્શિયલની માત્રા........ $\times 10^{-2} V$ છે. (નજીકનો પૂર્ણાક)
$(i)\, Cu^{2+} + 2e^- \rightarrow Cu\,,$ $ 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.